Check if it is possible to construct an Array of size N having sum as S and XOR value as X
Last Updated :
23 Jul, 2025
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 <bits/stdc++.h>
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 Code
int main()
{
int N = 3, S = 10, X = 4;
cout << findIfPossible(N, S, X);
return 0;
}
// C program for the above approach
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
// 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 Code
int main()
{
int N = 3, S = 10, X = 4;
printf("%s\n", findIfPossible(N, S, X));
return 0;
}
// This code is contributed by phalasi.
// Java program for the above approach
import 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.
# 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 Code
N = 3
S = 10
X = 4
print(findIfPossible(N, S, X))
# This code is contributed by shivanisinghss2110
// C# program for the above approach
using 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
<script>
// JavaScript Program to implement
// the above approach
// Function to find if any sequence is
// possible or not.
function findIfPossible(N, S, 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) {
let C = (S - X) / 2;
let A = C;
let B = C;
A = A + X;
if (((A ^ B) == X)) {
return "Yes";
}
else {
return "No";
}
}
}
else {
return "No";
}
}
// Driver Code
let N = 3, S = 10, X = 4;
document.write(findIfPossible(N, S, X));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(1)
Auxiliary Space: O(1)
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