# 1534. Count Good Triplets

## Description

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

## Solutions

• class Solution {
public int countGoodTriplets(int[] arr, int a, int b, int c) {
int n = arr.length;
int 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 (Math.abs(arr[i] - arr[j]) <= a && Math.abs(arr[j] - arr[k]) <= b
&& Math.abs(arr[i] - arr[k]) <= c) {
++ans;
}
}
}
}
return ans;
}
}

• class Solution {
public:
int countGoodTriplets(vector<int>& arr, int a, int b, int c) {
int n = arr.size();
int 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) {
ans += abs(arr[i] - arr[j]) <= a && abs(arr[j] - arr[k]) <= b && abs(arr[i] - arr[k]) <= c;
}
}
}
return ans;
}
};

• class Solution:
def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
ans, n = 0, len(arr)
for i in range(n):
for j in range(i + 1, n):
for k in range(j + 1, n):
ans += (
abs(arr[i] - arr[j]) <= a
and abs(arr[j] - arr[k]) <= b
and abs(arr[i] - arr[k]) <= c
)
return ans


• func countGoodTriplets(arr []int, a int, b int, c int) (ans int) {
n := len(arr)
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
for k := j + 1; k < n; k++ {
if abs(arr[i]-arr[j]) <= a && abs(arr[j]-arr[k]) <= b && abs(arr[i]-arr[k]) <= c {
ans++
}
}
}
}
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}