Add solution for maximizing partitions with bit manipulation

Implement a dynamic programming solution to maximize the number of partitions after operations on a string based on bit manipulation.

Author: Arunanshu655

bit_manipulation/Maximize _he_Number_of_Partitions_After_Operations.py

@@ -0,0 +1,57 @@
+# Topics: Bit Manuputation, Bit-Mask, DP
+
+
+class Solution:
+    def __init__(self):
+        self.dp = {}
+    
+    def countSetBits(self, n):
+        count = 0
+        while n > 0:
+            count += n & 1
+            n >>= 1
+        return count
+    
+    def helper(self, s, k, i, mask, isChange):
+        n = len(s)
+        if i == n:
+            return 1
+        
+        currState = (i << 27) | (mask << 1) | isChange
+        
+        if currState in self.dp:
+            return self.dp[currState]
+        
+        result = 0
+        val = ord(s[i]) - ord('a')
+        
+        count = self.countSetBits(mask | (1 << val))
+        temp = 0
+        if count > k:
+            temp = 1 + self.helper(s, k, i + 1, 1 << val, isChange)
+        else:
+            temp = self.helper(s, k, i + 1, mask | (1 << val), isChange)
+        result = max(result, temp)
+        
+        if isChange == 0:
+            for j in range(26):
+                count = self.countSetBits(mask | (1 << j))
+                if count > k:
+                    temp = 1 + self.helper(s, k, i + 1, 1 << j, 1)
+                else:
+                    temp = self.helper(s, k, i + 1, mask | (1 << j), 1)
+                result = max(result, temp)
+        
+        self.dp[currState] = result
+        return result
+    
+    def maxPartitionsAfterOperations(self, s, k):
+        """
+        :type s: str
+        :type k: int
+        :rtype: int
+        """
+        return self.helper(s, k, 0, 0, 0)
+
+ #  Time Complexity: O(n × 2²⁷)
+ # Auxiliary Space: O(n × 2²⁷)