Find the Maximum Divisibility Score

Find the Maximum Divisibility Score - Problem

Array Easy

You are given two integer arrays nums and divisors.

The divisibility score of divisors[i] is the number of indices j such that nums[j] is divisible by divisors[i].

Return the integer divisors[i] with the maximum divisibility score. If multiple integers have the maximum score, return the smallest one.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,7,9,3,9], divisors = [5,2,3]

› Output: 3

💡 Note: Divisor 5: no numbers divisible (score 0). Divisor 2: only 4 is divisible (score 1). Divisor 3: numbers 9, 3, 9 are divisible (score 3). Maximum score is 3, so return 3.

Example 2 — Tie Case

$ Input: nums = [20,14,21,10], divisors = [5,7,5]

› Output: 5

💡 Note: Divisor 5: numbers 20, 10 are divisible (score 2). Divisor 7: numbers 14, 21 are divisible (score 2). Both have score 2, but 5 < 7, so return 5.

Example 3 — Single Element

$ Input: nums = [12], divisors = [10,16]

› Output: 10

💡 Note: Divisor 10: 12 is not divisible by 10 (score 0). Divisor 16: 12 is not divisible by 16 (score 0). Both have score 0, but 10 < 16, so return 10.

Constraints

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

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 18

The key insight is to count divisibility for each divisor and track the maximum score. For ties, return the smallest divisor. The optimal approach uses O(n×m) time to check each divisor against all numbers, with O(1) space complexity.

Common Approaches

✓ Brute Force

⏱️ Time: O(n × m) Space: O(1)

Iterate through each divisor and for each one, count how many numbers in the nums array are divisible by it. Keep track of the maximum score and the corresponding divisor.

Single Pass Optimization

⏱️ Time: O(n × m) Space: O(1)

Instead of storing scores for all divisors, we track the maximum score and best divisor as we iterate. This saves space and is slightly more efficient.

Brute Force — Algorithm Steps

For each divisor in divisors array
Count numbers in nums divisible by this divisor
Track maximum score and corresponding divisor
Return divisor with highest score (smallest if tie)

Visualization

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Step-by-Step Walkthrough

Check First Divisor

Count how many numbers are divisible by divisor 3

Check Second Divisor

Count how many numbers are divisible by divisor 2

Compare Scores

Return divisor with highest score (smallest if tied)

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int* divisors, int divisorsSize) {
    int maxScore = -1;
    int bestDivisor = INT_MAX;
    
    for (int i = 0; i < divisorsSize; i++) {
        int divisor = divisors[i];
        int score = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] % divisor == 0) {
                score++;
            }
        }
        
        if (score > maxScore || (score == maxScore && divisor < bestDivisor)) {
            maxScore = score;
            bestDivisor = divisor;
        }
    }
    
    return bestDivisor;
}

int main() {
    char line1[10000], line2[10000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    // Parse arrays manually
    int nums[1000], divisors[1000];
    int numsSize = 0, divisorsSize = 0;
    
    // Parse nums
    char* token = strtok(line1 + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse divisors
    token = strtok(line2 + 1, ",]");
    while (token != NULL) {
        divisors[divisorsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(nums, numsSize, divisors, divisorsSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

For each of m divisors, we check all n numbers in nums

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track maximum score and best divisor

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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