# Find two numbers from their sum and OR

• Difficulty Level : Hard
• Last Updated : 12 Aug, 2021

Given two integers X and Y, the task is to find two numbers whose Bitwise OR is X and their sum is Y. If there exist no such integers, then print “-1”.

Examples:

Input: X = 7, Y = 11
Output: 4 7
Explanation:
The Bitwise OR of 4 and 7 is 7 and the sum of two integers is 4 + 7 = 11, satisfy the given conditions.

Input: X = 11, Y = 7
Output: -1

Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs of the array and if there exist any pairs that satisfy the given condition then print that pairs. Otherwise, print “-1”.

Time Complexity: O(Y)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the properties of Bitwise Operators. Consider any two integers A and B, then the sum of two integers can be represented as A + B = (A & B) + (A | B). Now, place the variables X and Y and change the equation as:

=> Y = (A & B) + X
=> (A & B) = Y – X
Therefore the above observations can be deduced with this equation.

Follow the steps below to solve the given problem:

• If the value of Y is less than X, then there will be no solution as Bitwise AND operations are always non-negative.
• Now for the Kth bit in the bit-wise representation of X and Y, if this bit is ‘1’ in (A&B) and “0” in (A | B) then there will be no possible solutions. This is because if the bit-wise AND of two numbers is 1 then is necessary that bit-wise OR should also be 1.
• Otherwise, it is always possible to choose two integers A and B which can be calculated as A = Y – X and B = X.

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach` `#include ``#define MaxBit 32``using` `namespace` `std;` `// Function to find the two integers from``// the given sum and Bitwise OR value``int` `possiblePair(``int` `X, ``int` `Y)``{``    ``int` `Z = Y - X;` `    ``// Check if Z is non negative``    ``if` `(Z < 0) {``        ``cout << ``"-1"``;``        ``return` `0;``    ``}` `    ``// Iterate through all the bits``    ``for` `(``int` `k = 0; k < MaxBit; k++) {` `        ``// Find the kth bit of A & B``        ``int` `bit1 = (Z >> k) & 1;` `        ``// Find the kth bit of A | B``        ``int` `bit2 = (Z >> k) & 1;` `        ``// If bit1 = 1 and bit2 = 0, then``        ``// there will be no possible pairs``        ``if` `(bit1 && !bit2) {``            ``cout << ``"-1"``;``            ``return` `0;``        ``}``    ``}` `    ``// Print the possible pairs``    ``cout << Z << ``' '` `<< X;` `    ``return` `0;``}` `// Driver Code``int` `main()``{``    ``int` `X = 7, Y = 11;``    ``possiblePair(X, Y);` `    ``return` `0;``}`

## Java

 `// Java code for above approach``import` `java.util.*;` `class` `GFG{` `static` `int` `MaxBit = ``32``;` `// Function to find the two integers from``// the given sum and Bitwise OR value``static` `void` `possiblePair(``int` `X, ``int` `Y)``{``    ``int` `Z = Y - X;` `    ``// Check if Z is non negative``    ``if` `(Z < ``0``) {``        ``System.out.print(``"-1"``);``    ``}` `    ``// Iterate through all the bits``    ``for` `(``int` `k = ``0``; k < MaxBit; k++) {` `        ``// Find the kth bit of A & B``        ``int` `bit1 = (Z >> k) & ``1``;` `        ``// Find the kth bit of A | B``        ``int` `bit2 = (Z >> k) & ``1``;` `        ``// If bit1 = 1 and bit2 = 0, then``        ``// there will be no possible pairs``        ``if` `(bit1 != ``0` `&& bit2 == ``0``) {``            ``System.out.print(``"-1"``);``        ``}``    ``}` `    ``// Print the possible pairs``    ``System.out.print( Z + ``" "` `+ X);` `}` `// Driver Code``public` `static` `void` `main(String[] args)` `{``    ``int` `X = ``7``, Y = ``11``;``    ``possiblePair(X, Y);``}``}` `// This code is contributed by avijitmondal1998.`

## Python3

 `# Python 3 program for the above approach``MaxBit ``=` `32` `# Function to find the two integers from``# the given sum and Bitwise OR value``def` `possiblePair(X, Y):``    ``Z ``=` `Y ``-` `X` `    ``# Check if Z is non negative``    ``if` `(Z < ``0``):``        ``print``(``"-1"``)``        ``return` `0` `    ``# Iterate through all the bits``    ``for` `k ``in` `range``(MaxBit):``      ` `        ``# Find the kth bit of A & B``        ``bit1 ``=` `(Z >> k) & ``1` `        ``# Find the kth bit of A | B``        ``bit2 ``=` `(Z >> k) & ``1` `        ``# If bit1 = 1 and bit2 = 0, then``        ``# there will be no possible pairs``        ``if` `(bit1 ``=``=` `1` `and` `bit2 ``=``=` `0``):``            ``print``(``"-1"``)``            ``return` `0` `    ``# Print the possible pairs``    ``print``(Z, X)` `    ``return` `0` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``X ``=` `7``    ``Y ``=` `11``    ``possiblePair(X, Y)``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `//C# code for the above approach``using` `System;` `public` `class` `GFG{``  ` `static` `int` `MaxBit = 32;` `// Function to find the two integers from``// the given sum and Bitwise OR value``static` `void` `possiblePair(``int` `X, ``int` `Y)``{``    ``int` `Z = Y - X;` `    ``// Check if Z is non negative``    ``if` `(Z < 0) {``        ``Console.Write(``"-1"``);``    ``}` `    ``// Iterate through all the bits``    ``for` `(``int` `k = 0; k < MaxBit; k++) {` `        ``// Find the kth bit of A & B``        ``int` `bit1 = (Z >> k) & 1;` `        ``// Find the kth bit of A | B``        ``int` `bit2 = (Z >> k) & 1;` `        ``// If bit1 = 1 and bit2 = 0, then``        ``// there will be no possible pairs``        ``if` `(bit1 != 0 && bit2 == 0) {``            ``Console.Write(``"-1"``);``        ``}``    ``}` `    ``// Print the possible pairs``    ``Console.Write( Z + ``" "` `+ X);` `}` `// Driver Code` `    ``static` `public` `void` `Main (){` `        ``// Code``      ``int` `X = 7, Y = 11;``    ``possiblePair(X, Y);``    ``}``}``// This code is contributed by Potta Lokesh`

## Javascript

 ``
Output:
`4 7`

Time Complexity: O(1)
Auxiliary Space: O(1)

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