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Code Implementation

Dice Roll Simulation Implementation

Below is the implementation of the dice roll simulation:

solution.js

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/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (3D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP array
  // dp[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const dp = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j][1] = 1;
  }
 
  // Fill DP array
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Case 1: Starting a new consecutive sequence (k = 1)
      // We can come from any face j' != j with any valid consecutive count k'
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          for (let k_prev = 1; k_prev <= rollMax[j_prev - 1]; k_prev++) {
            dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j_prev][k_prev]) % MOD;
          }
        }
      }
 
      // Case 2: Extending a consecutive sequence (k > 1)
      // We can only come from the same face j with consecutive count k-1
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        dp[i][j][k] = dp[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    for (let k = 1; k <= rollMax[j - 1]; k++) {
      result = (result + dp[n][j][k]) % MOD;
    }
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator2D(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays
  // dp[i][j] = number of sequences of length i ending with face j
  const dp = Array(n + 1).fill().map(() => Array(FACES + 1).fill(0));
 
  // consecutive[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const consecutive = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j] = 1;
    consecutive[1][j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[i][j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[i][j] = (dp[i][j] + dp[i - 1][j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[i][j] = (dp[i][j] + consecutive[i - 1][j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[i][j][1] = sequences that switch to face j
      consecutive[i][j][1] = 0;
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[i][j][1] = (consecutive[i][j][1] + dp[i - 1][j_prev]) % MOD;
        }
      }
 
      // consecutive[i][j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[i][j][k] = consecutive[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[n][j]) % MOD;
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming with space optimization.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulatorOptimized(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays for current and previous rolls
  let dp = Array(FACES + 1).fill(0);
  let consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[j] = 1;
    consecutive[j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    // Save the previous DP arrays
    const prevDp = [...dp];
    const prevConsecutive = consecutive.map(row => [...row]);
 
    // Reset current DP arrays
    dp = Array(FACES + 1).fill(0);
    consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[j] = (dp[j] + prevDp[j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[j] = (dp[j] + prevConsecutive[j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[j][1] = sequences that switch to face j
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[j][1] = (consecutive[j][1] + prevDp[j_prev]) % MOD;
        }
      }
 
      // consecutive[j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[j][k] = prevConsecutive[j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[j]) % MOD;
  }
 
  return result;
}
 
// Test cases
console.log(dieSimulator(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulator2D(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator2D(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator2D(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulatorOptimized(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulatorOptimized(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulatorOptimized(3, [1, 1, 1, 2, 2, 3]));  // 181

Step-by-Step Explanation

Let's break down the implementation:

Define the States: Define the states for the dynamic programming approach, considering the current roll number, face value, and consecutive occurrences.
Establish Base Cases: Set up the base cases for the first roll, where there is one way to roll each face value.
Derive Recurrence Relations: Derive the recurrence relations by considering all possible previous states that can lead to the current state.
Fill DP Arrays: Use dynamic programming to fill the arrays from the base cases up to the target number of rolls.
Calculate Final Answer: Sum up all valid sequences of the target length to get the final answer, applying the modulo operation to avoid integer overflow.

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Learning Objective

Implement the dice roll simulation solution in different programming languages.

Dice Roll Simulation Implementation

Below is the implementation of the dice roll simulation in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (3D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP array
  // dp[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const dp = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j][1] = 1;
  }
 
  // Fill DP array
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Case 1: Starting a new consecutive sequence (k = 1)
      // We can come from any face j' != j with any valid consecutive count k'
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          for (let k_prev = 1; k_prev <= rollMax[j_prev - 1]; k_prev++) {
            dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j_prev][k_prev]) % MOD;
          }
        }
      }
 
      // Case 2: Extending a consecutive sequence (k > 1)
      // We can only come from the same face j with consecutive count k-1
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        dp[i][j][k] = dp[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    for (let k = 1; k <= rollMax[j - 1]; k++) {
      result = (result + dp[n][j][k]) % MOD;
    }
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator2D(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays
  // dp[i][j] = number of sequences of length i ending with face j
  const dp = Array(n + 1).fill().map(() => Array(FACES + 1).fill(0));
 
