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1534. Count Good Triplets

SourceWeekly Contest 200 Q1DifficultyEasyRating1279

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: Enumeration

Thinking

Count index triples that satisfy three absolute-value bounds. \(n\le 100\), so a triple loop is about \(10^6\) comparisons and fits the limit.

Enumerate \(i<j<k\) and apply the three inequalities directly. The constraints are independent; there is no monotone structure that would justify a heavier data structure.

We can enumerate all \(i\), \(j\), and \(k\) where \(i \lt j \lt k\), and check if they simultaneously satisfy \(|\textit{arr}[i] - \textit{arr}[j]| \le a\), \(|\textit{arr}[j] - \textit{arr}[k]| \le b\), and \(|\textit{arr}[i] - \textit{arr}[k]| \le c\). If they do, we increment the answer by one.

After enumerating all possible triplets, we get the answer.

The time complexity is \(O(n^3)\), where \(n\) is the length of the array \(\textit{arr}\). The space complexity is \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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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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;
    }
}

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