LeetCode 3162: Find the Number of Good Pairs I Solution

Master LeetCode problem 3162 (Find the Number of Good Pairs I), a easy 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.

3162. Find the Number of Good Pairs I

Easy

Array Hash Table

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible pairs of indices (i, j) and check if nums1[i] is divisible by nums2[j]*k. If it is divisible, we increment a counter to keep track of the number of good pairs. We can do this by using nested loops to iterate through both arrays.

Algorithm:

Initialize a counter variable goodPairs to 0.
Iterate through each pair of indices (i, j) using nested loops.
Check if nums1[i] is divisible by nums2[j]*k.
If it is divisible, increment goodPairs.
Return the total number of good pairs.

Time Complexity: O(n*m), where n and m are the lengths of nums1 and nums2 respectively. Space Complexity: O(1)

Solution Code

class Solution {
    public int countGoodPairs(int[] nums1, int[] nums2, int k) {
        int goodPairs = 0;
        for (int i = 0; i < nums1.length; i++) {
            for (int j = 0; j < nums2.length; j++) {
                if (nums1[i] % (nums2[j] * k) == 0) {
                    goodPairs++;
                }
            }
        }
        return goodPairs;
    }
}

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