Question

1534. Count Good Triplets

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Description

Given an array of integers arr, and three integers a, b 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

Solution 1

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

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;
  }
};

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
}