Given an integer A, the task is to find the maximum value possible(B) which is less than A, such that xor of these two numbers A and B are equal to their sum, that is A ^ B = A + B.
Examples:
Input: A = 4
Output: 3
Explanation:
There are many such integers, such that A ^ B = A + B
Some of these integers are –
4 ^ 3 = 4 + 3 = 7
4 ^ 2 = 4 + 2 = 6
4 ^ 1 = 4 + 1 = 5
4 ^ 0 = 4 + 0 = 4
The maximum of these values is 3Input: 7
Output: 0
There is no integer except 0 such that A + B = A ^ B
Approach: The idea is to use the fact that
and to get the value of 
, the value of (A & B) must be equal to 0.
=> A & B = 0 => B = ~A
For Example:
A = 4 (1 0 0) B = ~ A = (0 1 1) = 3
Below is the implementation of the above approach:
C++
// C++ implementation to find // maximum value of B such that // A ^ B = A + B #include <bits/stdc++.h>using namespace std;// Function to find the maximum // value of B such that A^B = A+B void maxValue(int a) { // Binary Representation of A string c = bitset<3>(a).to_string(); string b = ""; // Loop to find the negation // of the integer A for(int i = 0; i < c.length(); i++) { if ((c[i] - '0') == 1) b += '0'; else b += '1'; } // Output cout << bitset<3>(b).to_ulong(); } // Driver codeint main(){ int a = 4; // Function Call maxValue(a); return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java implementation to find// maximum value of B such that// A ^ B = A + B // Function to find the maximum// value of B such that A^B = A+Bclass GFG{static void maxValue(int a){ // Binary Representation of A String c = Integer.toBinaryString(a); String b = ""; // Loop to find the negation // of the integer A for (int i = 0; i < c.length(); i++) { if((c.charAt(i)-'0')==1) b +='0'; else b+='1'; } // output System.out.print(Integer.parseInt(b, 2)); } // Driver Codepublic static void main(String []args){ int a = 4; // Function Call maxValue(a);}}// This code is contributed by chitranayal |
Python3
# Python3 implementation to find# maximum value of B such that# A ^ B = A + B# Function to find the maximum# value of B such that A^B = A+Bdef maxValue(a): # Binary Representation of A a = bin(a)[2:] b = '' # Loop to find the negation # of the integer A for i in list(a): b += str(int(not int(i))) # output print(int(b, 2)) return int(b, 2)# Driver Code if __name__ == '__main__': a = 4 # Function Call maxValue(a) |
C#
// C# implementation to find// maximum value of B such that// A ^ B = A + B // Function to find the maximum// value of B such that A^B = A+Busing System;using System.Collections.Generic;class GFG{ static void maxValue(int a){ // Binary Representation of A String c = Convert.ToString(a, 2); String b = ""; // Loop to find the negation // of the integer A for (int i = 0; i < c.Length; i++) { if((c[i] - '0') == 1) b += '0'; else b += '1'; } // output Console.Write(Convert.ToInt32(b, 2)); } // Driver Codepublic static void Main(String []args){ int a = 4; // Function Call maxValue(a);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation to find // maximum value of B such that // A ^ B = A + B // Function to find the maximum // value of B such that A^B = A+B function maxValue(a) { // Binary Representation of A var c = a.toString(2); var b = ""; // Loop to find the negation // of the integer A for(var i = 0; i < c.length; i++) { if ((c[i] - '0') == 1) b += '0'; else b += '1'; } // Output document.write(parseInt(b,2));} // Driver codevar a = 4; // Function Call maxValue(a);</script> |
Output:
3
Performance Analysis:
- Time Complexity: In the above-given approach, there is the conversion from decimal to binary which takes O(logN) time in the worst case. Therefore, the time complexity for this approach will be O(logN).
- Auxiliary Space Complexity: In the above-given approach, there is no extra space used. Therefore, the auxiliary space complexity for the above approach will be O(1)
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