Number of Subarrays Having Even Product - Problem

Given a 0-indexed integer array nums, return the number of subarrays of nums having an even product.

A subarray is a contiguous non-empty sequence of elements within an array. The product of a subarray is even if it contains at least one even number.

Input & Output

Example 1 — Mixed Even/Odd

$ Input: nums = [2,1,4,3,6]

› Output: 13

💡 Note: Total subarrays: 15. Odd-only subarrays: [1] and [3] = 2. Even product subarrays: 15 - 2 = 13

Example 2 — All Odd

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

› Output: 0

💡 Note: All numbers are odd, so no subarray can have an even product. Result is 0

Example 3 — All Even

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

› Output: 6

💡 Note: All subarrays contain at least one even number, so all 6 subarrays have even product

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that it's easier to count subarrays with ALL odd numbers and subtract from total. Best approach uses mathematical counting: total subarrays n×(n+1)/2 minus odd-only subarrays. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n³) Space: O(1)

For each starting position, extend the subarray one element at a time and check if the current product contains any even number. A product is even if at least one number in the subarray is even.

Count Odd-Only Subarrays

⏱️ Time: O(n) Space: O(1)

Instead of checking every subarray for even numbers, we count subarrays that contain ONLY odd numbers, then subtract from the total number of possible subarrays. This is much more efficient.

Brute Force - Check All Subarrays — Algorithm Steps

Iterate through each starting position i
For each i, extend subarray to position j
Check if subarray [i...j] contains any even number
Count subarrays with even product

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible contiguous subarrays

Check Each

Scan each subarray for even numbers

Count

Count subarrays containing at least one even

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            // Check if subarray [i...j] has even product
            int hasEven = 0;
            for (int k = i; k <= j; k++) {
                if (nums[k] % 2 == 0) {
                    hasEven = 1;
                    break;
                }
            }
            if (hasEven) {
                count++;
            }
        }
    }
    
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: O(n) starting positions × O(n) ending positions × O(n) to check each subarray

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

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Number of Subarrays Having Even Product - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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