Maximum value of B less than A such that A ^ B = A + B

• Difficulty Level : Easy
• Last Updated : 21 May, 2021

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:
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 3

Input:
Output:
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 using namespace std; // Function to find the maximum// value of B such that A^B = A+Bvoid 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 Codeif __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


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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