3Sum Smaller - Practice Coding Problems

3Sum Smaller - Problem

Given an array of n integers nums and an integer target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

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

Input & Output

Example 1 — Basic Case

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

› Output: 2

💡 Note: Two triplets sum to less than 2: [-2,0,1] with sum -1, and [-2,0,3] with sum 1. Both are < 2.

Example 2 — No Valid Triplets

$ Input: nums = [0], target = 0

› Output: 0

💡 Note: Array has only one element, cannot form any triplets (need at least 3 elements).

Example 3 — All Triplets Valid

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

› Output: 1

💡 Note: Only one possible triplet: [-1,-1,-1] with sum -3, which is less than 0.

Constraints

1 ≤ nums.length ≤ 3000
-100 ≤ nums[i] ≤ 100
-100 ≤ target ≤ 100

Visualization

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Understanding the Visualization

Input

Array [-2,0,1,3] and target 2

Process

Find all triplets with sum < target

Output

Count of valid triplets: 2

Key Takeaway

🎯 Key Insight: Sort first, then use two pointers to efficiently count all valid triplet combinations

Asked in

G Google 25 a Amazon 20 f Facebook 15 M Microsoft 18

The key insight is to sort the array first, then use two pointers to efficiently count valid triplets. When we find a valid triplet, all combinations between the pointers also work. Best approach is Sort + Two Pointers with Time: O(n²), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n³)	O(1)	Check all possible triplets with three nested loops
Sort + Two Pointers	O(n²)	O(1)	Sort array first, then use two pointers to efficiently count valid triplets

Brute Force — Algorithm Steps

Iterate through all i from 0 to n-3
For each i, iterate j from i+1 to n-2
For each j, iterate k from j+1 to n-1
Count triplets where nums[i] + nums[j] + nums[k] < target

Visualization

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Step-by-Step Walkthrough

Triple Loop

Use i, j, k to check all combinations

Sum Check

Count when nums[i] + nums[j] + nums[k] < target

Result

Return total count of valid triplets

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int target) {
    int count = 0;
    
    for (int i = 0; i < numsSize - 2; i++) {
        for (int j = i + 1; j < numsSize - 1; j++) {
            for (int k = j + 1; k < numsSize; k++) {
                if (nums[i] + nums[j] + nums[k] < target) {
                    count++;
                }
            }
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int target;
    scanf("%d", &target);
    
    printf("%d\n", solution(nums, numsSize, target));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops iterate through all possible triplet combinations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting and iteration

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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