# Find N distinct numbers whose bitwise Or is equal to K in C++

## Concept

With respect of given two integers N and K, our task is to determine N distinct integers whose bitwise OR is equal to K. It has been seen that if there does not exist any possible answer then print -1.

## Input

N = 4, K = 6

## Output

6 0 1 2

## Input

N = 11, K = 6

## Output

-1

It is not possible to find any solution.

## Method

• We have knowledge that if bit-wise OR of a sequence of numbers is K then all the bit indexes which are 0 in K must also be zero in all the numbers.

• As a result of this, we only have those positions to change where bit is 1 in K. Let that count is Bit_K.

• At present, we can create pow(2, Bit_K) distinct numbers with Bit_K bits. As a result of this, if, we treat one number to be K itself, then remaining N – 1 numbers can be built by setting 0 all the bits in each and every number which are0 in K and for other bit positions any permutation of Bit_K bits other than number K.

• It has been seen that if pow(2, Bit_K) < N then we cannot determine any possible answer.

## Example

Live Demo

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define MAX1 32
ll pow2[MAX1];
bool visited1[MAX1];
vector<int> ans1;
// Shows function to pre-calculate
// all the powers of 2 upto MAX
void power_2(){
ll ans1 = 1;
for (int i = 0; i < MAX1; i++) {
pow2[i] = ans1;
ans1 *= 2;
}
}
// Shows function to return the
// count of set bits in x
int countSetBits(ll x1){
// Used to store the count
// of set bits
int setBits1 = 0;
while (x1 != 0) {
x1 = x1 & (x1 - 1);
setBits1++;
}
return setBits1;
}
// by placing all bit positions as 0
// which are also 0 in K
int point1 = 0;
ll value1 = 0;
for (ll i = 0; i < MAX1; i++) {
// Bit i is 0 in K
if (visited1[i])
continue;
else {
if (num1 & 1) {
value1 += (1 << i);
}
num1 /= 2;
}
}
ans1.push_back(value1);
}
// Shows function to find and print N distinct
// numbers whose bitwise OR is K
void solve(ll n1, ll k1){
// Choosing K itself as one number
ans1.push_back(k1);
// Find the count of set bits in K
int countk1 = countSetBits(k1);
// It is not possible to get N
// distinct integers
if (pow2[countk1] < n1) {
cout << -1;
return;
}
int count1 = 0;
for (ll i = 0; i < pow2[countk1] - 1; i++) {
// placing all the bits as 0
// which are 0 in K
count1++;
// Now if N distinct numbers are generated
if (count1 == n1)
break;
}
// Now print the generated numbers
for (int i = 0; i < n1; i++) {
cout << ans1[i] << " ";
}
}
// Driver code
int main(){
ll n1 = 4, k1 = 6;
// Pre-calculate all
// the powers of 2
power_2();
solve(n1, k1);
return 0;
}

## Output

6 0 1 2