LeetCode #373 — MEDIUM

Find K Pairs with Smallest Sums

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given two integer arrays nums1 and nums2 sorted in non-decreasing order and an integer k.

Define a pair (u, v) which consists of one element from the first array and one element from the second array.

Return the k pairs (u₁, v₁), (u₂, v₂), ..., (u_k, v_k) with the smallest sums.

Example 1:

Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]]
Explanation: The first 3 pairs are returned from the sequence: [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]

Example 2:

Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [[1,1],[1,1]]
Explanation: The first 2 pairs are returned from the sequence: [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]

Constraints:

1 <= nums1.length, nums2.length <= 10⁵
-10⁹ <= nums1[i], nums2[i] <= 10⁹
nums1 and nums2 both are sorted in non-decreasing order.
1 <= k <= 10⁴
k <= nums1.length * nums2.length

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays nums1 and nums2 sorted in non-decreasing order and an integer k. Define a pair (u, v) which consists of one element from the first array and one element from the second array. Return the k pairs (u1, v1), (u2, v2), ..., (uk, vk) with the smallest sums.

Baseline thinking

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

Pattern signal: Array

Example 1

[1,7,11]
[2,4,6]
3

Example 2

[1,1,2]
[1,2,3]
2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Upper-end input sizes

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 #373: Find K Pairs with Smallest Sums
class Solution {
    public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        for (int i = 0; i < Math.min(nums1.length, k); ++i) {
            q.offer(new int[] {nums1[i] + nums2[0], i, 0});
        }
        List<List<Integer>> ans = new ArrayList<>();
        while (!q.isEmpty() && k > 0) {
            int[] e = q.poll();
            ans.add(Arrays.asList(nums1[e[1]], nums2[e[2]]));
            --k;
            if (e[2] + 1 < nums2.length) {
                q.offer(new int[] {nums1[e[1]] + nums2[e[2] + 1], e[1], e[2] + 1});
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #373: Find K Pairs with Smallest Sums
func kSmallestPairs(nums1, nums2 []int, k int) (ans [][]int) {
	m, n := len(nums1), len(nums2)
	h := hp{nil, nums1, nums2}
	for i := 0; i < k && i < m; i++ {
		h.data = append(h.data, pair{i, 0})
	}
	for h.Len() > 0 && len(ans) < k {
		p := heap.Pop(&h).(pair)
		i, j := p.i, p.j
		ans = append(ans, []int{nums1[i], nums2[j]})
		if j+1 < n {
			heap.Push(&h, pair{i, j + 1})
		}
	}
	return
}

type pair struct{ i, j int }
type hp struct {
	data         []pair
	nums1, nums2 []int
}

func (h hp) Len() int { return len(h.data) }
func (h hp) Less(i, j int) bool {
	a, b := h.data[i], h.data[j]
	return h.nums1[a.i]+h.nums2[a.j] < h.nums1[b.i]+h.nums2[b.j]
}
func (h hp) Swap(i, j int) { h.data[i], h.data[j] = h.data[j], h.data[i] }
func (h *hp) Push(v any)   { h.data = append(h.data, v.(pair)) }
func (h *hp) Pop() any     { a := h.data; v := a[len(a)-1]; h.data = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #373: Find K Pairs with Smallest Sums
class Solution:
    def kSmallestPairs(
        self, nums1: List[int], nums2: List[int], k: int
    ) -> List[List[int]]:
        q = [[u + nums2[0], i, 0] for i, u in enumerate(nums1[:k])]
        heapify(q)
        ans = []
        while q and k > 0:
            _, i, j = heappop(q)
            ans.append([nums1[i], nums2[j]])
            k -= 1
            if j + 1 < len(nums2):
                heappush(q, [nums1[i] + nums2[j + 1], i, j + 1])
        return ans

// Accepted solution for LeetCode #373: Find K Pairs with Smallest Sums
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;
use std::collections::HashSet;

impl Solution {
    fn k_smallest_pairs(nums1: Vec<i32>, nums2: Vec<i32>, mut k: i32) -> Vec<Vec<i32>> {
        let n1 = nums1.len();
        let n2 = nums2.len();
        let mut visited: HashSet<(usize, usize)> = HashSet::new();
        let mut queue: BinaryHeap<(Reverse<i32>, usize, usize)> = BinaryHeap::new();
        if 0 < n1 && 0 < n2 && visited.insert((0, 0)) {
            queue.push((Reverse(nums1[0] + nums2[0]), 0, 0));
        } else {
            return vec![];
        }
        let mut res = vec![];
        while k > 0 {
            if let Some((_, i, j)) = queue.pop() {
                res.push(vec![nums1[i], nums2[j]]);
                if i + 1 < n1 && visited.insert((i + 1, j)) {
                    queue.push((Reverse(nums1[i + 1] + nums2[j]), i + 1, j));
                }
                if j + 1 < n2 && visited.insert((i, j + 1)) {
                    queue.push((Reverse(nums1[i] + nums2[j + 1]), i, j + 1));
                }
                k -= 1;
            } else {
                break;
            }
        }
        res
    }
}

#[test]
fn test() {
    let nums1 = vec![1, 7, 11];
    let nums2 = vec![2, 4, 6];
    let k = 3;
    let mut res = vec_vec_i32![[1, 2], [1, 4], [1, 6]];
    let mut ans = Solution::k_smallest_pairs(nums1, nums2, k);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let nums1 = vec![1, 1, 2];
    let nums2 = vec![1, 2, 3];
    let k = 2;
    let mut res = vec_vec_i32![[1, 1], [1, 1]];
    let mut ans = Solution::k_smallest_pairs(nums1, nums2, k);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let nums1 = vec![1, 2];
    let nums2 = vec![3];
    let k = 3;
    let mut res = vec_vec_i32![[1, 3], [2, 3]];
    let mut ans = Solution::k_smallest_pairs(nums1, nums2, k);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #373: Find K Pairs with Smallest Sums
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #373: Find K Pairs with Smallest Sums
// class Solution {
//     public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
//         PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
//         for (int i = 0; i < Math.min(nums1.length, k); ++i) {
//             q.offer(new int[] {nums1[i] + nums2[0], i, 0});
//         }
//         List<List<Integer>> ans = new ArrayList<>();
//         while (!q.isEmpty() && k > 0) {
//             int[] e = q.poll();
//             ans.add(Arrays.asList(nums1[e[1]], nums2[e[2]]));
//             --k;
//             if (e[2] + 1 < nums2.length) {
//                 q.offer(new int[] {nums1[e[1]] + nums2[e[2] + 1], e[1], e[2] + 1});
//             }
//         }
//         return ans;
//     }
// }

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(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.