Minimum Path Sum LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 200

 Minimum Path Sum Solutions

Time: O(mn)
Space: O(1)

C++

class Solution {
 public:
  int minPathSum(vector<vector<int>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (i > 0 && j > 0)
          grid[i][j] += min(grid[i - 1][j], grid[i][j - 1]);
        else if (i > 0)
          grid[i][0] += grid[i - 1][0];
        else if (j > 0)
          grid[0][j] += grid[0][j - 1];

    return grid[m - 1][n - 1];
  }
};

Java

 class Solution {
  public int minPathSum(int[][] grid) {
    final int m = grid.length;
    final int n = grid[0].length;

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (i > 0 && j > 0)
          grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
        else if (i > 0)
          grid[i][0] += grid[i - 1][0];
        else if (j > 0)
          grid[0][j] += grid[0][j - 1];

    return grid[m - 1][n - 1];
  }
}

Python

  class Solution:
  def minPathSum(self, grid: List[List[int]]) -> int:
    m = len(grid)
    n = len(grid[0])

    for i in range(m):
      for j in range(n):
        if i > 0 and j > 0:
          grid[i][j] += min(grid[i - 1][j], grid[i][j - 1])
        elif i > 0:
          grid[i][0] += grid[i - 1][0]
        elif j > 0:
          grid[0][j] += grid[0][j - 1]

    return grid[m - 1][n - 1]

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