Find K-th Smallest Pair Distance

Find K-th Smallest Pair Distance - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,3,1], k = 1

› Output: 0

💡 Note: All pairs and their distances: (1,3)→2, (1,1)→0, (3,1)→2. Sorted distances: [0,2,2]. The 1st smallest is 0.

Example 2 — Second Smallest

$ Input: nums = [1,1,1], k = 2

› Output: 0

💡 Note: All pairs have the same values, so all distances are 0. The 2nd smallest distance is still 0.

Example 3 — Larger Array

$ Input: nums = [1,6,1], k = 3

› Output: 5

💡 Note: Pairs: (1,6)→5, (1,1)→0, (6,1)→5. Sorted: [0,5,5]. The 3rd smallest is 5.

Constraints

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

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is to use binary search on the answer space rather than generating all pairs. Instead of creating O(n²) pairs, we binary search on possible distance values and use two pointers to efficiently count pairs with distance ≤ candidate. Best approach is binary search with two pointers: Time O(n log n + n log W), Space O(1).

Common Approaches

✓ Binary Search on Answer Space

⏱️ Time: O(n log n + n log W) Space: O(1)

Instead of generating all pairs, we binary search on the answer. For each candidate distance, we count how many pairs have distance ≤ candidate using two pointers on sorted array.

Brute Force - Generate All Pairs

⏱️ Time: O(n² log n) Space: O(n²)

Calculate the absolute difference for every possible pair (i,j) where i < j, store all distances in an array, sort the array, and return the element at index k-1.

Binary Search on Answer Space — Algorithm Steps

Step 1: Sort the array to enable two-pointer counting
Step 2: Binary search on distance range [0, max-min]
Step 3: For each mid, count pairs with distance ≤ mid using two pointers
Step 4: Adjust search range based on count vs k

Visualization

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Step-by-Step Walkthrough

Sort Array

Sort nums to enable two-pointer counting

Binary Search

Search on distance range [0, max-min]

Count Pairs

Use two pointers to count pairs ≤ mid distance

Adjust Range

Narrow search based on count vs k

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int countPairs(int* nums, int numsSize, int maxDist) {
    int count = 0;
    int left = 0;
    for (int right = 0; right < numsSize; right++) {
        while (nums[right] - nums[left] > maxDist) {
            left++;
        }
        count += right - left;
    }
    return count;
}

int solution(int* nums, int numsSize, int k) {
    qsort(nums, numsSize, sizeof(int), compare);
    
    int left = 0, right = nums[numsSize-1] - nums[0];
    
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (countPairs(nums, numsSize, mid) >= k) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    
    return left;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char *token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '[' && token[0] != ']') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int k;
    scanf("%d", &k);
    
    printf("%d\n", solution(nums, numsSize, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n + n log W)

O(n log n) for sorting plus O(n log W) for binary search with counting, where W is the range of values

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables, sorting can be in-place

⚠ Quadratic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search on Answer Space — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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