create find non repeating elements

Author: ayushgupta98

find_non_repeating_elements.py

@@ -0,0 +1,77 @@
+arr = [1,2,3,7,3,2]
+
+"""
+# Approach 1: Brute-force approach
+# Complexity: n-squared
+"""
+def ispresent(arr, a):
+    if a in arr:
+        return 1
+    else:
+        return 0
+
+passed_entries = []
+entries_count = []
+for i in range(0, len(arr)):
+    if ispresent(passed_entries, arr[i]) == 0:
+        passed_entries.append(arr[i])
+        entries_count.append(1)
+    else:
+        entries_count[passed_entries.index(arr[i])] += 1 
+
+
+for i in range(len(entries_count)):
+    if entries_count[i] == 1:
+        print("Methid 1: Non-repeating entry ", passed_entries[i])
+
+
+"""
+# Approach 2: Hash Map
+# Complexity: O(n)
+# For this approach to work the pre-requisite is to have all positive number in the array. The downside of using this approach is to have extra memory requirement.
+"""
+entries_count = [0]*(max(arr)+1)
+for i in range(len(arr)):
+    entries_count[arr[i]] += 1
+
+for i in range(len(entries_count)):
+    if entries_count[i] == 1:
+        print("Method 2: Non-repeating entry ", i)
+        
+    
+"""
+# Approach 3: Using XOR operation
+# Complexity: O(n)
+# This is the best possible method to find a non-repeated entry in an array.
+# Downside of this method is that this method can find atmost 2 non-repeated enteries in a list.
+"""
+import math
+def getFirstSetBitPos(n):
+    return int(math.log2(n&-n)+1)
+
+xor = 0
+for i in range(len(arr)):
+    xor ^= arr[i]
+
+index = getFirstSetBitPos(xor)
+
+grp1 =[]
+grp2 = []
+
+for i in range(len(arr)):
+    x = (arr[i])>>(index-1)    
+    if x%2 == 1:
+        grp1.append(arr[i])
+    else:
+        grp2.append(arr[i])
+
+result = []
+r = 0
+for i in range(len(grp1)):
+    r ^= grp1[i]
+print("Method 3: Non-repeating entry is ", r)
+
+r = 0
+for i in range(len(grp2)):
+    r ^= grp2[i]
+print("Method 3: Non-repeating entry is ", r)