Question

Formatted question description: https://leetcode.ca/all/64.html

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:

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


@tag-array

Algorithm

Code

Java

• Java
• C++
• Python

• 
  public class Minimum_Path_Sum {
  
  	public static void main (String[] args) {
  		Minimum_Path_Sum out = new Minimum_Path_Sum();
  		Solution_recursion s = out.new Solution_recursion();
  
  		System.out.println(s.minPathSum(new int[][]{ {1,3,1}, {1,5,1}, {4,2,1} }));
  	}
  
  	public class Solution {
  	    public int minPathSum(int[][] grid) {
  
  	        if (grid.length == 0 || grid[0].length == 0) {
  	            return 0;
  	        }
  
  	        int m = grid.length;
  	        int n = grid[0].length;
  
  	        // dp[i][j] meaning min-sum at (i,j)
  	        int[][] dp = new int[m][n];
  
  	        dp[0][0] = grid[0][0];
  
  	        // pre-process, first row and col
  	        for (int j = 1; j < n; j++) {
  	            dp[0][j] = dp[0][j - 1] + grid[0][j];
  	        }
  	        for (int i = 1; i < m; i++) {
  	            dp[i][0] = dp[i - 1][0] + grid[i][0];
  	        }
  
  	        for (int i = 1; i < m; i++) {
  	            for (int 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];
  	    }
  
  	}
  
  
  	class Solution_recursion {
  
  		int min = Integer.MAX_VALUE;
  
  		public int minPathSum(int[][] grid) {
  
  			if (grid == null || grid.length == 0 || grid[0].length == 0) {
  				return -1;
  			}
  
  			dfs (grid, 0, 0, 0);
  			return min;
  		}
  
  		private void dfs(int[][] grid, int i, int j, int sum) {
  
  			if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) {
  				// out of boundary
  				return;
  			}
  
  			if (i == grid.length - 1 && j == grid[0].length - 1) {
  				// reaching termination position
  				min = Math.min(sum + grid[i][j], min);
  			}
  
  			dfs(grid, i + 1, j, sum + grid[i][j]);
  			dfs(grid, i, j + 1, sum + grid[i][j]);
  
  			// actually, if detour is allowed, then should probe for all 4 directions, since maybe there is a Ingeter.MIN value
  //			dfs(grid, i + 1, j, sum + grid[i][j]);
  //			dfs(grid, i + 1, j, sum + grid[i][j]);
  		}
  
  	}
  
  }
  

• // OJ: https://leetcode.com/problems/minimum-path-sum/
  // Time: O(MN)
  // Space: O(1)
  class Solution {
  public:
      int minPathSum(vector<vector<int>>& A) {
          int M = A.size(), N = A[0].size();
          for (int i = 0; i < M; ++i) {
              for (int j = 0; j < N; ++j) {
                  int sum = min(i - 1 >= 0 ? A[i - 1][j] : INT_MAX, j - 1 >= 0 ? A[i][j - 1] : INT_MAX);
                  A[i][j] += sum == INT_MAX ? 0 : sum;
              }
          }
          return A[M - 1][N - 1];
      }
  };
  

• class Solution(object):
    def minPathSum(self, grid):
      """
      :type grid: List[List[int]]
      :rtype: int
      """
      if len(grid) == 0:
        return 0
      dp = [0 for _ in range(0, len(grid[0]))]
      dp[0] = grid[0][0]
  
      for j in range(1, len(grid[0])):
        dp[j] = dp[j - 1] + grid[0][j]
  
      for i in range(1, len(grid)):
        pre = dp[0] + grid[i][0]
        for j in range(1, len(grid[0])):
          dp[j] = min(dp[j], pre) + grid[i][j]
          pre = dp[j]
        dp[0] += grid[i][0]
  
      return dp[-1]
  
  

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