1289

1289. Minimum Falling Path Sum II (Hard)

Given a square grid of integers arr, a falling path with non-zero shifts is a choice of exactly one element from each row of arr, such that no two elements chosen in adjacent rows are in the same column.

Return the minimum sum of a falling path with non-zero shifts.

 

Example 1:

Input: arr = [[1,2,3],[4,5,6],[7,8,9]]
Output: 13
Explanation: 
The possible falling paths are:
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
The falling path with the smallest sum is [1,5,7], so the answer is 13.

 

Constraints:

• 1 <= arr.length == arr[i].length <= 200
• -99 <= arr[i][j] <= 99

Related Topics:
Dynamic Programming

Similar Questions:

• Minimum Falling Path Sum (Medium)

Solution 1. DP

// OJ: https://leetcode.com/problems/minimum-falling-path-sum-ii/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
    pair<int, int> getSmallestTwo(vector<int> &A) {
        auto p = make_pair(-1, -1);
        for (int i = 0; i < A.size(); ++i) {
            if (p.first == -1 || A[i] < A[p.first]) {
                p.second = p.first;
                p.first = i;
            } else if (p.second == -1 || A[i] < A[p.second]) p.second = i;
        }
        return p;
    }
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        int N = A.size();
        if (N == 1) return A[0][0];
        for (int i = 1; i < N; ++i) {
            auto p = getSmallestTwo(A[i - 1]);
            for (int j = 0; j < N; ++j) {
                A[i][j] += A[i - 1][p.first == j ? p.second : p.first];
            }
        }
        return *min_element(A.back().begin(), A.back().end());
    }
};

Solution 2. DP

In case it's not allowed to change input array.

// OJ: https://leetcode.com/problems/minimum-falling-path-sum-ii/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
    pair<int, int> getSmallestTwo(vector<int> &A) {
        auto p = make_pair(-1, -1);
        for (int i = 0; i < A.size(); ++i) {
            if (p.first == -1 || A[i] < A[p.first]) {
                p.second = p.first;
                p.first = i;
            } else if (p.second == -1 || A[i] < A[p.second]) p.second = i;
        }
        return p;
    }
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        int N = A.size();
        if (N == 1) return A[0][0];
        vector<vector<int>> dp(2, vector<int>(N));
        for (int i = 0; i < N; ++i) dp[1][i] = A[0][i];
        for (int i = 1; i < N; ++i) {
            auto p = getSmallestTwo(dp[i % 2]);
            for (int j = 0; j < N; ++j) {
                dp[(i + 1) % 2][j] = A[i][j] + dp[i % 2][p.first == j ? p.second : p.first];
            }
        }
        return *min_element(dp[N % 2].begin(), dp[N % 2].end());
    }
};