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: 3

💡 Note: Valid pairs: (0,1): -4+1=-3, (0,2): -4+1=-3, (0,3): -4+3=-1. All have sums < 2. Total: 3 pairs.

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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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

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

Use nested loops to examine all pairs (i,j) where i < j and count those where nums[i] + nums[j] < target.

Two Pointers (After Sorting)

⏱️ Time: O(n log n) Space: O(1)

First sort the array, then use left and right pointers. If sum < target, all pairs from left to right are valid. Move pointers accordingly.

Brute Force — Algorithm Steps

Use outer loop for first element (i)
Use inner loop for second element (j > i)
Check if sum is less than target
Increment counter if condition met

Visualization

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Step-by-Step Walkthrough

Outer Loop

Fix first element at index i

Inner Loop

Check all j > i elements

Count

Increment if sum < target

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);
    
    int nums[1000];
    int numsSize = 0;
    char *p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        nums[numsSize++] = (int)strtol(p, &p, 10);
    }
    
    int target;
    scanf("%d", &target);
    printf("%d\n", solution(nums, numsSize, target));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

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

⚡ Linearithmic

Space Complexity

O(1)

Only using 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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