LeetCode #2040 — HARD

Kth Smallest Product of Two Sorted Arrays

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

The Problem

Problem Statement

Given two sorted 0-indexed integer arrays nums1 and nums2 as well as an integer k, return the k^th (1-based) smallest product of nums1[i] * nums2[j] where 0 <= i < nums1.length and 0 <= j < nums2.length.

Example 1:

Input: nums1 = [2,5], nums2 = [3,4], k = 2
Output: 8
Explanation: The 2 smallest products are:
- nums1[0] * nums2[0] = 2 * 3 = 6
- nums1[0] * nums2[1] = 2 * 4 = 8
The 2^nd smallest product is 8.

Example 2:

Input: nums1 = [-4,-2,0,3], nums2 = [2,4], k = 6
Output: 0
Explanation: The 6 smallest products are:
- nums1[0] * nums2[1] = (-4) * 4 = -16
- nums1[0] * nums2[0] = (-4) * 2 = -8
- nums1[1] * nums2[1] = (-2) * 4 = -8
- nums1[1] * nums2[0] = (-2) * 2 = -4
- nums1[2] * nums2[0] = 0 * 2 = 0
- nums1[2] * nums2[1] = 0 * 4 = 0
The 6^th smallest product is 0.

Example 3:

Input: nums1 = [-2,-1,0,1,2], nums2 = [-3,-1,2,4,5], k = 3
Output: -6
Explanation: The 3 smallest products are:
- nums1[0] * nums2[4] = (-2) * 5 = -10
- nums1[0] * nums2[3] = (-2) * 4 = -8
- nums1[4] * nums2[0] = 2 * (-3) = -6
The 3^rd smallest product is -6.

Constraints:

1 <= nums1.length, nums2.length <= 5 * 10⁴
-10⁵ <= nums1[i], nums2[j] <= 10⁵
1 <= k <= nums1.length * nums2.length
nums1 and nums2 are sorted.

Step 01

Brute Force Baseline

Problem summary: Given two sorted 0-indexed integer arrays nums1 and nums2 as well as an integer k, return the kth (1-based) smallest product of nums1[i] * nums2[j] where 0 <= i < nums1.length and 0 <= j < nums2.length.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[2,5]
[3,4]
2

Example 2

[-4,-2,0,3]
[2,4]
6

Example 3

[-2,-1,0,1,2]
[-3,-1,2,4,5]
3

Core Insight

What unlocks the optimal approach

Can we split this problem into four cases depending on the sign of the numbers?
Can we binary search the value?

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 #2040: Kth Smallest Product of Two Sorted Arrays
class Solution {
    private int[] nums1;
    private int[] nums2;

    public long kthSmallestProduct(int[] nums1, int[] nums2, long k) {
        this.nums1 = nums1;
        this.nums2 = nums2;
        int m = nums1.length;
        int n = nums2.length;
        int a = Math.max(Math.abs(nums1[0]), Math.abs(nums1[m - 1]));
        int b = Math.max(Math.abs(nums2[0]), Math.abs(nums2[n - 1]));
        long r = (long) a * b;
        long l = (long) -a * b;
        while (l < r) {
            long mid = (l + r) >> 1;
            if (count(mid) >= k) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }

    private long count(long p) {
        long cnt = 0;
        int n = nums2.length;
        for (int x : nums1) {
            if (x > 0) {
                int l = 0, r = n;
                while (l < r) {
                    int mid = (l + r) >> 1;
                    if ((long) x * nums2[mid] > p) {
                        r = mid;
                    } else {
                        l = mid + 1;
                    }
                }
                cnt += l;
            } else if (x < 0) {
                int l = 0, r = n;
                while (l < r) {
                    int mid = (l + r) >> 1;
                    if ((long) x * nums2[mid] <= p) {
                        r = mid;
                    } else {
                        l = mid + 1;
                    }
                }
                cnt += n - l;
            } else if (p >= 0) {
                cnt += n;
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #2040: Kth Smallest Product of Two Sorted Arrays
func kthSmallestProduct(nums1 []int, nums2 []int, k int64) int64 {
	m := len(nums1)
	n := len(nums2)
	a := max(abs(nums1[0]), abs(nums1[m-1]))
	b := max(abs(nums2[0]), abs(nums2[n-1]))
	r := int64(a) * int64(b)
	l := -r

