[LeetCode] 1534. Count Good Triplets

轉自LeetCode 

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

Solution

原本的想法是要用 DFS 去找出所有的組合

但寫起來又怪怪的，條件檢查有點複雜，這題又是 easy

後來發現，就真的照題意做就可以了

kotlin(參考解答)

	class Solution {
	fun countGoodTriplets(arr: IntArray, a: Int, b: Int, c: Int): Int {
	var ans = 0
	for(i in 0 until arr.lastIndex-1) {
	for(j in i+1 until arr.lastIndex) {
	if(Math.abs(arr[i]-arr[j]) <= a) {
	for(k in j+1 until arr.size) {
	if(Math.abs(arr[j]-arr[k]) <= b && Math.abs(arr[i]-arr[k]) <= c) {
	++ans
	}
	}
	}
	}
	}
	return ans
	}
	}

view raw count_good_triplets.kt hosted with ❤ by GitHub