_1534. Count Good Triplets.md

1534. Count Good Triplets

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First completed : April 14, 2025

Last updated : April 14, 2025

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Related Topics : Array, Enumeration

Acceptance Rate : 85.54 %

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Solutions

• e1534_v1.c
• m1534_v1_compressed_lol.c
• e1534_v1_brute_force.py
• e1534_v2_brute_force_oneliner.py

C

e1534_v1.c

int countGoodTriplets(int* arr, int arrSize, int a, int b, int c){
    int cnt = 0;
    
    for (int i = 0; i < arrSize - 2; i++) {
        for (int j = i + 1; j < arrSize - 1; j++) {
            for (int k = j + 1; k < arrSize; k++) {
                if (
                    abs(arr[i] - arr[j]) <= a &&
                    abs(arr[j] - arr[k]) <= b &&
                    abs(arr[i] - arr[k]) <= c
                ) {
                    cnt++;
                }
            }
        }
    }

    return cnt;
}

m1534_v1_compressed_lol.c

int countGoodTriplets(int* arr, int arrSize, int a, int b, int c){
    int cnt = 0;
    for (int i = 0; i < arrSize - 2; i++) { for (int j = i + 1; j < arrSize - 1; j++) { for (int k = j + 1; k < arrSize; k++) {
        if (abs(arr[i] - arr[j]) <= a && abs(arr[j] - arr[k]) <= b && abs(arr[i] - arr[k]) <= c) { cnt++; }
    }}}
    return cnt;
}

Python

e1534_v1_brute_force.py

class Solution:
    def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
        output = 0
        for i in range(len(arr) - 2) :
            for j in range(i + 1, len(arr) - 1) :
                for k in range(j + 1, len(arr)) :
                    if abs(arr[i] - arr[j]) <= a and \
                       abs(arr[j] - arr[k]) <= b and \
                       abs(arr[i] - arr[k]) <= c :
                        output += 1
        return output

e1534_v2_brute_force_oneliner.py

class Solution:
    def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
        return [
            abs(arr[i] - arr[j]) <= a and abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c
            for i in range(len(arr) - 2) for j in range(i + 1, len(arr) - 1) for k in range(j + 1, len(arr))
        ].count(True)