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Code Implementation

Minimum Path Sum Implementation

Below is the implementation of the minimum path sum:

solution.js

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/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (2D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSum(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array
  const dp = Array(m).fill().map(() => Array(n).fill(0));
 
  // Set base case for top-left cell
  dp[0][0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = grid[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = grid[i][0] + dp[i - 1][0];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = grid[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (1D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumOptimized(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array for first row
  const dp = Array(n).fill(0);
  dp[0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[j] = grid[0][j] + dp[j - 1];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    // Update first column
    dp[0] = dp[0] + grid[i][0];
 
    // Update rest of the row
    for (let j = 1; j < n; j++) {
      dp[j] = grid[i][j] + Math.min(dp[j], dp[j - 1]);
    }
  }
 
  return dp[n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * In-place Dynamic Programming approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumInPlace(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Create a copy of the grid to avoid modifying the input
  const dp = grid.map(row => [...row]);
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = dp[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = dp[i][0] + dp[i - 1][0];
  }
 
  // Fill grid
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
// Test cases
const grid1 = [
  [1, 3, 1],
  [1, 5, 1],
  [4, 2, 1]
];
console.log(minPathSum(grid1));  // 7
console.log(minPathSumOptimized(grid1));  // 7
console.log(minPathSumInPlace(grid1));  // 7
 
const grid2 = [
  [1, 2, 3],
  [4, 5, 6]
];
console.log(minPathSum(grid2));  // 12
console.log(minPathSumOptimized(grid2));  // 12
console.log(minPathSumInPlace(grid2));  // 12

Step-by-Step Explanation

Let's break down the implementation:

Initialize DP Array: Set up a dynamic programming array to keep track of the minimum path sum to reach each cell in the grid.
Set Base Cases: Define the base cases for the top-left cell, the first row, and the first column.
Fill DP Array: For each cell, calculate the minimum path sum by adding the value of the cell to the minimum of the path sums from the cell above and the cell to the left.
Return Final Answer: Return the value in the bottom-right corner of the DP array, which represents the minimum path sum to reach the destination.
Optimize Space Complexity: Reduce the space complexity by using a 1D array instead of a 2D array, or by modifying the input grid in-place.

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 minimum path sum solution in different programming languages.

Minimum Path Sum Implementation

Below is the implementation of the minimum path sum in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (2D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSum(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array
  const dp = Array(m).fill().map(() => Array(n).fill(0));
 
  // Set base case for top-left cell
  dp[0][0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = grid[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = grid[i][0] + dp[i - 1][0];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = grid[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (1D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumOptimized(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array for first row
  const dp = Array(n).fill(0);
  dp[0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[j] = grid[0][j] + dp[j - 1];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    // Update first column
    dp[0] = dp[0] + grid[i][0];
 
    // Update rest of the row
    for (let j = 1; j < n; j++) {
      dp[j] = grid[i][j] + Math.min(dp[j], dp[j - 1]);
    }
  }
 
  return dp[n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * In-place Dynamic Programming approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumInPlace(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Create a copy of the grid to avoid modifying the input
  const dp = grid.map(row => [...row]);
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = dp[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = dp[i][0] + dp[i - 1][0];
  }
 
  // Fill grid
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
// Test cases
const grid1 = [
  [1, 3, 1],
  [1, 5, 1],
  [4, 2, 1]
];
console.log(minPathSum(grid1));  // 7
console.log(minPathSumOptimized(grid1));  // 7
console.log(minPathSumInPlace(grid1));  // 7
 
const grid2 = [
  [1, 2, 3],
  [4, 5, 6]
];
console.log(minPathSum(grid2));  // 12
console.log(minPathSumOptimized(grid2));  // 12
console.log(minPathSumInPlace(grid2));  // 12

Step-by-Step Explanation

1Initialize DP Array

Set up a dynamic programming array to keep track of the minimum path sum to reach each cell in the grid.

2Set Base Cases

Define the base cases for the top-left cell, the first row, and the first column.

3Fill DP Array

For each cell, calculate the minimum path sum by adding the value of the cell to the minimum of the path sums from the cell above and the cell to the left.

4Return Final Answer

Return the value in the bottom-right corner of the DP array, which represents the minimum path sum to reach the destination.

5Optimize Space Complexity

Reduce the space complexity by using a 1D array instead of a 2D array, or by modifying the input grid in-place.

