Find the Maximum Divisibility Score - Practice Coding Problems

Find the Maximum Divisibility Score - Problem

Array Easy

You are given two integer arrays: nums and divisors. Your task is to find which divisor has the highest divisibility score.

The divisibility score of a divisor is simply the count of how many numbers in nums are divisible by that divisor. For example, if nums = [4, 7, 9, 3, 9] and we're checking divisor 3, we count how many numbers (4, 7, 9, 3, 9) are divisible by 3. In this case, both 9s and 3 are divisible by 3, giving us a score of 3.

Goal: Return the divisor with the maximum divisibility score. If there's a tie (multiple divisors have the same highest score), return the smallest divisor among them.

Example: nums = [4, 7, 9, 3, 9], divisors = [5, 2, 3]
• Divisor 5: score = 0 (no numbers divisible by 5)
• Divisor 2: score = 1 (only 4 is divisible by 2)
• Divisor 3: score = 3 (numbers 9, 3, 9 are divisible by 3)
Answer: 3 (highest score)

Input & Output

example_1.py — Basic Case

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

› Output: 3

💡 Note: Divisor 5: 0 numbers divisible (score=0). Divisor 2: 1 number divisible [4] (score=1). Divisor 3: 3 numbers divisible [9, 3, 9] (score=3). Maximum score is 3.

example_2.py — Tie Breaker

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

› Output: 5

💡 Note: Divisor 10: 2 numbers divisible [20, 10] (score=2). Divisor 7: 2 numbers divisible [14, 21] (score=2). Divisor 5: 2 numbers divisible [20, 10] (score=2). All have same score, return smallest divisor which is 5.

example_3.py — No Matches

$ Input: nums = [3], divisors = [2]

› Output: 2

💡 Note: Only one divisor 2, and 3 is not divisible by 2, so score is 0. Still return the only available divisor which is 2.

Constraints

1 ≤ nums.length, divisors.length ≤ 1000
1 ≤ nums[i], divisors[i] ≤ 10⁹
All divisors are positive integers
Return the smallest divisor in case of tie

Visualization

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3Allows maximum students (3) to pass

Understanding the Visualization

Set Up Thresholds

Teacher has multiple grade thresholds to test: 5, 2, and 3

Test First Threshold

Check threshold 5: No students meet this high standard (score=0)

Test Second Threshold

Check threshold 2: One student meets this standard (score=1)

Test Final Threshold

Check threshold 3: Three students meet this standard (score=3) - Best choice!

Key Takeaway

🎯 Key Insight: Just like a teacher finding the optimal grade threshold, we systematically test each divisor to find the one that 'passes' the most numbers, always preferring smaller divisors when there's a tie.

Asked in

a Amazon 12 G Google 8 ⊞ Microsoft 6 f Meta 4

The optimal approach uses a single pass through the divisors, tracking the best divisor and maximum score on-the-fly. This achieves O(n×m) time complexity with O(1) space complexity, avoiding the need for extra arrays to store all scores.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Optimal Solution	O(n × m)	O(1)	Track the best divisor and score while iterating, avoiding extra storage
Brute Force (Nested Loops)	O(n × m)	O(m)	Check each divisor against all numbers, store all scores, then find maximum

One-Pass Optimal Solution — Algorithm Steps

Initialize variables for the best divisor and maximum score
For each divisor, calculate its divisibility score
If current score is better than max, update both max score and best divisor
If scores are equal, keep the smaller divisor
Return the best divisor found

Visualization

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Step-by-Step Walkthrough

Initialize Tracking

Set up variables to track best divisor and max score

Process Each Divisor

Calculate score and immediately compare with current best

Update Best Result

Update tracking variables when better score found

Return Best

Return the tracked best divisor

Code -

solution.c — C

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

int maxDivScore(int* nums, int numsSize, int* divisors, int divisorsSize) {
    int maxScore = -1;
    int bestDivisor = divisors[0];
    
    for (int i = 0; i < divisorsSize; i++) {
        int score = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] % divisors[i] == 0) {
                score++;
            }
        }
        
        // Update best if current score is better, or same score but smaller divisor
        if (score > maxScore || (score == maxScore && divisors[i] < bestDivisor)) {
            maxScore = score;
            bestDivisor = divisors[i];
        }
    }
    
    return bestDivisor;
}

int main() {
    int nums[] = {4, 7, 9, 3, 9};
    int divisors[] = {5, 2, 3};
    printf("%d\n", maxDivScore(nums, 5, divisors, 3));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

n is length of nums, m is length of divisors. Still need to check each divisor against each number

✓ Linear Growth

Space Complexity

O(1)

Only use constant extra space for tracking best result

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Optimal Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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