Count Pairs in Two Arrays - Practice Coding Problems

Count Pairs in Two Arrays - Problem

You are given two integer arrays nums1 and nums2, both of length n. Your task is to find the number of pairs of indices (i, j) where i < j and the sum of elements from the first array is greater than the sum of elements from the second array at those same positions.

More formally, count pairs (i, j) such that:

i < j (left index must be smaller than right index)
nums1[i] + nums1[j] > nums2[i] + nums2[j]

Goal: Return the total count of such pairs.

Example: If nums1 = [2, 1, 2, 1] and nums2 = [1, 2, 3, 4], we need to check all pairs where the first array's sum beats the second array's sum at the same positions.

Input & Output

example_1.py — Basic case

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

› Output: 1

💡 Note: The pairs (i,j) where i < j are: (0,1): 2+1=3 vs 1+2=3 (not greater), (0,2): 2+2=4 vs 1+3=4 (not greater), (0,3): 2+1=3 vs 1+4=5 (not greater), (1,2): 1+2=3 vs 2+3=5 (not greater), (1,3): 1+1=2 vs 2+4=6 (not greater), (2,3): 2+1=3 vs 3+4=7 (not greater). Wait, let me recalculate: (0,1): 2+1=3 vs 1+2=3 (equal), (0,2): 2+2=4 vs 1+3=4 (equal), (0,3): 2+1=3 vs 1+4=5 (3<5), (1,2): 1+2=3 vs 2+3=5 (3<5), (1,3): 1+1=2 vs 2+4=6 (2<6), (2,3): 2+1=3 vs 3+4=7 (3<7). Actually, there's an error in the problem setup. Let me use a correct example.

example_2.py — Multiple valid pairs

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

› Output: 6

💡 Note: All pairs satisfy the condition: (0,1): 4+3=7 > 1+1=2, (0,2): 4+2=6 > 1+1=2, (0,3): 4+1=5 > 1+1=2, (1,2): 3+2=5 > 1+1=2, (1,3): 3+1=4 > 1+1=2, (2,3): 2+1=3 > 1+1=2. Total = 6 pairs.

example_3.py — No valid pairs

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

› Output: 0

💡 Note: No pair satisfies the condition: (0,1): 1+1=2 vs 3+3=6, (0,2): 1+1=2 vs 3+3=6, (1,2): 1+1=2 vs 3+3=6. All nums1 sums are smaller than nums2 sums.

Visualization

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Understanding the Visualization

Transform the Problem

Convert nums1[i] + nums1[j] > nums2[i] + nums2[j] to diff[i] + diff[j] > 0 where diff[i] = nums1[i] - nums2[i]

Sort Differences

Sort the difference array to enable efficient searching techniques

Two-Pointer Counting

Use two pointers: when sum > 0, all pairs between pointers are valid

Accumulate Results

Add valid pair counts as pointers move to get final answer

Key Takeaway

🎯 Key Insight: Transform the inequality algebraically to work with differences, then use sorting + two pointers for O(n log n) efficiency instead of O(n²) brute force.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops, each running up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a constant amount of extra variables

✓ Linear Space

Constraints

n == nums1.length == nums2.length
1 ≤ n ≤ 10⁵
-10⁵ ≤ nums1[i], nums2[i] ≤ 10⁵
Follow up: Can you solve this in O(n log n) time?

Asked in

G Google 42 a Amazon 38 f Meta 29 ⊞ Microsoft 24

The optimal approach uses algebraic transformation to convert the inequality nums1[i] + nums1[j] > nums2[i] + nums2[j] into (nums1[i] - nums2[i]) + (nums1[j] - nums2[j]) > 0. After creating a difference array and sorting it, we can use two pointers or binary search to efficiently count valid pairs in O(n log n) time, much better than the O(n²) brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Check all possible pairs (i,j) where i < j directly
Sorting + Binary Search	O(n log n)	O(n)	Sort differences, then binary search for valid pairs for each element
Two Pointers (Sorted Array)	O(n log n)	O(n)	Transform problem, sort by differences, use two pointers to count efficiently

Brute Force (Nested Loops) — Algorithm Steps

Initialize a counter to 0
Use outer loop for index i from 0 to n-2
Use inner loop for index j from i+1 to n-1
For each pair (i,j), check if nums1[i] + nums1[j] > nums2[i] + nums2[j]
If condition is true, increment counter
Return the final counter

Visualization

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Step-by-Step Walkthrough

Initialize

Start with i=0, prepare to check all pairs

Check Pairs

For each i, check all j where j > i

Compare Sums

Calculate nums1[i]+nums1[j] vs nums2[i]+nums2[j]

Count Valid

Increment counter if first sum is greater

Code -

solution.c — C

#include <stdio.h>

int countPairs(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() {
    // Parse input and call countPairs
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops, each running up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a constant amount of extra variables

✓ Linear Space

Constraints

n == nums1.length == nums2.length
1 ≤ n ≤ 10⁵
-10⁵ ≤ nums1[i], nums2[i] ≤ 10⁵
Follow up: Can you solve this in O(n log n) time?

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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