Formatted question description: https://leetcode.ca/all/1534.html

# 1534. Count Good Triplets (Easy)

Given an array of integers arr, and three integers ab and c. You need to find the number of good triplets.

A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true:

• 0 <= i < j < k < arr.length
• |arr[i] - arr[j]| <= a
• |arr[j] - arr[k]| <= b
• |arr[i] - arr[k]| <= c

Where |x| denotes the absolute value of x.

Return the number of good triplets.

Example 1:

Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3
Output: 4
Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].


Example 2:

Input: arr = [1,1,2,2,3], a = 0, b = 0, c = 1
Output: 0
Explanation: No triplet satisfies all conditions.


Constraints:

• 3 <= arr.length <= 100
• 0 <= arr[i] <= 1000
• 0 <= a, b, c <= 1000

Related Topics:
Array

## Solution 1. Brute force

// OJ: https://leetcode.com/problems/count-good-triplets/

// Time: O(N^3)
// Space: O(1)
class Solution {
public:
int countGoodTriplets(vector<int>& A, int a, int b, int c) {
int N = A.size(), ans = 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 (abs(A[i] - A[j]) <= a && abs(A[j] - A[k]) <= b && abs(A[i] - A[k]) <= c) ++ans;
}
}
}
return ans;
}
};


Java

class Solution {
public int countGoodTriplets(int[] arr, int a, int b, int c) {
int triplets = 0;
int length = arr.length;
for (int i = 0; i < length; i++) {
int num1 = arr[i];
for (int j = i + 1; j < length; j++) {
int num2 = arr[j];
if (Math.abs(num1 - num2) > a)
continue;
for (int k = j + 1; k < length; k++) {
int num3 = arr[k];
if (Math.abs(num2 - num3) <= b && Math.abs(num1 - num3) <= c)
triplets++;
}
}
}
return triplets;
}
}