Number of Pairs Satisfying Inequality - Problem

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set Hard

You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an integer diff. Find the number of pairs (i, j) such that:

0 <= i < j <= n - 1 and
nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

Return the number of pairs that satisfy the conditions.

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [3,2,5], nums2 = [2,1,1], diff = 1

› Output: 3

💡 Note: Check all pairs: (0,1): 3-2 ≤ 2-1+1 → 1 ≤ 2 ✓, (0,2): 3-5 ≤ 2-1+1 → -2 ≤ 2 ✓, (1,2): 2-5 ≤ 1-1+1 → -3 ≤ 1 ✓. All 3 pairs satisfy the condition.

Example 2 — No Valid Pairs

$ Input: nums1 = [3,2,5], nums2 = [2,1,1], diff = -1

› Output: 2

💡 Note: With diff = -1, check pairs: (0,1): 1 ≤ 0 ✗, (0,2): -2 ≤ 0 ✓, (1,2): -3 ≤ -1 ✓. Two pairs satisfy the condition.

Example 3 — Two Elements

$ Input: nums1 = [1,2], nums2 = [3,1], diff = 5

› Output: 1

💡 Note: Only one pair (0,1): 1-2 ≤ 3-1+5 → -1 ≤ 7 ✓. The pair satisfies the condition.

Constraints

n == nums1.length == nums2.length
1 ≤ n ≤ 10⁵
-10⁴ ≤ nums1[i], nums2[i] ≤ 10⁴
-10⁴ ≤ diff ≤ 10⁴

Visualization

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Asked in

G Google 15 M Microsoft 12 f Facebook 8 a Amazon 6

The key insight is to transform the inequality by rearranging terms to create a difference array, then use merge sort to efficiently count valid pairs. The optimal approach uses merge sort with counting to achieve O(n log n) time complexity instead of the brute force O(n²) approach.

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each pair of indices (i,j) where i < j, check if the inequality nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff holds. Count all valid pairs.

Merge Sort with Counting

⏱️ Time: O(n log n) Space: O(n)

Transform the inequality by rearranging terms to (nums1[i] - nums2[i]) - (nums1[j] - nums2[j]) <= diff. Create a new array of differences and count inversions during merge sort.

Brute Force — Algorithm Steps

Iterate through all pairs (i,j) with i < j
Check inequality condition for each pair
Count pairs that satisfy the condition

Visualization

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Step-by-Step Walkthrough

Setup

Initialize arrays and difference value

Check Pairs

For each pair (i,j), evaluate inequality

Count

Count valid pairs that satisfy condition

Code -

solution.c — C

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

int solution(int* nums1, int* nums2, int n, int diff) {
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    int nums1[1000], nums2[1000];
    int n1 = 0, n2 = 0;
    
    fgets(line, sizeof(line), stdin);
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            nums1[n1++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    fgets(line, sizeof(line), stdin);
    token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            nums2[n2++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int diff;
    scanf("%d", &diff);
    
    printf("%d\n", solution(nums1, nums2, n1, diff));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n(n-1)/2 pairs

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for counters

✓ Linear Space

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Number of Pairs Satisfying Inequality - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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