Count Number of Pairs With Absolute Difference K - Problem

Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.

The value of |x| is defined as:

x if x >= 0
-x if x < 0

Input & Output

Example 1 — Basic Case

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

› Output: 4

💡 Note: Pairs with absolute difference 1: (0,1), (0,2), (1,3), (2,3). All have |nums[i] - nums[j]| = 1.

Example 2 — Larger Difference

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

› Output: 0

💡 Note: No pairs exist where |nums[i] - nums[j]| = 3. Only pair is (0,1) with |1-3| = 2.

Example 3 — Zero Difference

$ Input: nums = [3,2,1,5,1,4], k = 2

› Output: 4

💡 Note: Pairs with absolute difference 2: (0,2) |3-1|=2, (0,3) |3-5|=2, (0,4) |3-1|=2, (1,5) |2-4|=2

Constraints

1 ≤ nums.length ≤ 200
1 ≤ nums[i] ≤ 100
0 ≤ k ≤ 99

Visualization

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Asked in

a Amazon 15 M Microsoft 12

The key insight is that for each number, we only need to find how many times (num+k) and (num-k) appear. The frequency count approach with HashMap is optimal: count occurrences of each number, then for each unique number, check if its partner (num+k) exists and multiply their frequencies. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

Iterate through all possible pairs of indices and check if the absolute difference equals k. For each element at index i, check all elements at indices j > i.

Frequency Count with HashMap

⏱️ Time: O(n) Space: O(n)

Count frequency of each number, then for each unique number, check how many times (num+k) and (num-k) appear in the map. This avoids checking all pairs explicitly.

Brute Force — Algorithm Steps

Iterate through each element at index i
For each i, check all elements at indices j > i
Count pairs where |nums[i] - nums[j]| == k

Visualization

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Step-by-Step Walkthrough

Compare Pairs

Check all pairs (i,j) where i < j

Check Difference

Calculate |nums[i] - nums[j]| for each pair

Count Matches

Increment counter when difference equals k

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (abs(nums[i] - nums[j]) == k) {
                count++;
            }
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all pairs - outer loop runs n times, inner loop runs up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a counter variable, no extra data structures

✓ Linear Space

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Count Number of Pairs With Absolute Difference K - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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