LeetCode – Minimum Path Sum (Java)

 

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.

Java Solution 1: Depth-First Search

A native solution would be depth-first search. It's time is too expensive and fails the online judgement.

public int minPathSum(int[][] grid) {
    return dfs(0,0,grid);
}
 
public int dfs(int i, int j, int[][] grid){
    if(i==grid.length-1 && j==grid[0].length-1){
        return grid[i][j];
    }
 
    if(i<grid.length-1 && j<grid[0].length-1){
        int r1 = grid[i][j] + dfs(i+1, j, grid);
        int r2 = grid[i][j] + dfs(i, j+1, grid);
        return Math.min(r1,r2);
    }
 
    if(i<grid.length-1){
        return grid[i][j] + dfs(i+1, j, grid);
    }
 
    if(j<grid[0].length-1){
        return grid[i][j] + dfs(i, j+1, grid);
    }
 
    return 0;
}

Java Solution 2: Dynamic Programming

public int minPathSum(int[][] grid) {
    if(grid == null || grid.length==0)
        return 0;
 
    int m = grid.length;
    int n = grid[0].length;
 
    int[][] dp = new int[m][n];
    dp[0][0] = grid[0][0];    
 
    // initialize top row
    for(int i=1; i<n; i++){
        dp[0][i] = dp[0][i-1] + grid[0][i];
    }
 
    // initialize left column
    for(int j=1; j<m; j++){
        dp[j][0] = dp[j-1][0] + grid[j][0];
    }
 
    // fill up the dp table
    for(int i=1; i<m; i++){
        for(int j=1; j<n; j++){
            if(dp[i-1][j] > dp[i][j-1]){
                dp[i][j] = dp[i][j-1] + grid[i][j];
            }else{
                dp[i][j] = dp[i-1][j] + grid[i][j];
            }
        }
    }
 
    return dp[m-1][n-1];
}

Related posts:

  1. LeetCode – Number of Islands (Java)
  2. LeetCode – Longest Increasing Path in a Matrix (Java)
  3. LeetCode – Unique Paths II (Java)
  4. LeetCode – Path Sum II (Java)
Category >> Algorithms >> Interview  
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  1. jason zhang on 2015-5-31

    The DP assumes that we can only walk down, but not up. If we can walk up, this is a a shortest ptah problem

  2. Sharan Srinivasan on 2015-12-3

    The recursive dfs method can be made more efficient with memoization. Store each i,j value previously computed and call it each it you need it – following the first computation

  3. Ryan Shaw on 2015-12-30

    Dijkstra algorithm is the solution.

  4. Vasyl Grygoryev on 2016-1-3

    For 2nd solution we can use only one row for temp datas – O(n) extra space:

    public int minPathSum(int[][] grid) {
    if (grid == null || grid.length == 0 || grid[0].length == 0)
    return 0;

    int[] temp = new int[grid[0].length];

    for (int i = 0; i < grid.length; i++)
    for (int j = 0; j 0)
    if (i > 0)
    temp[j] = Math.min(temp[j], temp[j - 1]);
    else
    temp[j] = temp[j - 1];
    temp[j] += grid[i][j];
    }

    return temp[temp.length - 1];
    }

  5. AJ on 2016-10-15

    static int minPath(int[][] a) {
    if (a == null || a.length == 0) {
    return 0;
    }
    int m = a.length;
    int n = a[0].length;
    for (int i = 1; i < m; i++) {
    a[i][0] = a[i – 1][0] + a[i][0];
    }

    for (int j = 1; j < n; j++) {
    a[0][j] = a[0][j – 1] + a[0][j];
    }

    for (int i = 1; i < m; i++) {
    for (int j = 1; j < n; j++) {
    a[i][j] = Math.min(a[i – 1][j], a[i][j – 1]) + a[i][j];
    }
    }
    return a[m – 1][n – 1];
    }

  6. ABK on 2016-12-6

    Exactly

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