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Code Implementation

Number of Ways to Stay in the Same Place After Some Steps Implementation

Below is the implementation of the number of ways to stay in the same place after some steps:

solution.js

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/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWays(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize DP array
  const dp = Array(steps + 1).fill().map(() => Array(maxPosition + 1).fill(0));
 
  // Base case: one way to be at position 0 after 0 steps
  dp[0][0] = 1;
 
  // Fill DP array
  for (let i = 1; i <= steps; i++) {
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      dp[i][j] = dp[i - 1][j];
 
      // Move right from position j-1
      if (j > 0) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
      }
    }
  }
 
  return dp[steps][0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (1D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming with Further Optimization approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysFurtherOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, Math.floor(steps / 2));
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
// Test cases
console.log(numWays(3, 2));  // 4
console.log(numWays(2, 4));  // 2
console.log(numWays(4, 2));  // 8
 
console.log(numWaysOptimized(3, 2));  // 4
console.log(numWaysOptimized(2, 4));  // 2
console.log(numWaysOptimized(4, 2));  // 8
 
console.log(numWaysFurtherOptimized(3, 2));  // 4
console.log(numWaysFurtherOptimized(2, 4));  // 2
console.log(numWaysFurtherOptimized(4, 2));  // 8

Step-by-Step Explanation

Let's break down the implementation:

Define Maximum Position: Determine the maximum position we need to consider, which is the minimum of arrLen - 1 and steps (or steps / 2 for further optimization).
Initialize DP Array: Set up a dynamic programming array to keep track of the number of ways to be at each position after each step.
Set Base Case: Define the base case: there is one way to be at position 0 after 0 steps.
Fill DP Array: For each step and position, calculate the number of ways by considering three possible moves: stay, move left, or move right.
Return Final Answer: Return the value in the DP array that represents the number of ways to be at position 0 after the given number of steps.

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 number of ways to stay in the same place after some steps solution in different programming languages.

Number of Ways to Stay in the Same Place After Some Steps Implementation

Below is the implementation of the number of ways to stay in the same place after some steps in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWays(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize DP array
  const dp = Array(steps + 1).fill().map(() => Array(maxPosition + 1).fill(0));
 
  // Base case: one way to be at position 0 after 0 steps
  dp[0][0] = 1;
 
  // Fill DP array
  for (let i = 1; i <= steps; i++) {
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      dp[i][j] = dp[i - 1][j];
 
      // Move right from position j-1
      if (j > 0) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
      }
    }
  }
 
  return dp[steps][0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (1D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming with Further Optimization approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysFurtherOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, Math.floor(steps / 2));
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
// Test cases
console.log(numWays(3, 2));  // 4
console.log(numWays(2, 4));  // 2
console.log(numWays(4, 2));  // 8
 
console.log(numWaysOptimized(3, 2));  // 4
console.log(numWaysOptimized(2, 4));  // 2
console.log(numWaysOptimized(4, 2));  // 8
 
console.log(numWaysFurtherOptimized(3, 2));  // 4
console.log(numWaysFurtherOptimized(2, 4));  // 2
console.log(numWaysFurtherOptimized(4, 2));  // 8

Step-by-Step Explanation

1Define Maximum Position

Determine the maximum position we need to consider, which is the minimum of arrLen - 1 and steps (or steps / 2 for further optimization).

2Initialize DP Array

Set up a dynamic programming array to keep track of the number of ways to be at each position after each step.

3Set Base Case

Define the base case: there is one way to be at position 0 after 0 steps.

4Fill DP Array

For each step and position, calculate the number of ways by considering three possible moves: stay, move left, or move right.

5Return Final Answer

Return the value in the DP array that represents the number of ways to be at position 0 after the given number of steps.

Language-Specific Notes

javascript

JavaScript's Math.min() is used to find the minimum of two values when determining the maximum position.
Array(n).fill().map() is used to create a 2D array for the DP approach.
Array(n).fill(0) is used to create and initialize a 1D array with all elements set to 0.
The % operator is used for the modulo operation to avoid integer overflow.
Destructuring assignment [prev, curr] = [curr, prev] is used to swap arrays efficiently.

python

Python's min() function is used to find the minimum of two values when determining the maximum position.
List comprehension [[0] * (max_position + 1) for _ in range(steps + 1)] is used to create a 2D array for the DP approach.
[0] * (max_position + 1) is used to create and initialize a 1D array with all elements set to 0.
The % operator is used for the modulo operation to avoid integer overflow.
Multiple assignment prev, curr = curr, prev is used to swap arrays efficiently.

java

Java's Math.min() is used to find the minimum of two values when determining the maximum position.
new int[steps + 1][maxPosition + 1] is used to create a 2D array for the DP approach.
new int[maxPosition + 1] is used to create a 1D array, which is initialized with 0s by default.
The % operator is used for the modulo operation to avoid integer overflow.
A temporary variable (int[] temp) is used to swap arrays in the optimized approaches.

cpp

C++'s std::min() is used to find the minimum of two values when determining the maximum position.
std::vector> with initialization is used to create a 2D array for the DP approach.
std::vector(maxPosition + 1, 0) is used to create and initialize a 1D vector with all elements set to 0.
std::fill(curr.begin(), curr.end(), 0) is used to reset the current array in each iteration.
prev.swap(curr) is used to swap vectors efficiently in the optimized approaches.