  // consecutive[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const consecutive = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j] = 1;
    consecutive[1][j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[i][j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[i][j] = (dp[i][j] + dp[i - 1][j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[i][j] = (dp[i][j] + consecutive[i - 1][j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[i][j][1] = sequences that switch to face j
      consecutive[i][j][1] = 0;
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[i][j][1] = (consecutive[i][j][1] + dp[i - 1][j_prev]) % MOD;
        }
      }
 
      // consecutive[i][j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[i][j][k] = consecutive[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[n][j]) % MOD;
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming with space optimization.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulatorOptimized(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays for current and previous rolls
  let dp = Array(FACES + 1).fill(0);
  let consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[j] = 1;
    consecutive[j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    // Save the previous DP arrays
    const prevDp = [...dp];
    const prevConsecutive = consecutive.map(row => [...row]);
 
    // Reset current DP arrays
    dp = Array(FACES + 1).fill(0);
    consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[j] = (dp[j] + prevDp[j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[j] = (dp[j] + prevConsecutive[j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[j][1] = sequences that switch to face j
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[j][1] = (consecutive[j][1] + prevDp[j_prev]) % MOD;
        }
      }
 
      // consecutive[j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[j][k] = prevConsecutive[j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[j]) % MOD;
  }
 
  return result;
}
 
// Test cases
console.log(dieSimulator(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulator2D(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator2D(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator2D(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulatorOptimized(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulatorOptimized(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulatorOptimized(3, [1, 1, 1, 2, 2, 3]));  // 181

Step-by-Step Explanation

1Define the States

Define the states for the dynamic programming approach, considering the current roll number, face value, and consecutive occurrences.

2Establish Base Cases

Set up the base cases for the first roll, where there is one way to roll each face value.

3Derive Recurrence Relations

Derive the recurrence relations by considering all possible previous states that can lead to the current state.

4Fill DP Arrays

Use dynamic programming to fill the arrays from the base cases up to the target number of rolls.

5Calculate Final Answer

Sum up all valid sequences of the target length to get the final answer, applying the modulo operation to avoid integer overflow.

Language-Specific Notes

javascript

JavaScript's Array.fill().map() is used to create multi-dimensional arrays.
The !== operator is used for strict inequality comparison.
The % operator is used for the modulo operation to avoid integer overflow.
The spread operator (...) is used to create a copy of an array in the optimized approach.
The map() method with a callback function is used to create deep copies of arrays.

python

Python's list comprehension [[[0 for _ in range(16)] for _ in range(FACES + 1)] for _ in range(n + 1)] is used to create multi-dimensional arrays.
The != operator is used for inequality comparison.
The ** operator is used for exponentiation (10**9 + 7).
The copy() method and list comprehension [row.copy() for row in consecutive] are used to create deep copies of arrays.
The sum() function with slicing (sum(dp[1:])) is used to sum all elements in an array except the first one.

java

Java's multi-dimensional arrays are created using the new keyword.
The final keyword is used to define constants (final int MOD = 1000000007).
The != operator is used for inequality comparison.
The clone() method is used to create shallow copies of arrays.
Type casting (int) is used to convert the final result to an integer.

cpp

C++'s std::vector is used to create multi-dimensional arrays.
The const keyword is used to define constants (const int MOD = 1000000007).
The != operator is used for inequality comparison.
The reference operator (&) is used to pass vectors by reference to avoid copying.
The std::vector constructor with size and initial value is used to create and initialize vectors.

Key Takeaways

This problem requires defining multiple states to represent different configurations of the dice rolls.
The recurrence relations are derived by considering all possible previous states that can lead to the current state.
Dynamic programming is an efficient way to solve problems with overlapping subproblems.
Space optimization can reduce the memory usage by only keeping track of the previous and current states.
The modulo operation needs to be applied at each step to avoid integer overflow.

Try It Yourself

Modify the code to implement an alternative approach and test it with the same examples.

Challenge:

Implement a function that solves the dice roll simulation problem using a different approach than shown above.

