Simple Yet Effective Approach Explained Using Java
IntuitionThe problem requires finding the divisor that maximizes the number of elements it divides in nums. If multiple divisors have the same count, we return
# Intuition
The problem requires finding the divisor that maximizes the number of elements it divides in `nums`. If multiple divisors have the same count, we return the smallest one. A straightforward way to solve this is by iterating through each divisor and counting how many numbers in `nums` are divisible by it.

# Approach
1. Initialize `divisionScore` to store the maximum count of divisible numbers and `value` to track the best divisor.
2. Iterate through each divisor in `divisors`:
   - Count how many numbers in `nums` are divisible by the current divisor.
   - If this count is higher than `divisionScore`, update `divisionScore` and set `value` to the current divisor.
   - If the count is equal to `divisionScore`, update `value` to the smaller divisor.
3. Return the `value` with the highest division score.

# Complexity
- **Time complexity:**  
  - We iterate through each divisor (`m` times) and check divisibility for each number in `nums` (`n` times).  
  - This results in a total complexity of **O(m * n)**.

- **Space complexity:**  
  - We use only a few integer variables (`O(1)` constant space).  
# Code
```java []
class Solution {
    public int maxDivScore(int[] nums, int[] divisors) {
        int divisionScore = Integer.MIN_VALUE;
        int value = 0;

        for(int i = 0; i < divisors.length; ++i) {
            int currScore = 0;
            for(int j = 0; j < nums.length; ++j) {
                if(nums[j] % divisors[i] == 0) {
                    currScore++;
                }
            }

            if(currScore > divisionScore) {
                value = divisors[i];
                divisionScore = currScore;
            } else if(currScore == divisionScore) {
                if(divisors[i] < value) {
                    value = divisors[i];
                }
            }
        }
        return value;
    }
}
```

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['Java']