Language-Specific Notes

javascript

JavaScript's Array.fill().map() is used to create a 2D array.
Math.min() is used to find the minimum of two values.
The map() method with spread operator ([...row]) is used to create a deep copy of the grid in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The array indexing (grid[i][j]) is used to access the value of each cell in the grid.

python

Python's list comprehension [[0] * n for _ in range(m)] is used to create a 2D array.
The built-in min() function is used to find the minimum of two values.
The row[:] syntax is used to create a copy of a list in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The range() function is used to iterate through the grid from 1 to m-1 and 1 to n-1.

java

Java's new int[m][n] is used to create a 2D array.
Math.min() is used to find the minimum of two values.
A nested for loop is used to create a deep copy of the grid in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The array indexing (grid[i][j]) is used to access the value of each cell in the grid.

cpp

C++'s std::vector> with initialization is used to create a 2D array.
std::min() is used to find the minimum of two values.
The std::vector> dp = grid syntax is used to create a copy of the grid in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The reference operator (&) is used to pass vectors by reference to avoid copying.

Key Takeaways

This problem can be solved using dynamic programming by building up the solution from smaller subproblems.
The minimum path sum for each cell is the value of the cell plus the minimum of the path sums from the cell above and the cell to the left.
The base cases are crucial for the correctness of the solution: the top-left cell, the first row, and the first column.
The space complexity can be optimized by using a 1D array instead of a 2D array, as we only need the values from the current row and the row above.
We can further optimize by modifying the input grid in-place, but this approach should be used with caution as it modifies the input.

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 minimum path sum problem using a different approach than shown above.

Common Edge Cases

Single Cell Grid

Handle the case where the grid has only one cell.

grid = [[5]]

Single Row Grid

Handle the case where the grid has only one row.

grid = [[1, 2, 3, 4, 5]]

Single Column Grid

Handle the case where the grid has only one column.

grid = [[1], [2], [3], [4], [5]]

Large Grid

Handle large grid sizes efficiently to avoid timeout issues.

grid of size 200 × 200

Solution Approach Cherry Pickup

Problem Solution Code

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Minimum Path Sum - Implementation

Code Implementation

Minimum Path Sum Implementation

Below is the implementation of the minimum path sum:

solution.js

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/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (2D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSum(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array
  const dp = Array(m).fill().map(() => Array(n).fill(0));
 
  // Set base case for top-left cell
  dp[0][0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = grid[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = grid[i][0] + dp[i - 1][0];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = grid[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (1D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumOptimized(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array for first row
  const dp = Array(n).fill(0);
  dp[0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[j] = grid[0][j] + dp[j - 1];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    // Update first column
    dp[0] = dp[0] + grid[i][0];
 
    // Update rest of the row
    for (let j = 1; j < n; j++) {
      dp[j] = grid[i][j] + Math.min(dp[j], dp[j - 1]);
    }
  }
 
  return dp[n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * In-place Dynamic Programming approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumInPlace(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Create a copy of the grid to avoid modifying the input
  const dp = grid.map(row => [...row]);
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = dp[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = dp[i][0] + dp[i - 1][0];
  }
 
  // Fill grid
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
// Test cases
const grid1 = [
  [1, 3, 1],
  [1, 5, 1],
  [4, 2, 1]
];
console.log(minPathSum(grid1));  // 7
console.log(minPathSumOptimized(grid1));  // 7
console.log(minPathSumInPlace(grid1));  // 7
 
const grid2 = [
  [1, 2, 3],
  [4, 5, 6]
];
console.log(minPathSum(grid2));  // 12
console.log(minPathSumOptimized(grid2));  // 12
console.log(minPathSumInPlace(grid2));  // 12

Step-by-Step Explanation

Let's break down the implementation:

Initialize DP Array: Set up a dynamic programming array to keep track of the minimum path sum to reach each cell in the grid.
Set Base Cases: Define the base cases for the top-left cell, the first row, and the first column.
Fill DP Array: For each cell, calculate the minimum path sum by adding the value of the cell to the minimum of the path sums from the cell above and the cell to the left.
Return Final Answer: Return the value in the bottom-right corner of the DP array, which represents the minimum path sum to reach the destination.
Optimize Space Complexity: Reduce the space complexity by using a 1D array instead of a 2D array, or by modifying the input grid in-place.

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 minimum path sum solution in different programming languages.

Minimum Path Sum Implementation

Below is the implementation of the minimum path sum in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (2D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSum(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array
  const dp = Array(m).fill().map(() => Array(n).fill(0));
 
  // Set base case for top-left cell
  dp[0][0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = grid[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = grid[i][0] + dp[i - 1][0];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = grid[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * Dynamic Programming (1D) approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumOptimized(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Initialize DP array for first row
  const dp = Array(n).fill(0);
  dp[0] = grid[0][0];
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[j] = grid[0][j] + dp[j - 1];
  }
 
  // Fill DP array
  for (let i = 1; i < m; i++) {
    // Update first column
    dp[0] = dp[0] + grid[i][0];
 
