LeetCode 2183: Count Array Pairs Divisible by K Solution

Master LeetCode problem 2183 (Count Array Pairs Divisible by K), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2183. Count Array Pairs Divisible by K

Hard

Array Math Number Theory

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible pairs of indices (i, j) in the given array nums. For each pair, we calculate the product of the elements at those indices and check if the product is divisible by k. If it is divisible, we increment a counter to keep track of the number of valid pairs.

The key idea is to use a hash map to store the frequency of remainders when dividing the elements of the array by k. We iterate through the array and calculate the remainder of each element when divided by k, then store the frequency of each remainder in the hash map. After that, we iterate through the hash map and calculate the total count of pairs that are divisible by k.

Time Complexity

The time complexity of this approach is O(n), where n is the length of the input array nums.

Space Complexity

The space complexity of this approach is O(k), where k is the value of the integer k.

Solution Code

import java.util.*;

class Solution {
    public int countPairs(int[] nums, int k) {
        Map<Integer, Integer> remainders = new HashMap<>();
        int count = 0;

        for (int num : nums) {
            int remainder = num % k;
            int complement = (k - remainder) % k;
            count += remainders.getOrDefault(complement, 0);
            remainders.put(remainder, remainders.getOrDefault(remainder, 0) + 1);
        }

        return count;
    }
}

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