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());
}
};
Java
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;
}
}