# 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 the implementation of above steps.

## C++

 `// C++ program to find next permutation in a ` `// binary string. ` `#include ` `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; ` `} `

## Java

 `// Java program to find next permutation in a ` `// binary string. ` `class` `GFG  ` `{ ` ` `  `// Function to find the next greater number ` `// with same number of 1's and 0's ` `static` `String nextGreaterWithSameDigits(``char``[] bnum) ` `{ ` `    ``int` `l = bnum.length; ` `    ``int` `i; ` `    ``for` `(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[i] == ``'0'` `&& ` `        ``bnum[i+``1``] == ``'1'``) ` `        ``{ ` `            ``char` `ch = bnum[i]; ` `            ``bnum[i] = bnum[i+``1``]; ` `            ``bnum[i+``1``] = ch; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// if no swapping performed ` `    ``if` `(i == ``0``) ` `        ``System.out.println(``"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[j] == ``'1'` `&& bnum[k] == ``'0'``) ` `        ``{ ` `            ``char` `ch = bnum[j]; ` `            ``bnum[j] = bnum[k]; ` `            ``bnum[k] = ch; ` `            ``j++; ` `            ``k--; ` `        ``} ` ` `  `        ``// special case while swapping if '0' ` `        ``// occurs then break ` `        ``else` `if` `(bnum[i] == ``'0'``) ` `            ``break``; ` ` `  `        ``else` `            ``j++; ` ` `  `    ``} ` ` `  `    ``// required next greater number ` `    ``return` `String.valueOf(bnum); ` `} ` ` `  `// Driver program to test above ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``char``[] bnum = ``"10010"``.toCharArray(); ` `    ``System.out.println(``"Binary representation of next greater number = "` `        ``+ nextGreaterWithSameDigits(bnum)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python3 program to find next permutation in a ` `# binary string. ` ` `  `# Function to find the next greater number ` `# with same number of 1's and 0's ` `def` `nextGreaterWithSameDigits(bnum): ` `    ``l ``=` `len``(bnum) ` `    ``bnum ``=` `list``(bnum) ` `    ``for` `i ``in` `range``(l ``-` `2``, ``0``, ``-``1``): ` `         `  `        ``# locate first 'i' from end such that ` `        ``# bnum[i]=='0' and bnum[i+1]=='1' ` `        ``# swap these value and break ` `        ``if` `(bnum[i] ``=``=` `'0'` `and` `bnum[i ``+` `1``] ``=``=` `'1'``): ` `            ``ch ``=` `bnum[i] ` `            ``bnum[i] ``=` `bnum[i ``+` `1``] ` `            ``bnum[i ``+` `1``] ``=` `ch          ` `            ``break` `         `  `    ``# if no swapping performed ` `    ``if` `(i ``=``=` `0``): ` `        ``return` `"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' ` `    ``j ``=` `i ``+` `2` `    ``k ``=` `l ``-` `1` `    ``while` `(j < k): ` `        ``if` `(bnum[j] ``=``=` `'1'` `and` `bnum[k] ``=``=` `'0'``): ` `            ``ch ``=` `bnum[j] ` `            ``bnum[j] ``=` `bnum[k] ` `            ``bnum[k] ``=` `ch ` `            ``j ``+``=` `1` `            ``k ``-``=` `1` `             `  `        ``# special case while swapping if '0' ` `        ``# occurs then break ` `        ``elif` `(bnum[i] ``=``=` `'0'``): ` `            ``break` `        ``else``: ` `            ``j ``+``=` `1` `     `  `    ``# required next greater number ` `    ``return` `bnum ` ` `  `# Driver code ` `bnum ``=` `"10010"` `print``(``"Binary representation of next greater number = "``,``*``nextGreaterWithSameDigits(bnum),sep``=``"") ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# program to find next permutation in a ` `// binary string. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find the next greater number ` `// with same number of 1's and 0's ` `static` `String nextGreaterWithSameDigits(``char``[] bnum) ` `{ ` `    ``int` `l = bnum.Length; ` `    ``int` `i; ` `    ``for` `(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[i] == ``'0'` `&& ` `        ``bnum[i+1] == ``'1'``) ` `        ``{ ` `            ``char` `ch = bnum[i]; ` `            ``bnum[i] = bnum[i+1]; ` `            ``bnum[i+1] = ch; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// if no swapping performed ` `    ``if` `(i == 0) ` `        ``Console.WriteLine(``"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[j] == ``'1'` `&& bnum[k] == ``'0'``) ` `        ``{ ` `            ``char` `ch = bnum[j]; ` `            ``bnum[j] = bnum[k]; ` `            ``bnum[k] = ch; ` `            ``j++; ` `            ``k--; ` `        ``} ` ` `  `        ``// special case while swapping if '0' ` `        ``// occurs then break ` `        ``else` `if` `(bnum[i] == ``'0'``) ` `            ``break``; ` ` `  `        ``else` `            ``j++; ` ` `  `    ``} ` ` `  `    ``// required next greater number ` `    ``return` `String.Join(``""``,bnum); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``char``[] bnum = ``"10010"``.ToCharArray(); ` `    ``Console.WriteLine(``"Binary representation of next greater number = "` `        ``+ nextGreaterWithSameDigits(bnum)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

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

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

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