# Construct the smallest possible Array with given Sum and XOR

• Difficulty Level : Hard
• Last Updated : 07 Apr, 2021

Given two positive integers S and X which represents the sum and Bitwise XOR of all the elements of an array arr[]. The task is to find the elements of the array arr[]. If no such array can be generated, print -1.
Examples:

Input: Sum = 4, Xor = 2
Output: {3, 1}
Explanation:
Sum of 1 and 3 is 4.
Bitwise XOR of 1 and 3 is 2.
Input: Sum = 5, Xor = 8
Output: -1
Explanation:
There is no such array exists.

Approach: It can be proven that the maximum length of the array will be at most 3. Let us consider the following cases:

• Case 1: If the given Sum and Bitwise XOR are equal and non-zero, then, that element will be the only element of the required array.

• Case 2: For unequal Sum and Bitwise XOR, the shortest array length can be either 2 or 3.
If the given Bitwise XOR and Sum is a and b respectively, then the shortest array can be {a, (b – a)/2, (b-a)/2 } by using the following two properties of XOR:
• a Xor 0 = a
• a Xor a = 0
• Case 3: When length of array can be 2.
The array we took in the previous case can be reduced to two elements if it is possible.

One important formula helpful here is:-

•

p + q = (p ^ q) + 2*(p & q)

•

Substituting values of sum and xor in the above formula we get a very helpful relation.

•

b = a + 2*(p & q)
(p & q) = (b-a)/2
From previous case,
x = (b-a)/2 = (p & q)

•

So, now let’s see some relation between a (given xor) and x ((b-a)/2).

•
p  q  a=(p^q)  x=(p&q)  a&x
0  0    0        0       0
0  1    1        0       0
1  0    1        0       0
1  1    0        1       0
• Note: p and q represents all corresponding 32 bits of two numbers in array.

It is important to note that if a & x becomes zero then a + x = a ^ x which means the array will be reduced to {a+x, x} because a+x = a^x. From the above formula, which can still lead to overall XOR as a, and sum will still be b as x is (b-a)/2.

• Case 4: The only case left is to check whether array exists or not. In this case, there are only two condition to check as:
1. If Bitwise XOR is greater than the sum.
2. If sum and xor have different parities i.e., one is even and other is odd.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to find arrayvoid findArray(int sum, int xorr){    // array not possible    if (xorr > sum        || sum % 2 != xorr % 2) {        cout << "No Array Possible\n";        return;    }     // Array possible with exactly 1    // or no element    if (sum == xorr) {        if (sum == 0)            cout << "Array is empty"                 << " with size 0\n";        else            cout << "Array size is "                 << 1                 << "\n Array is "                 << sum << "\n";        return;    }     int mid = (sum - xorr) / 2;     // Checking array with two    // elements possible or not.    if (xorr & mid == 1) {        cout << "Array size is "             << 3 << "\n";         cout << "Array is "             << xorr << " "             << mid << " "             << mid << "\n";    }    else {         cout << "Array size is "             << 2 << "\n";         cout << "Array is "             << (xorr + mid)             << " "             << mid << "\n";    }} // Driver Codeint main(){    // Given sum and value    // of Bitwise XOR    int sum = 4, xorr = 2;     // Function Call    findArray(sum, xorr);    cout << "\n";    return 0;}

## Java

 // Java program implementation// of the approachimport java.util.*;import java.io.*; class GFG{     // Function to find arraystatic void findArray(int sum, int xorr){         // Array not possible    if (xorr > sum  || sum % 2 != xorr % 2)    {        System.out.println("No Array Possible");        return;    }     // Array possible with exactly 1    // or no element    if (sum == xorr)    {        if (sum == 0)            System.out.println("Array is empty " +                               "with size 0");        else            System.out.println("Array size is " + 1);            System.out.println("Array is " + sum);                         return;    }    int mid = (sum - xorr) / 2;     // Checking array with two    // elements possible or not.    if (xorr == 1 && mid == 1)    {        System.out.println("Array size is " + 3);        System.out.println("Array is " + xorr +                              " " + mid + " " + mid);    }    else    {        System.out.println("Array size is " + 2);        System.out.println("Array is " + (xorr + mid) +                                           " " + mid);    }} // Driver codepublic static void main(String[] args){         // Given sum and value    // of Bitwise XOR    int sum = 4, xorr = 2;     // Function call    findArray(sum, xorr);}} // This code is contributed by sanjoy_62

## Python3

 # Python3 program for the above approach # Function to find arraydef findArray(_sum, xorr):     # Array not possible    if (xorr > _sum or        _sum % 2 != xorr % 2):        print("No Array Possible")        return     # Array possible with exactly 1    # or no element    if (_sum == xorr):        if (_sum == 0):            print("Array is empty",                  " with size 0")                           else:            print("Array size is", 1)            print("Array is", _sum)        return     mid = (_sum - xorr) // 2     # Checking array with two    # elements possible or not.    if (xorr & mid == 1):        print("Array size is", 3)        print("Array is", xorr, mid, mid)     else:        print("Array size is", 2)         print("Array is" ,(xorr + mid), mid) # Driver Code # Given sum and value# of Bitwise XOR_sum = 4xorr = 2 # Function CallfindArray(_sum, xorr)     # This code is contributed by divyamohan123

## C#

 // C# program implementation// of the approachusing System; class GFG{     // Function to find arraystatic void findArray(int sum, int xorr){     // Array not possible    if (xorr > sum || sum % 2 != xorr % 2)    {        Console.WriteLine("No Array Possible");        return;    }     // Array possible with exactly 1    // or no element    if (sum == xorr)    {        if (sum == 0)            Console.WriteLine("Array is empty " +                              "with size 0");        else            Console.WriteLine("Array size is " + 1);            Console.WriteLine("Array is " + sum);            return;    }    int mid = (sum - xorr) / 2;     // Checking array with two    // elements possible or not.    if (xorr == 1 && mid == 1)    {        Console.WriteLine("Array size is " + 3);        Console.WriteLine("Array is " + xorr +                             " " + mid + " " + mid);    }    else    {        Console.WriteLine("Array size is " + 2);        Console.WriteLine("Array is " + (xorr + mid) +                                          " " + mid);    }} // Driver codepublic static void Main(){         // Given sum and value    // of Bitwise XOR    int sum = 4, xorr = 2;     // Function call    findArray(sum, xorr);}} // This code is contributed by sanjoy_62

## Javascript



Output:

Array size is 2
Array is 3 1

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

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