# Check if it is possible to construct an Array of size N having sum as S and XOR value as X

• Difficulty Level : Expert
• Last Updated : 05 Jul, 2022

Given three numbers N, S and X the task is to find if it is possible to construct a sequence A of length N, where each A[i] >= 0 for 1<=i<=N and the sum of all numbers in a sequence is equal to S, and the bit-wise XOR of sequence equals to X.

Examples:

Input: N = 3, S = 10, X = 4
Output: Yes
Explanation: One of the sequence possible is {4, 3, 3} where sum equals 10 and XOR equals 4

Input: N = 1, S = 5, X = 3
Output: No

Approach: Lets consider the following test cases.

Case-1: When N equals 1 it can be easily seen that when (S equals X) then only return “Yes” otherwise “No“.

Case-2: When N is greater than equal to 3, use the formula (a + b) = (a xor b) + 2(a and b) Here it can be seen that (a + b) = S and (a xor b) = X so equation becomes S = X + 2(a.b). Therefore, (S-X) should be even because on right side we have 2(a.b). So, it can be said that is S is odd then X is odd  and if S is even then X is even then only, (S-X) is also even which can be checked by (S%2 == X%2) also S >= X otherwise A.B turns negative Which is not possible.

Case-3: For the case N equals 3, it is something like  A + B + C =  S  and A^B^C = X. Use the property A^A = 0 and  0^A = A => X + ( S – X)/2 + (S – X)/2 =  X + (S-X) => X + ( S – X)/2 + (S – X)/2 =  S and also this way: X ^( (S – X)/2 ^ (S-X)/2 ) = X ^ 0 = X. Hence, it is proved that for N == 3 there will always be such sequence and we can just return “Yes“.

Case-4: When N == 2 and  (S%2 == X%2) and S >= X, assume  A + B == S  and (A^B) == X then ( A and B) == (S-X)/2  From the equation discussed above. Let C = A.B. On observing carefully it can be noticed that the bits of C are “1” only when A and B bits are “1” for that position and off otherwise. And X that is xor of A, B has on bit only when there are different bits that is at ith  position A has ‘0’ and B has ‘1’ or the just opposite: So looking at this sequence, assign every bits into variable A and B, C = ( S – X)/2. Assign A and B from C -> A = C, B = C

Now add the X into A or B to assign all the ones into A and all zero to B so when we XOR both numbers then the added ‘1’ bits into A will just be opposite to what we added into B that is ‘0’. The fun part is when set bits of C coincides with some set bits of X then it will not give the desired xor of X, Now, A = C  + X, B = C. Now A+B = ( C + X) + C =  S and when XOR A.B equals X then it can be sure that there exist such pair when A + B == S  and (A^B) == X;

Follow the steps below to solve the problem:

• If S is greater than equal to X, and S%2 is equal to X%2 then perform the following steps, else return No.
• If n is greater than equal to 3, then return Yes.
• If n equals 1, and if S equals X, then return Yes else return No.
• If n equals to 2, initialize the variable C as (S-X)/2 and set variables A and B as C and add the value X to the variable A and if A^B equals X, then print Yes else print No.

Below is the implementation of the above approach.

## C++

 // C++ program for the above approach#include using namespace std; // Function to find if any sequence is// possible or not.string findIfPossible(int N, int S, int X){    if (S >= X and S % 2 == X % 2) {         // Since, S is greater than equal to        // X, and either both are odd or even        // There always exists a sequence        if (N >= 3) {            return "Yes";        }        if (N == 1) {             // Only one case possible is            // S == X or NOT;            if (S == X) {                return "Yes";            }            else {                return "No";            }        }         // Considering the above conditions true,        // check if XOR of S^(S-X) is X or not        if (N == 2) {             int C = (S - X) / 2;            int A = C;            int B = C;            A = A + X;            if (((A ^ B) == X)) {                return "Yes";            }            else {                return "No";            }        }    }    else {        return "No";    }} // Driver Codeint main(){     int N = 3, S = 10, X = 4;     cout << findIfPossible(N, S, X);     return 0;}

