From c52a533882884f39f44b24d02bbba6a825e53c63 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Sezai=20Burak=20Kantarc=C4=B1?=
 <45287996+kantarcise@users.noreply.github.com>
Date: Wed, 17 Apr 2024 11:58:08 +0300
Subject: [PATCH] Brute Force - Fenwick

---
 Leet_Code/easy/1534_Count_Good_Triplets.py | 74 ++++++++++++++++++++++
 1 file changed, 74 insertions(+)
 create mode 100644 Leet_Code/easy/1534_Count_Good_Triplets.py

diff --git a/Leet_Code/easy/1534_Count_Good_Triplets.py b/Leet_Code/easy/1534_Count_Good_Triplets.py
new file mode 100644
index 0000000..1c66f9f
--- /dev/null
+++ b/Leet_Code/easy/1534_Count_Good_Triplets.py
@@ -0,0 +1,74 @@
+"""
+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
+
+Takeaway:
+
+    Brute force works, 
+    
+    there is also a Fenwick Tree Solution
+
+    Search for:
+    
+        -Roughly O(n log n) with Fenwick Tree-
+"""
+
+class Solution:
+    def countGoodTriplets(self, arr: list[int], a: int, b: int, c: int) -> int:
+        # just brute force!
+
+        result = 0
+        for i in range(len(arr)):
+            for j in range(i+1, len(arr)):
+                for k in range(j+1, len(arr)):
+                    print(i, j, k)
+                    if ((abs(arr[i] - arr[j]) <= a) and
+                        (abs(arr[j] - arr[k]) <= b) and
+                        (abs(arr[i] - arr[k]) <= c)):
+                        result += 1
+        
+sol = Solution()
+print(sol.countGoodTriplets(arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3))