64. Minimum Path Sum

LeetCode

Dynamic Programming

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

        int[][] dp = new int[m][n];

        // update the first row and column
        for (int i = 0; i < m; i++) {
            if (i == 0) {
                dp[i][0] = grid[0][0];
            } else {
                dp[i][0] = dp[i-1][0] + grid[i][0];
            }
        }

        for (int j = 0; j < n; j++) {
            if (j == 0) {
                dp[0][j] = grid[0][0];
            } else {
                dp[0][j] = dp[0][j-1] + grid[0][j];
            }
        }

        // calculate from (1,1)
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.min(dp[i-1][j] + grid[i][j], dp[i][j-1] + grid[i][j]);
            }
        }

        return dp[m-1][n-1];

    }
}

Remarks:

  1. Adapted from No. 62.
  2. TC: $O(m\times n)$, SC: $O(m\times n)$.

DP with pruning (top-buttom)

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

        int[][] dp = new int[m][n];
        for (int[] row : dp) {
            Arrays.fill(row, -1);
        }

        return rec(m - 1, n - 1, grid, dp);
    }

    private int rec(int i, int j, int[][] grid, int[][] dp) {
        if (i < 0 || j < 0) {
            return Integer.MAX_VALUE;  // out of range
        }

        if (i == 0 && j == 0) {
            return grid[0][0];  // start
        }

        if (dp[i][j] != -1) {
            return dp[i][j];  // memorize
        }

        int left = rec(i, j - 1, grid, dp);
        int up = rec(i - 1, j, grid, dp);

        dp[i][j] = grid[i][j] + Math.min(left, up);
        return dp[i][j];
    }
}

Remarks:

  1. TC: $O(m\times n)$, SC: $O(m\times n)$.
  2. Pruning: only calculate the blocks that are required to find the minimum path. Larger stack usage.