Find Greatest Common Divisor of Array

Find Greatest Common Divisor of Array - Problem

Given an integer array nums, return the greatest common divisor of the smallest number and largest number in nums.

The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.

Input & Output

Example 1 — Basic Array

$ Input: nums = [2,5,6,9,10]

› Output: 2

💡 Note: The smallest number is 2, the largest is 10. GCD(2, 10) = 2 since 2 divides both 2 and 10 evenly.

Example 2 — Small Array

$ Input: nums = [7,5,6,8,3]

› Output: 1

💡 Note: The smallest number is 3, the largest is 8. GCD(3, 8) = 1 since they share no common divisors other than 1.

Example 3 — Two Elements

$ Input: nums = [3,3]

› Output: 3

💡 Note: Both minimum and maximum are 3. GCD(3, 3) = 3.

Constraints

2 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 12 M Microsoft 8 a Amazon 6

The key insight is that we only need the GCD of the minimum and maximum elements. Find min/max in O(n), then use Euclidean algorithm for efficient GCD calculation. Time: O(n + log(min(a,b))), Space: O(1)

Common Approaches

✓ Brute Force GCD

⏱️ Time: O(n + min(a,b)) Space: O(1)

First find the minimum and maximum values in the array. Then check every number from the smaller value down to 1 to find the largest one that divides both numbers evenly.

Euclidean Algorithm

⏱️ Time: O(n + log(min(a,b))) Space: O(1)

First find the minimum and maximum values in the array. Then apply the Euclidean algorithm which repeatedly replaces the larger number with the remainder of dividing it by the smaller number until one becomes zero.

Brute Force GCD — Algorithm Steps

Find minimum and maximum values in array
Check divisors from min(a,b) down to 1
Return first divisor that divides both numbers

Visualization

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Step-by-Step Walkthrough

Find Min/Max

Scan array to find smallest and largest values

Check Divisors

Test each number from smaller value down to 1

First Match

Return first number that divides both evenly

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int minVal = nums[0];
    int maxVal = nums[0];
    
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] < minVal) minVal = nums[i];
        if (nums[i] > maxVal) maxVal = nums[i];
    }
    
    // Check all possible divisors from smaller number down to 1
    for (int i = minVal; i >= 1; i--) {
        if (minVal % i == 0 && maxVal % i == 0) {
            return i;
        }
    }
    
    return 1;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n + min(a,b))

O(n) to find min/max, then O(min(a,b)) to check all divisors

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for min, max, and loop counter

✓ Linear Space

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Medium Frequency

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Input & Output

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Related Problems

Common Approaches

Brute Force GCD — Algorithm Steps

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Code -

Time & Space Complexity

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