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:
<li><code>0 <= i < j < k < arr.length</code></li>
<li><code>|arr[i] - arr[j]| <= a</code></li>
<li><code>|arr[j] - arr[k]| <= b</code></li>
<li><code>|arr[i] - arr[k]| <= c</code></li>
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:
<li><code>3 <= arr.length <= 100</code></li>
<li><code>0 <= arr[i] <= 1000</code></li>
<li><code>0 <= a, b, c <= 1000</code></li>
Solutions
Solution 1: Enumeration
We can enumerate all i, j, and k where i \lt j \lt k, and check if they simultaneously satisfy |arr[i] - arr[j]| \le a, |arr[j] - arr[k]| \le b, and |arr[i] - 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(n3), where n is the length of the array arr. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustC#
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
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;
}
}(code-box)
