LeetCode 2179: Count Good Triplets in an Array Solution

Master LeetCode problem 2179 (Count Good Triplets in an Array), 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.

2179. Count Good Triplets in an Array

Hard

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

Problem Explanation

Explanation

To solve this problem, we iterate through both arrays nums1 and nums2.
For each value in nums1, we find its position in nums2 and store it in a hashmap.
We then iterate through all possible triplets in nums1 and check if they form a good triplet with increasing positions in nums2.
If a triplet is found to be good, we increment the count.
Finally, we return the total count of good triplets.

Time Complexity: O(n^2)
Space Complexity: O(n)

Solution Code

class Solution {
    public int countGoodTriplets(int[] nums1, int[] nums2) {
        int n = nums1.length;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            map.put(nums2[i], i);
        }
        
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                for (int k = j+1; k < n; k++) {
                    if (nums1[i] < nums1[j] && nums1[j] < nums1[k] && map.get(nums1[i]) < map.get(nums1[j]) && map.get(nums1[j]) < map.get(nums1[k])) {
                        count++;
                    }
                }
            }
        }
        
        return count;
    }
}

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