2644. Find the Maximum Divisibility Score

Explanation

To solve this problem, we can iterate over each divisor and count the number of elements in the nums array that are divisible by that divisor. We then track the divisor with the maximum count. If there are multiple divisors with the same maximum count, we return the smallest one.

Initialize a map to store the count of elements divisible by each divisor.
Iterate over the nums array and for each element, iterate over the divisors array to check if the element is divisible by the divisor. Increment the count in the map accordingly.
Finally, find the divisor with the maximum count. If there are multiple divisors with the same maximum count, return the smallest one.

Time Complexity: O(N * M), where N is the length of the nums array and M is the length of the divisors array.

Space Complexity: O(M), where M is the length of the divisors array.

View Leetcode Problem

class Solution {
    public int maxDivisor(int[] nums, int[] divisors) {
        Map<Integer, Integer> countMap = new HashMap<>();
        
        for (int num : nums) {
            for (int divisor : divisors) {
                if (num % divisor == 0) {
                    countMap.put(divisor, countMap.getOrDefault(divisor, 0) + 1);
                }
            }
        }
        
        int maxDivisor = Integer.MAX_VALUE;
        int maxCount = 0;
        
        for (int divisor : divisors) {
            if (countMap.containsKey(divisor) && countMap.get(divisor) > maxCount) {
                maxDivisor = divisor;
                maxCount = countMap.get(divisor);
            } else if (countMap.containsKey(divisor) && countMap.get(divisor) == maxCount) {
                maxDivisor = Math.min(maxDivisor, divisor);
            }
        }
        
        return maxDivisor;
    }
}

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