Count Pairs Whose Sum is Less than Target

Count Pairs Whose Sum is Less than Target - Problem

Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target.

You need to count all valid pairs where the sum of two elements is strictly less than the target value.

Input & Output

Example 1 — Basic Case

$ Input: nums = [-4,1,1,3], target = 2

› Output: 6

💡 Note: Valid pairs: (-4,1), (-4,1), (-4,3), (1,1), (1,3), (1,3). All have sums < 2.

Example 2 — No Valid Pairs

$ Input: nums = [-1,1,2,3,1], target = 2

› Output: 3

💡 Note: Valid pairs: (-1,1), (-1,1), (-1,2). Other sums are ≥ 2.

Example 3 — All Pairs Valid

$ Input: nums = [-5,-3,-2,-1], target = 0

› Output: 6

💡 Note: All negative numbers, so all 6 pairs have negative sums < 0.

Constraints

2 ≤ nums.length ≤ 50
-50 ≤ nums[i] ≤ 50
-50 ≤ target ≤ 50

Visualization

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Understanding the Visualization

Input

Array nums and target value

Process

Find all pairs (i,j) where i<j and nums[i]+nums[j]<target

Output

Count of valid pairs

Key Takeaway

🎯 Key Insight: Sorting allows us to use two pointers and count multiple valid pairs in one step

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that after sorting, when we find nums[left] + nums[right] < target, we know that nums[left] paired with any element from left+1 to right will also be valid. Best approach is two pointers after sorting. Time: O(n log n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Two Pointers (After Sorting)	O(n log n)	O(1)	Sort array then use two pointers to efficiently count pairs
Brute Force	O(n²)	O(1)	Check every possible pair of elements

Two Pointers (After Sorting) — Algorithm Steps

Sort the array in ascending order
Set left=0, right=n-1 pointers
If nums[left]+nums[right] < target, add (right-left) pairs
Move appropriate pointer based on sum comparison

Visualization

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Step-by-Step Walkthrough

Sort

Sort array in ascending order

Two Pointers

Use left and right pointers

Count Pairs

If sum < target, all pairs from left to right are valid

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 solution(int* nums, int numsSize, int target) {
    qsort(nums, numsSize, sizeof(int), compare);
    int count = 0;
    int left = 0, right = numsSize - 1;
    
    while (left < right) {
        if (nums[left] + nums[right] < target) {
            count += right - left;
            left++;
        } else {
            right--;
        }
    }
    
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Sorting takes O(n log n), two-pointer scan takes O(n)

⚡ Linearithmic

Space Complexity

O(1)

Only using two pointer variables (assuming in-place sort)

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two Pointers (After Sorting) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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