## C

 // C program for the above approach#include #include #include  // Function to find if any sequence is// possible or not.char* findIfPossible(int N, int S, int X){    if (S >= X && S % 2 == X % 2) {         // Since, S is greater than equal to        // X, and either both are odd or even        // There always exists a sequence        if (N >= 3) {            return "Yes";        }        if (N == 1) {             // Only one case possible is            // S == X or NOT;            if (S == X) {                return "Yes";            }            else {                return "No";            }        }         // Considering the above conditions true,        // check if XOR of S^(S-X) is X or not        if (N == 2) {             int C = (S - X) / 2;            int A = C;            int B = C;            A = A + X;            if (((A ^ B) == X)) {                return "Yes";            }            else {                return "No";            }        }    }    else {        return "No";    }} // Driver Codeint main(){     int N = 3, S = 10, X = 4;    printf("%s\n", findIfPossible(N, S, X));    return 0;} // This code is contributed by phalasi.

## Java

 // Java program for the above approachimport java.io.*;import java.util.*; class GFG {     // Function to find if any sequence is    // possible or not.    static void findIfPossible(int N, int S, int X)    {        if ((S >= X) && (S % 2 == X % 2)) {             // Since, S is greater than equal to            // X, and either both are odd or even            // There always exists a sequence            if (N >= 3) {                System.out.println("Yes");            }            if (N == 1) {                 // Only one case possible is                // S == X or NOT;                if (S == X) {                    System.out.println("Yes");                }                else {                    System.out.println("No");                }            }             // Considering the above conditions true,            // check if XOR of S^(S-X) is X or not            if (N == 2) {                 int C = (S - X) / 2;                int A = C;                int B = C;                A = A + X;                if (((A ^ B) == X)) {                    System.out.println("Yes");                }                else {                    System.out.println("No");                }            }        }        else {            System.out.println("No");        }    }     // Driver code    public static void main(String args[])    {        int N = 3, S = 10, X = 4;         findIfPossible(N, S, X);    }} // This code is contributed by code_hunt.

## Python3

 # Python program for the above approach# Function to find if any sequence is# possible or not.  def findIfPossible(N,  S,  X):     if (S >= X and S % 2 == X % 2):        # Since, S is greater than equal to        # X, and either both are odd or even        # There always exists a sequence        if (N >= 3):            return "Yes"         if (N == 1):             # Only one case possible is            # S == X or NOT            if (S == X):                return "Yes"            else:                return "No"         # Considering the above conditions true,        # check if XOR of S^(S-X) is X or not        if (N == 2):            C = (S - X) // 2            A = C            B = C            A = A + X            if (((A ^ B) == X)):                return "Yes"            else:                return "No"    else:        return "No"  # Driver CodeN = 3S = 10X = 4 print(findIfPossible(N, S, X)) # This code is contributed by shivanisinghss2110

## C#

 // C# program for the above approachusing System; public class GFG {     // Function to find if any sequence is    // possible or not.    static void findIfPossible(int N, int S, int X)    {        if ((S >= X) && (S % 2 == X % 2)) {             // Since, S is greater than equal to            // X, and either both are odd or even            // There always exists a sequence            if (N >= 3) {                Console.WriteLine("Yes");            }            if (N == 1) {                 // Only one case possible is                // S == X or NOT;                if (S == X) {                    Console.WriteLine("Yes");                }                else {                    Console.WriteLine("No");                }            }             // Considering the above conditions true,            // check if XOR of S^(S-X) is X or not            if (N == 2) {                 int C = (S - X) / 2;                int A = C;                int B = C;                A = A + X;                if (((A ^ B) == X)) {                    Console.WriteLine("Yes");                }                else {                    Console.WriteLine("No");                }            }        }        else {            Console.WriteLine("No");        }    }     // Driver code    public static void Main(String[] args)    {        int N = 3, S = 10, X = 4;         findIfPossible(N, S, X);    }} // This code is contributed by Princi Singh

## Javascript



Output

Yes

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

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