LeetCode #1439 — HARD

Find the Kth Smallest Sum of a Matrix With Sorted Rows

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

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 k^th 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, n^m)
mat[i] is a non-decreasing array.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[[1,3,11],[2,4,6]]
5

Example 2

[[1,3,11],[2,4,6]]
9

Example 3

[[1,10,10],[1,4,5],[2,3,6]]
7

Step 02

Core Insight

What unlocks the optimal approach

Save all visited sums and corresponding indexes in a priority queue. Then, once you pop the smallest sum so far, you can quickly identify the next m candidates for smallest sum by incrementing each row index by 1.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1439: Find the Kth Smallest Sum of a Matrix With Sorted Rows
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);
    }
}

// Accepted solution for LeetCode #1439: Find the Kth Smallest Sum of a Matrix With Sorted Rows
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]
}

# Accepted solution for LeetCode #1439: Find the Kth Smallest Sum of a Matrix With Sorted Rows
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]

// Accepted solution for LeetCode #1439: Find the Kth Smallest Sum of a Matrix With Sorted Rows
struct Solution;

use std::cmp::Reverse;
use std::collections::BinaryHeap;
use std::collections::HashSet;

impl Solution {
    fn kth_smallest(mat: Vec<Vec<i32>>, mut k: i32) -> i32 {
        let n = mat.len();
        let m = mat[0].len();
        let mut queue: BinaryHeap<(Reverse<i32>, Vec<usize>)> = BinaryHeap::new();
        let mut visited: HashSet<Vec<usize>> = HashSet::new();
        let mut sum = 0;
        let indexes = vec![0; n];
        for i in 0..n {
            sum += mat[i][0];
        }
        visited.insert(indexes.to_vec());
        queue.push((Reverse(sum), indexes));
        let mut res = 0;
        while k > 0 {
            if let Some((Reverse(sum), indexes)) = queue.pop() {
                res = sum;
                for i in 0..n {
                    let mut next_sum = sum;
                    let mut next_indexes = indexes.to_vec();
                    let j = indexes[i];
                    if j + 1 < m {
                        next_indexes[i] = j + 1;
                        if visited.insert(next_indexes.to_vec()) {
                            next_sum -= mat[i][j];
                            next_sum += mat[i][j + 1];
                            queue.push((Reverse(next_sum), next_indexes));
                        }
                    }
                }
            }
            k -= 1;
        }
        res
    }
}

#[test]
fn test() {
    let mat = vec_vec_i32![[1, 3, 11], [2, 4, 6]];
    let k = 5;
    let res = 7;
    assert_eq!(Solution::kth_smallest(mat, k), res);
    let mat = vec_vec_i32![[1, 3, 11], [2, 4, 6]];
    let k = 9;
    let res = 17;
    assert_eq!(Solution::kth_smallest(mat, k), res);
    let mat = vec_vec_i32![[1, 10, 10], [1, 4, 5], [2, 3, 6]];
    let k = 7;
    let res = 9;
    assert_eq!(Solution::kth_smallest(mat, k), res);
    let mat = vec_vec_i32![[1, 1, 10], [2, 2, 9]];
    let k = 7;
    let res = 12;
    assert_eq!(Solution::kth_smallest(mat, k), res);
}

// Accepted solution for LeetCode #1439: Find the Kth Smallest Sum of a Matrix With Sorted Rows
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];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.