	count := func(p int64) int64 {
		var cnt int64
		for _, x := range nums1 {
			if x > 0 {
				l, r := 0, n
				for l < r {
					mid := (l + r) >> 1
					if int64(x)*int64(nums2[mid]) > p {
						r = mid
					} else {
						l = mid + 1
					}
				}
				cnt += int64(l)
			} else if x < 0 {
				l, r := 0, n
				for l < r {
					mid := (l + r) >> 1
					if int64(x)*int64(nums2[mid]) <= p {
						r = mid
					} else {
						l = mid + 1
					}
				}
				cnt += int64(n - l)
			} else if p >= 0 {
				cnt += int64(n)
			}
		}
		return cnt
	}

	for l < r {
		mid := (l + r) >> 1
		if count(mid) >= k {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2040: Kth Smallest Product of Two Sorted Arrays
class Solution:
    def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
        def count(p: int) -> int:
            cnt = 0
            n = len(nums2)
            for x in nums1:
                if x > 0:
                    cnt += bisect_right(nums2, p / x)
                elif x < 0:
                    cnt += n - bisect_left(nums2, p / x)
                else:
                    cnt += n * int(p >= 0)
            return cnt

        mx = max(abs(nums1[0]), abs(nums1[-1])) * max(abs(nums2[0]), abs(nums2[-1]))
        return bisect_left(range(-mx, mx + 1), k, key=count) - mx

// Accepted solution for LeetCode #2040: Kth Smallest Product of Two Sorted Arrays
impl Solution {
    pub fn kth_smallest_product(nums1: Vec<i32>, nums2: Vec<i32>, k: i64) -> i64 {
        let m = nums1.len();
        let n = nums2.len();
        let a = nums1[0].abs().max(nums1[m - 1].abs()) as i64;
        let b = nums2[0].abs().max(nums2[n - 1].abs()) as i64;
        let mut l = -a * b;
        let mut r = a * b;

        let count = |p: i64| -> i64 {
            let mut cnt = 0i64;
            for &x in &nums1 {
                if x > 0 {
                    let mut left = 0;
                    let mut right = n;
                    while left < right {
                        let mid = (left + right) / 2;
                        if (x as i64) * (nums2[mid] as i64) > p {
                            right = mid;
                        } else {
                            left = mid + 1;
                        }
                    }
                    cnt += left as i64;
                } else if x < 0 {
                    let mut left = 0;
                    let mut right = n;
                    while left < right {
                        let mid = (left + right) / 2;
                        if (x as i64) * (nums2[mid] as i64) <= p {
                            right = mid;
                        } else {
                            left = mid + 1;
                        }
                    }
                    cnt += (n - left) as i64;
                } else if p >= 0 {
                    cnt += n as i64;
                }
            }
            cnt
        };

        while l < r {
            let mid = l + (r - l) / 2;
            if count(mid) >= k {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}

// Accepted solution for LeetCode #2040: Kth Smallest Product of Two Sorted Arrays
function kthSmallestProduct(nums1: number[], nums2: number[], k: number): number {
    const m = nums1.length;
    const n = nums2.length;

    const a = BigInt(Math.max(Math.abs(nums1[0]), Math.abs(nums1[m - 1])));
    const b = BigInt(Math.max(Math.abs(nums2[0]), Math.abs(nums2[n - 1])));

    let l = -a * b;
    let r = a * b;

    const count = (p: bigint): bigint => {
        let cnt = 0n;
        for (const x of nums1) {
            const bx = BigInt(x);
            if (bx > 0n) {
                let l = 0,
                    r = n;
                while (l < r) {
                    const mid = (l + r) >> 1;
                    const prod = bx * BigInt(nums2[mid]);
                    if (prod > p) {
                        r = mid;
                    } else {
                        l = mid + 1;
                    }
                }
                cnt += BigInt(l);
            } else if (bx < 0n) {
                let l = 0,
                    r = n;
                while (l < r) {
                    const mid = (l + r) >> 1;
                    const prod = bx * BigInt(nums2[mid]);
                    if (prod <= p) {
                        r = mid;
                    } else {
                        l = mid + 1;
                    }
                }
                cnt += BigInt(n - l);
            } else if (p >= 0n) {
                cnt += BigInt(n);
            }
        }
        return cnt;
    };

    while (l < r) {
        const mid = (l + r) >> 1n;
        if (count(mid) >= BigInt(k)) {
            r = mid;
        } else {
            l = mid + 1n;
        }
    }

    return Number(l);
}

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.