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

Solution: Queue

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[][]} mat
 * @param {number} k
 * @return {number}
 */
const kthSmallest = function (mat, k) {
  const m = mat.length;
  const n = mat[0].length;
  let currentSums = mat[0];

  const kSmallestSums = (nums1, nums2, k) => {
    const minHeap = new MinPriorityQueue({ priority: ({ sum }) => sum });
    const result = [];

    for (const [index, element] of nums1.entries()) {
      const sum = element + nums2[0];

      minHeap.enqueue({ a: index, b: 0, sum });
    }

    while (minHeap.size() && result.length < k) {
      const { a, b, sum } = minHeap.dequeue().element;

      result.push(sum);

      if (b + 1 < n) {
        const nextSum = nums1[a] + nums2[b + 1];

        minHeap.enqueue({ a, b: b + 1, sum: nextSum });
      }
    }

    return result;
  };

  for (let row = 1; row < m; row++) {
    currentSums = kSmallestSums(currentSums, mat[row], k);
  }

  return currentSums[k - 1];
};