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1534. Count Good Triplets

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

• Java
• C++
• Python
• Go
• TypeScript
• C#
• Rust

• 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
  }
  

• function countGoodTriplets(arr: number[], a: number, b: number, c: number): number {
      let n = arr.length;
      let ans = 0;
      for (let i = 0; i < n; ++i) {
          for (let j = i + 1; j < n; ++j) {
              for (let 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;
  }
  
  

• public 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;
      }
  }
  
  

• impl Solution {
      pub fn count_good_triplets(arr: Vec<i32>, a: i32, b: i32, c: i32) -> i32 {
          let n = arr.len();
          let mut ans = 0;
  
          for i in 0..n {
              for j in i + 1..n {
                  for k in j + 1..n {
                      if (arr[i] - arr[j]).abs() <= a
                          && (arr[j] - arr[k]).abs() <= b
                          && (arr[i] - arr[k]).abs() <= c
                      {
                          ans += 1;
                      }
                  }
              }
          }
  
          ans
      }
  }
  
  

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