Count Pairs in Two Arrays

Count Pairs in Two Arrays - Problem

Given two integer arrays nums1 and nums2 of length n, count the pairs of indices (i, j) such that i < j and nums1[i] + nums1[j] > nums2[i] + nums2[j].

Return the number of pairs satisfying the condition.

Input & Output

Example 1 — Basic Case

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

› Output: 0

💡 Note: Check all pairs (i,j) where i < j: (0,1): 2+1=3, 3+3=6, 3>6? No. (0,2): 2+3=5, 3+2=5, 5>5? No. (0,3): 2+1=3, 3+2=5, 3>5? No. (1,2): 1+3=4, 3+2=5, 4>5? No. (1,3): 1+1=2, 3+2=5, 2>5? No. (2,3): 3+1=4, 2+2=4, 4>4? No. Total: 0 valid pairs.

Example 2 — Clear Valid Pair

$ Input: nums1 = [3,4,1], nums2 = [1,1,2]

› Output: 3

💡 Note: Check pairs: (0,1): 3+4=7, 1+1=2, 7>2? Yes ✓. (0,2): 3+1=4, 1+2=3, 4>3? Yes ✓. (1,2): 4+1=5, 1+2=3, 5>3? Yes ✓. Total: 3 valid pairs.

Example 3 — No Valid Pairs

$ Input: nums1 = [1,1], nums2 = [2,2]

› Output: 0

💡 Note: Only one pair (0,1): 1+1=2, 2+2=4, 2>4? No. No valid pairs.

Constraints

2 ≤ nums1.length ≤ 10⁵
nums1.length == nums2.length
-10⁵ ≤ nums1[i], nums2[i] ≤ 10⁵

Visualization

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Asked in

M Meta 15 a Amazon 12 G Google 8

The key insight is to transform the inequality nums1[i] + nums1[j] > nums2[i] + nums2[j] into (nums1[i] - nums2[i]) + (nums1[j] - nums2[j]) > 0. Best approach is transform and sort with two pointers. Time: O(n log n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each pair of indices (i,j) where i < j, check if nums1[i] + nums1[j] > nums2[i] + nums2[j]. Count all pairs that satisfy this condition.

Transform and Sort

⏱️ Time: O(n log n) Space: O(n)

Transform nums1[i] + nums1[j] > nums2[i] + nums2[j] to (nums1[i] - nums2[i]) + (nums1[j] - nums2[j]) > 0. Create difference array, sort it, then use two pointers to count pairs efficiently.

Brute Force — Algorithm Steps

Iterate through all pairs (i,j) with i < j
For each pair, check if nums1[i] + nums1[j] > nums2[i] + nums2[j]
Count pairs that satisfy the condition

Visualization

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Step-by-Step Walkthrough

Initialize

Set up nested loops to check all pairs

Compare

For each pair, check if sum condition holds

Count

Increment counter for valid pairs

Code -

solution.c — C

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

int solution(int* nums1, int* nums2, int n) {
    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]) {
                count++;
            }
        }
    }
    return count;
}

int main() {
    char line[1000000];
    fgets(line, sizeof(line), stdin);
    
    int nums1[100000], nums2[100000];
    int n = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n' && token[0] != ']') {
            nums1[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    fgets(line, sizeof(line), stdin);
    int m = 0;
    token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n' && token[0] != ']') {
            nums2[m++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(nums1, nums2, n));
    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 uses a counter variable

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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