Common Edge Cases

Single Roll

Handle the case where n is 1 (there are 6 possible sequences, one for each face value).

n = 1, rollMax = [1, 1, 2, 2, 2, 3]

No Consecutive Restrictions

Handle the case where all values in rollMax are large enough to not restrict any sequences of length n.

n = 3, rollMax = [3, 3, 3, 3, 3, 3]

Strict Consecutive Restrictions

Handle the case where all values in rollMax are 1, meaning no face value can appear consecutively.

n = 2, rollMax = [1, 1, 1, 1, 1, 1]

Solution Approach All Problems

Problem Solution Code

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Dice Roll Simulation - Implementation

Code Implementation

Dice Roll Simulation Implementation

Below is the implementation of the dice roll simulation:

solution.js

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/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (3D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP array
  // dp[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const dp = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j][1] = 1;
  }
 
  // Fill DP array
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Case 1: Starting a new consecutive sequence (k = 1)
      // We can come from any face j' != j with any valid consecutive count k'
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          for (let k_prev = 1; k_prev <= rollMax[j_prev - 1]; k_prev++) {
            dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j_prev][k_prev]) % MOD;
          }
        }
      }
 
      // Case 2: Extending a consecutive sequence (k > 1)
      // We can only come from the same face j with consecutive count k-1
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        dp[i][j][k] = dp[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    for (let k = 1; k <= rollMax[j - 1]; k++) {
      result = (result + dp[n][j][k]) % MOD;
    }
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator2D(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays
  // dp[i][j] = number of sequences of length i ending with face j
  const dp = Array(n + 1).fill().map(() => Array(FACES + 1).fill(0));
 
  // consecutive[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const consecutive = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j] = 1;
    consecutive[1][j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[i][j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[i][j] = (dp[i][j] + dp[i - 1][j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[i][j] = (dp[i][j] + consecutive[i - 1][j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[i][j][1] = sequences that switch to face j
      consecutive[i][j][1] = 0;
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[i][j][1] = (consecutive[i][j][1] + dp[i - 1][j_prev]) % MOD;
        }
      }
 
      // consecutive[i][j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[i][j][k] = consecutive[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[n][j]) % MOD;
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming with space optimization.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulatorOptimized(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays for current and previous rolls
  let dp = Array(FACES + 1).fill(0);
  let consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[j] = 1;
    consecutive[j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    // Save the previous DP arrays
    const prevDp = [...dp];
    const prevConsecutive = consecutive.map(row => [...row]);
 
    // Reset current DP arrays
    dp = Array(FACES + 1).fill(0);
    consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[j] = (dp[j] + prevDp[j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[j] = (dp[j] + prevConsecutive[j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[j][1] = sequences that switch to face j
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[j][1] = (consecutive[j][1] + prevDp[j_prev]) % MOD;
        }
      }
 
      // consecutive[j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[j][k] = prevConsecutive[j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[j]) % MOD;
  }
 
  return result;
}
 
// Test cases
console.log(dieSimulator(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulator2D(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator2D(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator2D(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulatorOptimized(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulatorOptimized(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulatorOptimized(3, [1, 1, 1, 2, 2, 3]));  // 181

Step-by-Step Explanation

Let's break down the implementation:

Define the States: Define the states for the dynamic programming approach, considering the current roll number, face value, and consecutive occurrences.
Establish Base Cases: Set up the base cases for the first roll, where there is one way to roll each face value.
Derive Recurrence Relations: Derive the recurrence relations by considering all possible previous states that can lead to the current state.
Fill DP Arrays: Use dynamic programming to fill the arrays from the base cases up to the target number of rolls.
Calculate Final Answer: Sum up all valid sequences of the target length to get the final answer, applying the modulo operation to avoid integer overflow.

Previous Next

String Problems

Palindrome Checker

Determine if a string reads the same forward and backward.

EasyString Manipulation

Message Encryption

Reverse a string to create a simple encryption system.

EasyString Manipulation

Word Puzzle Challenge

Check if two strings are anagrams of each other.

EasyString Manipulation

Learning Objective

Implement the dice roll simulation solution in different programming languages.

