LeetCode-in-Java
1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows
Hard
You are given an m x n matrix mat that has its rows sorted in non-decreasing order and an integer k.
You are allowed to choose exactly one element from each row to form an array.
Return the kth smallest array sum among all possible arrays.
Example 1:
Input: mat = [[1,3,11],[2,4,6]], k = 5
Output: 7
Explanation: Choosing one element from each row, the first k smallest sum are: [1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
Example 2:
Input: mat = [[1,3,11],[2,4,6]], k = 9
Output: 17
Example 3:
Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
Output: 9
Explanation: Choosing one element from each row, the first k smallest sum are: [1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.
Constraints:
m == mat.lengthn == mat.length[i]1 <= m, n <= 401 <= mat[i][j] <= 50001 <= k <= min(200, nm)mat[i]is a non-decreasing array.
Solution
import java.util.Arrays;
import java.util.Objects;
import java.util.TreeSet;
public class Solution {
public int kthSmallest(int[][] mat, int k) {
TreeSet<int[]> treeSet =
new TreeSet<>(
(o1, o2) -> {
if (o1[0] != o2[0]) {
return o1[0] - o2[0];
} else {
for (int i = 1; i < o1.length; i++) {
if (o1[i] != o2[i]) {
return o1[i] - o2[i];
}
}
return 0;
}
});
int m = mat.length;
int n = mat[0].length;
int sum = 0;
int[] entry = new int[m + 1];
for (int[] ints : mat) {
sum += ints[0];
}
entry[0] = sum;
treeSet.add(entry);
int count = 0;
while (count < k) {
int[] curr = treeSet.pollFirst();
count++;
if (count == k) {
return Objects.requireNonNull(curr)[0];
}
for (int i = 0; i < m; i++) {
int[] next = Arrays.copyOf(Objects.requireNonNull(curr), curr.length);
if (curr[i + 1] + 1 < n) {
next[0] -= mat[i][curr[i + 1]];
next[0] += mat[i][curr[i + 1] + 1];
next[i + 1]++;
treeSet.add(next);
}
}
}
return -1;
}
}