Number of Subarrays With GCD Equal to K - Practice Coding Problems

Number of Subarrays With GCD Equal to K - Problem

Given an integer array nums and an integer k, return the number of subarrays of nums where the greatest common divisor of the subarray's elements is k.

A subarray is a contiguous non-empty sequence of elements within an array.

The greatest common divisor of an array is the largest integer that evenly divides all the array elements.

Input & Output

Example 1 — Basic Case

$ Input: nums = [9,3,1,2,6,3], k = 3

› Output: 4

💡 Note: The subarrays with GCD = 3 are: [9,3] (GCD=3), [3] at index 1 (GCD=3), [6,3] (GCD=3), [3] at index 5 (GCD=3). Total = 4 subarrays.

Example 2 — Single Element

$ Input: nums = [4], k = 7

› Output: 0

💡 Note: Only one subarray [4] exists, but GCD(4) = 4 ≠ 7, so no valid subarrays.

Example 3 — All Elements Same

$ Input: nums = [6,6,6], k = 6

› Output: 6

💡 Note: All subarrays have GCD = 6: [6], [6], [6], [6,6], [6,6], [6,6,6]. Total = 6 subarrays.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i], k ≤ 10⁹

Visualization

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Understanding the Visualization

Input

Array nums and target GCD value k

Process

Check all subarrays and calculate their GCD

Output

Count of subarrays with GCD = k

Key Takeaway

🎯 Key Insight: GCD can only decrease when extending subarrays, enabling early termination optimization

Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that GCD can only decrease or stay same when adding elements to a subarray. The optimized approach uses incremental GCD calculation with early termination when GCD becomes smaller than k. Time: O(n² log(max)), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Optimized with Early GCD Calculation	O(n² log(max))	O(1)	Calculate GCD incrementally to avoid recomputation
Brute Force - Check All Subarrays	O(n³)	O(1)	Generate all subarrays and compute GCD for each

Optimized with Early GCD Calculation — Algorithm Steps

For each starting position
Extend subarray incrementally
Update GCD with new element
Break early when GCD < k

Visualization

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Step-by-Step Walkthrough

Start Position

Begin with first element

Extend & Update

Add elements and update GCD incrementally

Early Break

Stop when GCD < k (cannot recover)

Code -

solution.c — C

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

int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    
    // For each starting position
    for (int i = 0; i < numsSize; i++) {
        int currentGcd = 0;
        // Extend subarray from position i
        for (int j = i; j < numsSize; j++) {
            // Update GCD incrementally
            currentGcd = gcd(currentGcd, nums[j]);
            
            // Early termination: if GCD < k, it will never increase
            if (currentGcd < k) {
                break;
            }
                
            if (currentGcd == k) {
                count++;
            }
        }
    }
    
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n² log(max))

O(n²) nested loops, O(log max) for each GCD calculation

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized with Early GCD Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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