Count Special Quadruplets

Count Special Quadruplets - Problem

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

nums[a] + nums[b] + nums[c] == nums[d], and
a < b < c < d

You need to find all valid combinations where three numbers sum to equal a fourth number, with all indices in strictly increasing order.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,0,1,0,2]

› Output: 1

💡 Note: The only valid quadruplet is (0,1,2,4): nums[0] + nums[1] + nums[2] = 1 + 0 + 1 = 2 = nums[4], and 0 < 1 < 2 < 4

Example 2 — No Valid Quadruplets

$ Input: nums = [2,2,2,2]

› Output: 0

💡 Note: No valid quadruplet exists because 2 + 2 + 2 = 6 ≠ 2. All elements are the same but the sum of any three doesn't equal the fourth

Example 3 — Multiple Quadruplets

$ Input: nums = [1,1,1,3,5]

› Output: 2

💡 Note: Two valid quadruplets: (0,1,2,3) where 1+1+1=3, and (0,1,3,4) where 1+1+3=5

Constraints

4 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 100

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to convert the quadruplet search into a hash map lookup problem. Instead of four nested loops, we can use three loops with hash map optimization. The best approach uses a hash map to store pair sums and check for matches in O(1) time. Time: O(n³), Space: O(n²)

Common Approaches

✓ Brute Force - Four Nested Loops

⏱️ Time: O(n⁴) Space: O(1)

Use four nested loops to check every possible quadruplet (a,b,c,d) where a < b < c < d. For each combination, verify if nums[a] + nums[b] + nums[c] equals nums[d].

Hash Map Optimization

⏱️ Time: O(n³) Space: O(n²)

Instead of the innermost loop, use a hash map to store sums of pairs along with their indices. For each triplet (a,b,c), lookup if the required sum exists in positions after c.

Reverse Hash Map Approach

⏱️ Time: O(n³) Space: O(n²)

Instead of rebuilding the hash map for each c, build it incrementally. For each position c, add it to existing pairs to form new sums, then check against future d values.

Brute Force - Four Nested Loops — Algorithm Steps

Iterate through all possible values of a
For each a, iterate through all possible b > a
For each (a,b), iterate through all possible c > b
For each (a,b,c), iterate through all possible d > c
Check if nums[a] + nums[b] + nums[c] == nums[d]

Visualization

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Step-by-Step Walkthrough

Four Nested Loops

Check all combinations where a < b < c < d

Sum Check

For each combo, verify nums[a] + nums[b] + nums[c] == nums[d]

Count Matches

Increment counter for valid quadruplets

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int count = 0;
    
    for (int a = 0; a < numsSize - 3; a++) {
        for (int b = a + 1; b < numsSize - 2; b++) {
            for (int c = b + 1; c < numsSize - 1; c++) {
                for (int d = c + 1; d < numsSize; d++) {
                    if (nums[a] + nums[b] + nums[c] == nums[d]) {
                        count++;
                    }
                }
            }
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[200];
    int numsSize = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n⁴)

Four nested loops, each potentially iterating n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

⚠ Quadratic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Four Nested Loops — Algorithm Steps

Visualization

Code -

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