Sync LeetCode submission Runtime - 300 ms (40.32%), Memory - 17.9 MB (24.73%)

Author: jimit105

1656-count-good-triplets/README.md

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+<p>Given an array of integers <code>arr</code>, and three integers&nbsp;<code>a</code>,&nbsp;<code>b</code>&nbsp;and&nbsp;<code>c</code>. You need to find the number of good triplets.</p>
+
+<p>A triplet <code>(arr[i], arr[j], arr[k])</code>&nbsp;is <strong>good</strong> if the following conditions are true:</p>
+
+<ul>
+	<li><code>0 &lt;= i &lt; j &lt; k &lt;&nbsp;arr.length</code></li>
+	<li><code>|arr[i] - arr[j]| &lt;= a</code></li>
+	<li><code>|arr[j] - arr[k]| &lt;= b</code></li>
+	<li><code>|arr[i] - arr[k]| &lt;= c</code></li>
+</ul>
+
+<p>Where <code>|x|</code> denotes the absolute value of <code>x</code>.</p>
+
+<p>Return<em> the number of good triplets</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3
+<strong>Output:</strong> 4
+<strong>Explanation:</strong>&nbsp;There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr = [1,1,2,2,3], a = 0, b = 0, c = 1
+<strong>Output:</strong> 0
+<strong>Explanation: </strong>No triplet satisfies all conditions.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= arr.length &lt;= 100</code></li>
+	<li><code>0 &lt;= arr[i] &lt;= 1000</code></li>
+	<li><code>0 &lt;= a, b, c &lt;= 1000</code></li>
+</ul>

1656-count-good-triplets/solution.py

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+# Approach 1: Enumeration
+
+# Time: O(n ^ 3)
+# Space: O(1)
+
+class Solution:
+    def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
+        n = len(arr)
+        count = 0
+        
+        for i in range(n):
+            for j in range(i + 1, n):
+                for k in range(j + 1, n):
+                    if (
+                        abs(arr[i] - arr[j]) <= a and
+                        abs(arr[j] - arr[k]) <= b and
+                        abs(arr[i] - arr[k]) <= c
+                    ):
+                        count += 1
+        return count
+
+