Key Takeaways

This problem can be solved using dynamic programming by building up the solution from smaller subproblems.
The key insight is to define dp[i][j] as the number of ways to be at position j after i steps.
We can optimize the space complexity by using a 1D array instead of a 2D array, as we only need the values from the previous step.
We can further optimize by observing that we only need to consider positions up to min(arrLen-1, steps/2), as we can't go further than steps/2 positions away from the starting point if we want to return to position 0.
The modulo operation is crucial to avoid integer overflow, as the number of ways can be very large.

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 number of ways to stay in the same place after some steps problem using a different approach than shown above.

Common Edge Cases

Small Steps and Array Length

Handle the case where both steps and arrLen are small.

steps = 2, arrLen = 2

Large Array Length

Handle the case where arrLen is much larger than steps.

steps = 3, arrLen = 1000000

Maximum Steps

Handle the case where steps is at its maximum value.

steps = 500, arrLen = 1000000

Solution Approach All Problems

Problem Solution Code

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Number of Ways to Stay in the Same Place After Some Steps - Implementation

Code Implementation

Number of Ways to Stay in the Same Place After Some Steps Implementation

Below is the implementation of the number of ways to stay in the same place after some steps:

solution.js

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/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWays(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize DP array
  const dp = Array(steps + 1).fill().map(() => Array(maxPosition + 1).fill(0));
 
  // Base case: one way to be at position 0 after 0 steps
  dp[0][0] = 1;
 
  // Fill DP array
  for (let i = 1; i <= steps; i++) {
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      dp[i][j] = dp[i - 1][j];
 
      // Move right from position j-1
      if (j > 0) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
      }
    }
  }
 
  return dp[steps][0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (1D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming with Further Optimization approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysFurtherOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, Math.floor(steps / 2));
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
// Test cases
console.log(numWays(3, 2));  // 4
console.log(numWays(2, 4));  // 2
console.log(numWays(4, 2));  // 8
 
console.log(numWaysOptimized(3, 2));  // 4
console.log(numWaysOptimized(2, 4));  // 2
console.log(numWaysOptimized(4, 2));  // 8
 
console.log(numWaysFurtherOptimized(3, 2));  // 4
console.log(numWaysFurtherOptimized(2, 4));  // 2
console.log(numWaysFurtherOptimized(4, 2));  // 8

Step-by-Step Explanation

Let's break down the implementation:

Define Maximum Position: Determine the maximum position we need to consider, which is the minimum of arrLen - 1 and steps (or steps / 2 for further optimization).
Initialize DP Array: Set up a dynamic programming array to keep track of the number of ways to be at each position after each step.
Set Base Case: Define the base case: there is one way to be at position 0 after 0 steps.
Fill DP Array: For each step and position, calculate the number of ways by considering three possible moves: stay, move left, or move right.
Return Final Answer: Return the value in the DP array that represents the number of ways to be at position 0 after the given number of steps.

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 number of ways to stay in the same place after some steps solution in different programming languages.

Number of Ways to Stay in the Same Place After Some Steps Implementation

Below is the implementation of the number of ways to stay in the same place after some steps in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (2D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWays(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize DP array
  const dp = Array(steps + 1).fill().map(() => Array(maxPosition + 1).fill(0));
 
  // Base case: one way to be at position 0 after 0 steps
  dp[0][0] = 1;
 
  // Fill DP array
  for (let i = 1; i <= steps; i++) {
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      dp[i][j] = dp[i - 1][j];
 
      // Move right from position j-1
      if (j > 0) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
      }
    }
  }
 
  return dp[steps][0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming (1D) approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, steps);
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
/**
 * Calculate the number of ways to stay at index 0 after exactly steps steps.
 * Dynamic Programming with Further Optimization approach.
 *
 * @param {number} steps - Number of steps
 * @param {number} arrLen - Length of the array
 * @return {number} - Number of ways
 */
function numWaysFurtherOptimized(steps, arrLen) {
  const MOD = 1000000007;
 
  // Define the maximum position we need to consider
  const maxPosition = Math.min(arrLen - 1, Math.floor(steps / 2));
 
  // Initialize arrays for previous and current steps
  let prev = Array(maxPosition + 1).fill(0);
  let curr = Array(maxPosition + 1).fill(0);
 
  // Base case: one way to be at position 0 after 0 steps
  prev[0] = 1;
 
  // Fill arrays
  for (let i = 1; i <= steps; i++) {
    // Reset current array
    curr = Array(maxPosition + 1).fill(0);
 
    for (let j = 0; j <= maxPosition; j++) {
      // Stay at position j
      curr[j] = prev[j];
 
