# Construct the smallest possible Array with given Sum and XOR

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

a + x + x = b
2*x = b-a
x = (b-a)/2

Hence, if we take x as (b-a)/2 then xor will be a, and the sum will also be b.

• 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 array ` `void` `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 Code ` `int` `main() ` `{ ` `    ``// Given sum and value ` `    ``// of Bitwise XOR ` `    ``int` `sum = 4, xorr = 2; ` ` `  `    ``// Function Call ` `    ``findArray(sum, xorr); ` `    ``cout << ``"\n"``; ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program for the above approach  ` ` `  `# Function to find array  ` `def` `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 ``=` `4` `xorr ``=` `2` ` `  `# Function Call  ` `findArray(_sum, xorr)  ` `     `  `# This code is contributed by divyamohan123 `

Output:

```Array size is 2
Array is 3 1
```

Time Complexity: O(1)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : divyamohan123