    // Update rest of the row
    for (let j = 1; j < n; j++) {
      dp[j] = grid[i][j] + Math.min(dp[j], dp[j - 1]);
    }
  }
 
  return dp[n - 1];
}
 
/**
 * Find the minimum path sum from top-left to bottom-right.
 * In-place Dynamic Programming approach.
 *
 * @param {number[][]} grid - Grid filled with non-negative integers
 * @return {number} - Minimum path sum
 */
function minPathSumInPlace(grid) {
  const m = grid.length;
  const n = grid[0].length;
 
  // Create a copy of the grid to avoid modifying the input
  const dp = grid.map(row => [...row]);
 
  // Set base cases for first row
  for (let j = 1; j < n; j++) {
    dp[0][j] = dp[0][j] + dp[0][j - 1];
  }
 
  // Set base cases for first column
  for (let i = 1; i < m; i++) {
    dp[i][0] = dp[i][0] + dp[i - 1][0];
  }
 
  // Fill grid
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
    }
  }
 
  return dp[m - 1][n - 1];
}
 
// Test cases
const grid1 = [
  [1, 3, 1],
  [1, 5, 1],
  [4, 2, 1]
];
console.log(minPathSum(grid1));  // 7
console.log(minPathSumOptimized(grid1));  // 7
console.log(minPathSumInPlace(grid1));  // 7
 
const grid2 = [
  [1, 2, 3],
  [4, 5, 6]
];
console.log(minPathSum(grid2));  // 12
console.log(minPathSumOptimized(grid2));  // 12
console.log(minPathSumInPlace(grid2));  // 12

Step-by-Step Explanation

1Initialize DP Array

Set up a dynamic programming array to keep track of the minimum path sum to reach each cell in the grid.

2Set Base Cases

Define the base cases for the top-left cell, the first row, and the first column.

3Fill DP Array

For each cell, calculate the minimum path sum by adding the value of the cell to the minimum of the path sums from the cell above and the cell to the left.

4Return Final Answer

Return the value in the bottom-right corner of the DP array, which represents the minimum path sum to reach the destination.

5Optimize Space Complexity

Reduce the space complexity by using a 1D array instead of a 2D array, or by modifying the input grid in-place.

Language-Specific Notes

javascript

JavaScript's Array.fill().map() is used to create a 2D array.
Math.min() is used to find the minimum of two values.
The map() method with spread operator ([...row]) is used to create a deep copy of the grid in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The array indexing (grid[i][j]) is used to access the value of each cell in the grid.

python

Python's list comprehension [[0] * n for _ in range(m)] is used to create a 2D array.
The built-in min() function is used to find the minimum of two values.
The row[:] syntax is used to create a copy of a list in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The range() function is used to iterate through the grid from 1 to m-1 and 1 to n-1.

java

Java's new int[m][n] is used to create a 2D array.
Math.min() is used to find the minimum of two values.
A nested for loop is used to create a deep copy of the grid in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The array indexing (grid[i][j]) is used to access the value of each cell in the grid.

cpp

C++'s std::vector> with initialization is used to create a 2D array.
std::min() is used to find the minimum of two values.
The std::vector> dp = grid syntax is used to create a copy of the grid in the in-place approach.
The + operator is used to add the value of the cell to the minimum path sum from the previous cells.
The reference operator (&) is used to pass vectors by reference to avoid copying.

Key Takeaways

This problem can be solved using dynamic programming by building up the solution from smaller subproblems.
The minimum path sum for each cell is the value of the cell plus the minimum of the path sums from the cell above and the cell to the left.
The base cases are crucial for the correctness of the solution: the top-left cell, the first row, and the first column.
The space complexity can be optimized by using a 1D array instead of a 2D array, as we only need the values from the current row and the row above.
We can further optimize by modifying the input grid in-place, but this approach should be used with caution as it modifies the input.

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 minimum path sum problem using a different approach than shown above.

Common Edge Cases

Single Cell Grid

Handle the case where the grid has only one cell.

grid = [[5]]

Single Row Grid

Handle the case where the grid has only one row.

grid = [[1, 2, 3, 4, 5]]

Single Column Grid

Handle the case where the grid has only one column.

grid = [[1], [2], [3], [4], [5]]

Large Grid

Handle large grid sizes efficiently to avoid timeout issues.

grid of size 200 × 200

Solution Approach Cherry Pickup

Code Implementation

Minimum Path Sum Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Minimum Path Sum Implementation

Step-by-Step Explanation

1Initialize DP Array

2Set Base Cases

3Fill DP Array

4Return Final Answer

5Optimize Space Complexity

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

Single Cell Grid

Single Row Grid

Single Column Grid

Large Grid

Minimum Path Sum - Implementation

Code Implementation

Minimum Path Sum Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Minimum Path Sum Implementation

Step-by-Step Explanation

1Initialize DP Array

2Set Base Cases

3Fill DP Array

4Return Final Answer

5Optimize Space Complexity

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

Single Cell Grid

Single Row Grid

Single Column Grid

Large Grid

String Problems

String Problems