Dice Roll Simulation Implementation

Below is the implementation of the dice roll simulation in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (3D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP array
  // dp[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const dp = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j][1] = 1;
  }
 
  // Fill DP array
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Case 1: Starting a new consecutive sequence (k = 1)
      // We can come from any face j' != j with any valid consecutive count k'
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          for (let k_prev = 1; k_prev <= rollMax[j_prev - 1]; k_prev++) {
            dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j_prev][k_prev]) % MOD;
          }
        }
      }
 
      // Case 2: Extending a consecutive sequence (k > 1)
      // We can only come from the same face j with consecutive count k-1
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        dp[i][j][k] = dp[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    for (let k = 1; k <= rollMax[j - 1]; k++) {
      result = (result + dp[n][j][k]) % MOD;
    }
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulator2D(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays
  // dp[i][j] = number of sequences of length i ending with face j
  const dp = Array(n + 1).fill().map(() => Array(FACES + 1).fill(0));
 
  // consecutive[i][j][k] = number of sequences of length i ending with k consecutive occurrences of face j
  const consecutive = Array(n + 1).fill().map(() =>
    Array(FACES + 1).fill().map(() =>
      Array(16).fill(0)
    )
  );
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[1][j] = 1;
    consecutive[1][j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[i][j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[i][j] = (dp[i][j] + dp[i - 1][j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[i][j] = (dp[i][j] + consecutive[i - 1][j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[i][j][1] = sequences that switch to face j
      consecutive[i][j][1] = 0;
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[i][j][1] = (consecutive[i][j][1] + dp[i - 1][j_prev]) % MOD;
        }
      }
 
      // consecutive[i][j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[i][j][k] = consecutive[i - 1][j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[n][j]) % MOD;
  }
 
  return result;
}
 
/**
 * Calculate the number of distinct sequences that can be obtained with exact n rolls.
 * Dynamic Programming with space optimization.
 *
 * @param {number} n - Number of rolls
 * @param {number[]} rollMax - Maximum consecutive occurrences for each face value
 * @return {number} - Number of distinct sequences
 */
function dieSimulatorOptimized(n, rollMax) {
  const MOD = 1000000007;
  const FACES = 6;
 
  // Initialize DP arrays for current and previous rolls
  let dp = Array(FACES + 1).fill(0);
  let consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
  // Base cases: for the first roll, there is one way to roll each face
  for (let j = 1; j <= FACES; j++) {
    dp[j] = 1;
    consecutive[j][1] = 1;
  }
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    // Save the previous DP arrays
    const prevDp = [...dp];
    const prevConsecutive = consecutive.map(row => [...row]);
 
    // Reset current DP arrays
    dp = Array(FACES + 1).fill(0);
    consecutive = Array(FACES + 1).fill().map(() => Array(16).fill(0));
 
    for (let j = 1; j <= FACES; j++) {
      // Calculate dp[j]
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          // Add sequences ending with a different face
          dp[j] = (dp[j] + prevDp[j_prev]) % MOD;
        } else {
          // Add sequences ending with the same face, but not exceeding rollMax
          for (let k = 1; k < rollMax[j - 1]; k++) {
            dp[j] = (dp[j] + prevConsecutive[j][k]) % MOD;
          }
        }
      }
 
      // Update consecutive array
      // consecutive[j][1] = sequences that switch to face j
      for (let j_prev = 1; j_prev <= FACES; j_prev++) {
        if (j_prev !== j) {
          consecutive[j][1] = (consecutive[j][1] + prevDp[j_prev]) % MOD;
        }
      }
 
      // consecutive[j][k] for k > 1 = sequences that extend consecutive occurrences of face j
      for (let k = 2; k <= rollMax[j - 1]; k++) {
        consecutive[j][k] = prevConsecutive[j][k - 1];
      }
    }
  }
 
  // Calculate the final answer: sum of all valid sequences of length n
  let result = 0;
  for (let j = 1; j <= FACES; j++) {
    result = (result + dp[j]) % MOD;
  }
 
  return result;
}
 
// Test cases
console.log(dieSimulator(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulator2D(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulator2D(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulator2D(3, [1, 1, 1, 2, 2, 3]));  // 181
 
console.log(dieSimulatorOptimized(2, [1, 1, 2, 2, 2, 3]));  // 34
console.log(dieSimulatorOptimized(2, [1, 1, 1, 1, 1, 1]));  // 30
console.log(dieSimulatorOptimized(3, [1, 1, 1, 2, 2, 3]));  // 181

Step-by-Step Explanation

1Define the States

Define the states for the dynamic programming approach, considering the current roll number, face value, and consecutive occurrences.

