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Author: Manasi2001

Bit Manipulation/Bitwise_AND.py

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+'''
+
+Aim: Given set S = {1,2,3,....,N}. Find two integers, I and J (where I<J), 
+from set S such that the value of I & J is the maximum possible and also 
+less than a given integer, K. In this case, '&' represents the bitwise AND 
+operator.
+
+Input Format: 
+1. The first line contains an integer T, the number of test cases.
+2. Each of the T subsequent lines defines a test case as 2 space-separated 
+   integers, N and K, respectively.
+
+'''
+
+def max_bit(n,k):
+  maximum = 0              # we need a counter to keep a check on the value of I & J, it should not exceed K
+  for i in range(1,n+1):
+    for j in range(1,i):
+      h = i & j            # storing bitwise AND value in a variable h
+      if maximum < h < k:  # I & J < K
+        maximum = h
+      if maximum == k-1:   # if maximum becomes equal to k-1, we have found the maximum value of I & J which will be less than K
+        return maximum
+  return maximum
+
+if __name__ == '__main__':
+  t = int(input())
+  for t_itr in range(t):
+    nk = input().split()
+    n = int(nk[0])           # getting the value of N
+    k = int(nk[1])           # getting the value of K
+    print(max_bit(n,k))      # calling function to compute the answer
+
+'''   
+
+Sample Test Case:
+Input: 
+3       
+5 2     
+8 5     
+2 2 
+Output: 
+1
+4
+0
+Explaination:
+N = 5, K = 2, S = {1,2,3,4,5}
+All possible values of I and J are:
+ 1. I = 1, J = 2; I & J = 0 
+ 2. I = 1, J = 3; I & J = 1 
+ 3. I = 1, J = 4; I & J = 0 
+ 4. I = 1, J = 5; I & J = 1
+ 5. I = 2, J = 3; I & J = 2 
+ 6. I = 2, J = 4; I & J = 0 
+ 8. I = 2, J = 5; I & J = 0 
+ 9. I = 3, J = 5; I & J = 1
+10. I = 4, J = 5; I & J = 4
+The maximum possible value of I & J that is also < (K = 2) is 1, so the output will be 1.
+
+'''