Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

Notice

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

Example

Given the following triangle:

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Solution 1: Search

public int minimumTotal(int[][] triangle) {
    if(triangle == null || triangle.length == 0) {
        return 0;
    }
    int[] min = new int[1];
    return search(triangle, 0, 0, 0, min);
}

public int search(int[][] triangle, int i, int j, int sum, int[] min) {
    int n = triangle.length;
    if(i < 0 || j < 0 || i >= triangle.length || j >= triangle[i].length) {
        return 0;
    }
    if(i == triangle.length) {
        if(sum < min[0]) {
            min[0] = sum;
            return min[0];
        }
    }
    min[0] = triangle[i][j] + Math.min(search(triangle, i + 1, j, sum + triangle[i][j], min), search(triangle, i + 1, j + 1, sum + triangle[i][j], min));
    return min[0];    
}

Solution 2: Memory Search.

public int minimumTotal(int[][] triangle) {
    if(triangle == null || triangle.length == 0) {
        return 0;
    }
    int n = triangle.length;
    int[][] dp = new int[n][n];
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            dp[i][j] = -1;
        }
    }
    return memorySearch(triangle, 0, 0, 0, dp);
}

public int memorySearch(int[][] triangle, int i, int j, int sum, int[][] dp) {
    if(i < 0 || j < 0 || i >= triangle.length || j >= triangle.length) {
        return 0;
    }
    if(dp[i][j] != -1) {
        return dp[i][j];
    }
    if(i == triangle.length) {
        if(sum < dp[i][j]) {
            dp[i][j] = sum;
            return dp[i][j];
        }
    }
    dp[i][j] = triangle[i][j] + Math.min(memorySearch(triangle, i + 1, j, sum + triangle[i][j], dp),
                                         memorySearch(triangle, i + 1, j + 1, sum + triangle[i][j], dp));
    return dp[i][j];
}

Solution 3: DP

public int minimumTotal(int[][] triangle) {
    if(triangle == null || triangle.length == 0) {
        return 0;
    }
    int n = triangle.length;
    int[][] dp = new int[n][n];
    for(int i = 0; i < n; i++) {
        dp[n - 1][i] = triangle[n - 1][i];
    }
    for(int i = n - 2; i >= 0; i--) {
        for(int j = 0; j < triangle[i].length; j++) {
            dp[i][j] = triangle[i][j] + Math.min(dp[i + 1][j], dp[i + 1][j + 1]);
        }
    }
    return dp[0][0];
}