LeetCode #719 — HARD

Find K-th Smallest Pair Distance

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

The Problem

Problem Statement

The distance of a pair of integers a and b is defined as the absolute difference between a and b.

Given an integer array nums and an integer k, return the k^th smallest distance among all the pairs nums[i] and nums[j] where 0 <= i < j < nums.length.

Example 1:

Input: nums = [1,3,1], k = 1
Output: 0
Explanation: Here are all the pairs:
(1,3) -> 2
(1,1) -> 0
(3,1) -> 2
Then the 1^st smallest distance pair is (1,1), and its distance is 0.

Example 2:

Input: nums = [1,1,1], k = 2
Output: 0

Example 3:

Input: nums = [1,6,1], k = 3
Output: 5

Constraints:

n == nums.length
2 <= n <= 10⁴
0 <= nums[i] <= 10⁶
1 <= k <= n * (n - 1) / 2

Step 01

Brute Force Baseline

Problem summary: The distance of a pair of integers a and b is defined as the absolute difference between a and b. Given an integer array nums and an integer k, return the kth smallest distance among all the pairs nums[i] and nums[j] where 0 <= i < j < nums.length.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[1,3,1]
1

Example 2

[1,1,1]
2

Example 3

[1,6,1]
3

Core Insight

What unlocks the optimal approach

Binary search for the answer. How can you check how many pairs have distance <= X?

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 #719: Find K-th Smallest Pair Distance
class Solution {
    public int smallestDistancePair(int[] nums, int k) {
        Arrays.sort(nums);
        int left = 0, right = nums[nums.length - 1] - nums[0];
        while (left < right) {
            int mid = (left + right) >> 1;
            if (count(mid, nums) >= k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private int count(int dist, int[] nums) {
        int cnt = 0;
        for (int i = 0; i < nums.length; ++i) {
            int left = 0, right = i;
            while (left < right) {
                int mid = (left + right) >> 1;
                int target = nums[i] - dist;
                if (nums[mid] >= target) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            cnt += i - left;
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #719: Find K-th Smallest Pair Distance
func smallestDistancePair(nums []int, k int) int {
	sort.Ints(nums)
	n := len(nums)
	left, right := 0, nums[n-1]-nums[0]
	count := func(dist int) int {
		cnt := 0
		for i, v := range nums {
			target := v - dist
			left, right := 0, i
			for left < right {
				mid := (left + right) >> 1
				if nums[mid] >= target {
					right = mid
				} else {
					left = mid + 1
				}
			}
			cnt += i - left
		}
		return cnt
	}
	for left < right {
		mid := (left + right) >> 1
		if count(mid) >= k {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

# Accepted solution for LeetCode #719: Find K-th Smallest Pair Distance
class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        def count(dist):
            cnt = 0
            for i, b in enumerate(nums):
                a = b - dist
                j = bisect_left(nums, a, 0, i)
                cnt += i - j
            return cnt

        nums.sort()
        return bisect_left(range(nums[-1] - nums[0]), k, key=count)

// Accepted solution for LeetCode #719: Find K-th Smallest Pair Distance
struct Solution;

impl Solution {
    fn smallest_distance_pair(mut nums: Vec<i32>, k: i32) -> i32 {
        let n = nums.len();
        nums.sort_unstable();
        let mut lo = 0;
        let mut hi = nums[n - 1] - nums[0];
        while lo < hi {
            let mi = lo + (hi - lo) / 2;
            let mut count = 0;
            let mut j = 1;
            for i in 0..n - 1 {
                while j < n && nums[j] - nums[i] <= mi {
                    j += 1;
                }
                count += j - 1 - i;
            }
            if count >= k as usize {
                hi = mi;
            } else {
                lo = mi + 1;
            }
        }
        lo
    }
}

#[test]
fn test() {
    let nums = vec![1, 3, 1];
    let k = 1;
    let res = 0;
    assert_eq!(Solution::smallest_distance_pair(nums, k), res);
    let nums = vec![62, 100, 4];
    let k = 2;
    let res = 58;
    assert_eq!(Solution::smallest_distance_pair(nums, k), res);
}

// Accepted solution for LeetCode #719: Find K-th Smallest Pair Distance
function smallestDistancePair(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let left = 0,
        right = nums[n - 1] - nums[0];
    while (left < right) {
        let mid = (left + right) >> 1;
        let count = 0,
            i = 0;
        for (let j = 0; j < n; j++) {
            // 索引[i, j]距离nums[j]的距离<=mid
            while (nums[j] - nums[i] > mid) {
                i++;
            }
            count += j - i;
        }
        if (count >= k) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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 of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

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.