# Binary representation of next greater number with same number of 1’s and 0’s

Given a binary input that represents binary representation of positive number n, find binary representation of smallest number greater than n with same number of 1’s and 0’s as in binary representation of n. If no such number can be formed, print “no greater number”.

The binary input may be and may not fit even in unsigned long long int.

Examples:

```Input : 10010
Output : 10100
Here n = (18)10 = (10010)2
next greater = (20)10 = (10100)2
Binary representation of 20 contains same number of
1's and 0's as in 18.

Input : 111000011100111110
Output :  111000011101001111
```

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

This problem simply boils down to finding next permutation of a given string. We can find the next_permutation() of the input binary number.

Below is an algorithm to find next permutation in binary string.

1. Traverse the binary string bstr from the right.
2. While traversing find the first index i such that bstr[i] = ‘0’ and bstr[i+1] = ‘1’.
3. Exchange character of at index ‘i’ and ‘i+1’.
4. Since we need smallest next value, consider substring from index i+2 to end and move all 1’s in the substring in the end.

Below is C++ implementation of above steps.

```// C++ program to find next permutation in a
// binary string.
#include <bits/stdc++.h>
using namespace std;

// Function to find the next greater number
// with same number of 1's and 0's
string nextGreaterWithSameDigits(string bnum)
{
int l = bnum.size();
int i;
for (int i=l-2; i>=1; i--)
{
// locate first 'i' from end such that
// bnum[i]=='0' and bnum[i+1]=='1'
// swap these value and break;
if (bnum.at(i) == '0' &&
bnum.at(i+1) == '1')
{
char ch = bnum.at(i);
bnum.at(i) = bnum.at(i+1);
bnum.at(i+1) = ch;
break;
}
}

// if no swapping performed
if (i == 0)
"no greater number";

// Since we want the smallest next value,
// shift all 1's at the end in the binary
// substring starting from index 'i+2'
int j = i+2, k = l-1;
while (j < k)
{
if (bnum.at(j) == '1' && bnum.at(k) == '0')
{
char ch = bnum.at(j);
bnum.at(j) = bnum.at(k);
bnum.at(k) = ch;
j++;
k--;
}

// special case while swapping if '0'
// occurs then break
else if (bnum.at(i) == '0')
break;

else
j++;

}

// required next greater number
return bnum;
}

// Driver program to test above
int main()
{
string bnum = "10010";
cout << "Binary representation of next greater number = "
<< nextGreaterWithSameDigits(bnum);
return 0;
}

```

Output:

```Binary representation of next greater number = 10100
```

Time Complexity : O(n) where n is number of bits in input.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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