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: 1

💡 Note: Check triplets: [1,6,2] → 2*(1+2)=6, middle=6 ✓. [6,2,4] → 2*(6+4)=20, middle=2 ✗. [2,4,8] → 2*(2+8)=20, middle=4 ✗. [4,8,2] → 2*(4+2)=12, middle=8 ✗. One valid triplet.

Constraints

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

Visualization

Tap to expand

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

✓ Mathematical Optimization

⏱️ Time: O(n) Space: O(1)

Instead of checking if first + third = middle / 2, we rearrange to 2 * (first + third) = middle to avoid floating point operations and precision issues.

Single Pass Check

⏱️ Time: O(n) Space: O(1)

For each possible starting position, extract the triplet and check if first + third equals half of middle element. This is actually optimal since we must examine every triplet.

Mathematical Optimization — Algorithm Steps

Step 1: Iterate through valid starting positions
Step 2: Rearrange condition to 2 * (first + third) == middle
Step 3: Count triplets satisfying integer condition

Visualization

Tap to expand

Step-by-Step Walkthrough

Original

first + third = middle ÷ 2

Rearranged

2 × (first + third) = middle

Check

Compare using only integers

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++) {
        // Rearrange condition: first + third = middle/2
        // Equivalent to: 2 * (first + third) = middle
        if (2 * (nums[i] + nums[i + 2]) == nums[i + 1]) {
            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 pass through array examining each consecutive triplet

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

12.5K Views

Medium Frequency

~15 min Avg. Time

234 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Count Subarrays of Length Three With a Condition - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler