leetcode-2176-easy

Count Equal and Divisible Pairs in an Array

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.

Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.
  Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.
Constraints:

1 <= nums.length <= 100
1 <= nums[i], k <= 100

思路一：先存储值相等的数组下标，然后对这些下别进行遍历判断。不过运行效率不太行，看了一下题解，发现用暴力算法反而是最快的...

public int countPairs(int[] nums, int k) {
    Map<Integer, List<Integer>> map = new HashMap<>();

    for (int i = 0; i < nums.length; i++) {
        int finalI = i;
        map.compute(nums[i], (key, v) -> {
            if (v == null) {
                v = new ArrayList<>();
            }
            v.add(finalI);
            return v;
        });
    }

    int count = 0;
    for (Map.Entry<Integer, List<Integer>> e : map.entrySet()) {
        List<Integer> values = e.getValue();
        if (values.size() <= 1) continue;

        for (int i = 0; i < values.size(); i++) {
            for (int j = i + 1; j < values.size(); j++) {
                if ((values.get(i) * values.get(j)) % k == 0) count++;
            }
        }
    }

    return count;
}