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1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows

Description

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.length
  • n == mat.length[i]
  • 1 <= m, n <= 40
  • 1 <= mat[i][j] <= 5000
  • 1 <= k <= min(200, nm)
  • mat[i] is a non-decreasing array.

Solutions

  • class Solution {
    public int kthSmallest(int[][] mat, int k) {
        int m = mat.length, n = mat[0].length;
        List<Integer> pre = new ArrayList<>(k);
        List<Integer> cur = new ArrayList<>(n * k);
        pre.add(0);
        for (int[] row : mat) {
            cur.clear();
            for (int a : pre) {
                for (int b : row) {
                    cur.add(a + b);
                }
            }
            Collections.sort(cur);
            pre.clear();
            for (int i = 0; i < Math.min(k, cur.size()); ++i) {
                pre.add(cur.get(i));
            }
        }
        return pre.get(k - 1);
    }
}

  • class Solution {
public:
    int kthSmallest(vector<vector<int>>& mat, int k) {
        int pre[k];
        int cur[mat[0].size() * k];
        memset(pre, 0, sizeof pre);
        int size = 1;
        for (auto& row : mat) {
            int i = 0;
            for (int j = 0; j < size; ++j) {
                for (int& v : row) {
                    cur[i++] = pre[j] + v;
                }
            }
            sort(cur, cur + i);
            size = min(i, k);
            for (int j = 0; j < size; ++j) {
                pre[j] = cur[j];
            }
        }
        return pre[k - 1];
    }
};

  • class Solution:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        pre = [0]
        for cur in mat:
            pre = sorted(a + b for a in pre for b in cur[:k])[:k]
        return pre[-1]


  • func kthSmallest(mat [][]int, k int) int {
	pre := []int{0}
	for _, row := range mat {
		cur := []int{}
		for _, a := range pre {
			for _, b := range row {
				cur = append(cur, a+b)
			}
		}
		sort.Ints(cur)
		pre = cur[:min(k, len(cur))]
	}
	return pre[k-1]
}

  • function kthSmallest(mat: number[][], k: number): number {
    let pre: number[] = [0];
    for (const cur of mat) {
        const next: number[] = [];
        for (const a of pre) {
            for (const b of cur) {
                next.push(a + b);
            }
        }
        pre = next.sort((a, b) => a - b).slice(0, k);
    }
    return pre[k - 1];
}

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