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Formatted question description: https://leetcode.ca/all/1289.html
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:
Solution 1. DP
// OJ: https://leetcode.com/problems/minimum-falling-path-sum-ii/
// 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/
// 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());
}
};
-
class Solution { public int minFallingPathSum(int[][] arr) { int side = arr.length; if (side <= 1) return 0; int[][] dp = new int[side][side]; for (int i = 0; i < side; i++) dp[0][i] = arr[0][i]; for (int i = 1; i < side; i++) { int[] minPrevRow = new int[side]; System.arraycopy(dp[i - 1], 0, minPrevRow, 0, side); Arrays.sort(minPrevRow); int min1 = minPrevRow[0], min2 = minPrevRow[1]; if (min1 == min2) { for (int j = 0; j < side; j++) dp[i][j] = min1 + arr[i][j]; } else { int minIndex = 0; for (int j = 0; j < side; j++) { if (dp[i - 1][j] == min1) { minIndex = j; break; } } for (int j = 0; j < side; j++) { int prevMin = j == minIndex ? min2 : min1; dp[i][j] = prevMin + arr[i][j]; } } } int minSum = Integer.MAX_VALUE; for (int i = 0; i < side; i++) minSum = Math.min(minSum, dp[side - 1][i]); return minSum; } } ############ class Solution { public int minFallingPathSum(int[][] grid) { int f = 0, g = 0; int fp = -1; final int inf = 1 << 30; for (int[] row : grid) { int ff = inf, gg = inf; int ffp = -1; for (int j = 0; j < row.length; ++j) { int s = (j != fp ? f : g) + row[j]; if (s < ff) { gg = ff; ff = s; ffp = j; } else if (s < gg) { gg = s; } } f = ff; g = gg; fp = ffp; } return f; } } -
// OJ: https://leetcode.com/problems/minimum-falling-path-sum-ii/ // 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()); } }; -
class Solution: def minFallingPathSum(self, grid: List[List[int]]) -> int: f = g = 0 fp = -1 for row in grid: ff = gg = inf ffp = -1 for j, v in enumerate(row): s = (g if j == fp else f) + v if s < ff: gg = ff ff = s ffp = j elif s < gg: gg = s f, g, fp, = ( ff, gg, ffp, ) return f -
func minFallingPathSum(grid [][]int) int { const inf = 1 << 30 f, g := 0, 0 fp := -1 for _, row := range grid { ff, gg := inf, inf ffp := -1 for j, v := range row { s := f if j == fp { s = g } s += v if s < ff { ff, gg, ffp = s, ff, j } else if s < gg { gg = s } } f, g, fp = ff, gg, ffp } return f }