2Establish Base Cases

Set up the base cases for the first roll, where there is one way to roll each face value.

3Derive Recurrence Relations

Derive the recurrence relations by considering all possible previous states that can lead to the current state.

4Fill DP Arrays

Use dynamic programming to fill the arrays from the base cases up to the target number of rolls.

5Calculate Final Answer

Sum up all valid sequences of the target length to get the final answer, applying the modulo operation to avoid integer overflow.

Language-Specific Notes

javascript

JavaScript's Array.fill().map() is used to create multi-dimensional arrays.
The !== operator is used for strict inequality comparison.
The % operator is used for the modulo operation to avoid integer overflow.
The spread operator (...) is used to create a copy of an array in the optimized approach.
The map() method with a callback function is used to create deep copies of arrays.

python

Python's list comprehension [[[0 for _ in range(16)] for _ in range(FACES + 1)] for _ in range(n + 1)] is used to create multi-dimensional arrays.
The != operator is used for inequality comparison.
The ** operator is used for exponentiation (10**9 + 7).
The copy() method and list comprehension [row.copy() for row in consecutive] are used to create deep copies of arrays.
The sum() function with slicing (sum(dp[1:])) is used to sum all elements in an array except the first one.

java

Java's multi-dimensional arrays are created using the new keyword.
The final keyword is used to define constants (final int MOD = 1000000007).
The != operator is used for inequality comparison.
The clone() method is used to create shallow copies of arrays.
Type casting (int) is used to convert the final result to an integer.

cpp

C++'s std::vector is used to create multi-dimensional arrays.
The const keyword is used to define constants (const int MOD = 1000000007).
The != operator is used for inequality comparison.
The reference operator (&) is used to pass vectors by reference to avoid copying.
The std::vector constructor with size and initial value is used to create and initialize vectors.

Key Takeaways

This problem requires defining multiple states to represent different configurations of the dice rolls.
The recurrence relations are derived by considering all possible previous states that can lead to the current state.
Dynamic programming is an efficient way to solve problems with overlapping subproblems.
Space optimization can reduce the memory usage by only keeping track of the previous and current states.
The modulo operation needs to be applied at each step to avoid integer overflow.

Try It Yourself

Modify the code to implement an alternative approach and test it with the same examples.

Challenge:

Implement a function that solves the dice roll simulation problem using a different approach than shown above.

Common Edge Cases

Single Roll

Handle the case where n is 1 (there are 6 possible sequences, one for each face value).

n = 1, rollMax = [1, 1, 2, 2, 2, 3]

No Consecutive Restrictions

Handle the case where all values in rollMax are large enough to not restrict any sequences of length n.

n = 3, rollMax = [3, 3, 3, 3, 3, 3]

Strict Consecutive Restrictions

Handle the case where all values in rollMax are 1, meaning no face value can appear consecutively.

n = 2, rollMax = [1, 1, 1, 1, 1, 1]

Solution Approach All Problems

Code Implementation

Dice Roll Simulation Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Dice Roll Simulation Implementation

Step-by-Step Explanation

1Define the States

2Establish Base Cases

3Derive Recurrence Relations

4Fill DP Arrays

5Calculate Final Answer

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

Single Roll

No Consecutive Restrictions

Strict Consecutive Restrictions

Dice Roll Simulation - Implementation

Code Implementation

Dice Roll Simulation Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Dice Roll Simulation Implementation

Step-by-Step Explanation

1Define the States

2Establish Base Cases

3Derive Recurrence Relations

4Fill DP Arrays

5Calculate Final Answer

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

Single Roll

No Consecutive Restrictions

Strict Consecutive Restrictions

String Problems

String Problems