Count Subarrays of Length Three With a Condition

Count Subarrays of Length Three With a Condition - Problem

Array Easy

Given an integer array nums, return the number of subarrays of length 3 such that the sum of the first and third numbers equals exactly half of the second number.

A subarray is a contiguous part of an array. For each valid subarray [nums[i], nums[i+1], nums[i+2]], check if nums[i] + nums[i+2] == nums[i+1] / 2.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,4,5]

› Output: 0

💡 Note: Check each triplet: [1,2,3] → 1+3=4, 2/2=1, 4≠1. [2,3,4] → 2+4=6, 3/2=1.5, 6≠1.5. [3,4,5] → 3+5=8, 4/2=2, 8≠2. No valid triplets.

Example 2 — Valid Triplet

$ Input: nums = [2,10,3]

› Output: 1

💡 Note: Triplet [2,10,3]: 2+3=5, 10/2=5. Since 5=5, this triplet satisfies the condition. Count = 1.

Example 3 — Multiple Triplets

$ Input: nums = [1,6,2,4,8,2]

› Output: 2

💡 Note: Check triplets: [1,6,2] → 1+2=3, 6/2=3 ✓. [6,2,4] → 6+4=10, 2/2=1 ✗. [2,4,8] → 2+8=10, 4/2=2 ✗. [4,8,2] → 4+2=6, 8/2=4 ✗. Two valid triplets.

Constraints

3 ≤ nums.length ≤ 100
-100 ≤ nums[i] ≤ 100

Visualization

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Asked in

G Google 25 a Amazon 15

The key insight is to rearrange the condition from first + third = middle/2 to 2 × (first + third) = middle to avoid floating point precision issues. Best approach uses a single pass through the array checking each consecutive triplet. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass Check	O(n)	O(1)	Iterate through array and check each triplet of consecutive elements
Mathematical Optimization	O(n)	O(1)	Use integer arithmetic to avoid floating point precision issues

Single Pass Check — Algorithm Steps

Step 1: Iterate from index 0 to n-3
Step 2: For each triplet, check if nums[i] + nums[i+2] == nums[i+1] / 2
Step 3: Count valid triplets

Visualization

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Step-by-Step Walkthrough

Check Triplet

Examine nums[i], nums[i+1], nums[i+2]

Verify Condition

Check if first + third = middle / 2

Count Valid

Increment counter if condition holds

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int count = 0;
    
    for (int i = 0; i < n - 2; i++) {
        int first = nums[i];
        int middle = nums[i + 1];
        int third = nums[i + 2];
        
        // Check if first + third equals half of middle
        // To avoid floating point issues, check: 2 * (first + third) == middle
        if (2 * (first + third) == middle) {
            count++;
        }
    }
    
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through array, checking each triplet once

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for counting and checking

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Single Pass Check — Algorithm Steps

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Code -

Time & Space Complexity

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