      // Move right from position j-1
      if (j > 0) {
        curr[j] = (curr[j] + prev[j - 1]) % MOD;
      }
 
      // Move left from position j+1
      if (j < maxPosition) {
        curr[j] = (curr[j] + prev[j + 1]) % MOD;
      }
    }
 
    // Swap arrays for next step
    [prev, curr] = [curr, prev];
  }
 
  return prev[0];
}
 
// Test cases
console.log(numWays(3, 2));  // 4
console.log(numWays(2, 4));  // 2
console.log(numWays(4, 2));  // 8
 
console.log(numWaysOptimized(3, 2));  // 4
console.log(numWaysOptimized(2, 4));  // 2
console.log(numWaysOptimized(4, 2));  // 8
 
console.log(numWaysFurtherOptimized(3, 2));  // 4
console.log(numWaysFurtherOptimized(2, 4));  // 2
console.log(numWaysFurtherOptimized(4, 2));  // 8

Step-by-Step Explanation

1Define Maximum Position

Determine the maximum position we need to consider, which is the minimum of arrLen - 1 and steps (or steps / 2 for further optimization).

2Initialize DP Array

Set up a dynamic programming array to keep track of the number of ways to be at each position after each step.

3Set Base Case

Define the base case: there is one way to be at position 0 after 0 steps.

4Fill DP Array

For each step and position, calculate the number of ways by considering three possible moves: stay, move left, or move right.

5Return Final Answer

Return the value in the DP array that represents the number of ways to be at position 0 after the given number of steps.

Language-Specific Notes

javascript

JavaScript's Math.min() is used to find the minimum of two values when determining the maximum position.
Array(n).fill().map() is used to create a 2D array for the DP approach.
Array(n).fill(0) is used to create and initialize a 1D array with all elements set to 0.
The % operator is used for the modulo operation to avoid integer overflow.
Destructuring assignment [prev, curr] = [curr, prev] is used to swap arrays efficiently.

python

Python's min() function is used to find the minimum of two values when determining the maximum position.
List comprehension [[0] * (max_position + 1) for _ in range(steps + 1)] is used to create a 2D array for the DP approach.
[0] * (max_position + 1) is used to create and initialize a 1D array with all elements set to 0.
The % operator is used for the modulo operation to avoid integer overflow.
Multiple assignment prev, curr = curr, prev is used to swap arrays efficiently.

java

Java's Math.min() is used to find the minimum of two values when determining the maximum position.
new int[steps + 1][maxPosition + 1] is used to create a 2D array for the DP approach.
new int[maxPosition + 1] is used to create a 1D array, which is initialized with 0s by default.
The % operator is used for the modulo operation to avoid integer overflow.
A temporary variable (int[] temp) is used to swap arrays in the optimized approaches.

cpp

C++'s std::min() is used to find the minimum of two values when determining the maximum position.
std::vector> with initialization is used to create a 2D array for the DP approach.
std::vector(maxPosition + 1, 0) is used to create and initialize a 1D vector with all elements set to 0.
std::fill(curr.begin(), curr.end(), 0) is used to reset the current array in each iteration.
prev.swap(curr) is used to swap vectors efficiently in the optimized approaches.

Key Takeaways

This problem can be solved using dynamic programming by building up the solution from smaller subproblems.
The key insight is to define dp[i][j] as the number of ways to be at position j after i steps.
We can optimize the space complexity by using a 1D array instead of a 2D array, as we only need the values from the previous step.
We can further optimize by observing that we only need to consider positions up to min(arrLen-1, steps/2), as we can't go further than steps/2 positions away from the starting point if we want to return to position 0.
The modulo operation is crucial to avoid integer overflow, as the number of ways can be very large.

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 number of ways to stay in the same place after some steps problem using a different approach than shown above.

Common Edge Cases

Small Steps and Array Length

Handle the case where both steps and arrLen are small.

steps = 2, arrLen = 2

Large Array Length

Handle the case where arrLen is much larger than steps.

steps = 3, arrLen = 1000000

Maximum Steps

Handle the case where steps is at its maximum value.

steps = 500, arrLen = 1000000

Solution Approach All Problems

Code Implementation

Number of Ways to Stay in the Same Place After Some Steps Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Number of Ways to Stay in the Same Place After Some Steps Implementation

Step-by-Step Explanation

1Define Maximum Position

2Initialize DP Array

3Set Base Case

4Fill DP Array

5Return Final Answer

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

Small Steps and Array Length

Large Array Length

Maximum Steps

Number of Ways to Stay in the Same Place After Some Steps - Implementation

Code Implementation

Number of Ways to Stay in the Same Place After Some Steps Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Number of Ways to Stay in the Same Place After Some Steps Implementation

Step-by-Step Explanation

1Define Maximum Position

2Initialize DP Array

3Set Base Case

4Fill DP Array

5Return Final Answer

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

Small Steps and Array Length

Large Array Length

Maximum Steps

String Problems

String Problems