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
 

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 <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;
}

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
        else if (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

Javascript

<script>

// Javascript 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
function nextGreaterWithSameDigits(bnum)
{
    let l = bnum.length;
    let 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')
        {
            let ch = bnum[i];
            bnum[i] = bnum[i+1];
            bnum[i+1] = ch;
            break;
        }
    }

    // If no swapping performed
    if (i == 0)
        document.write("no greater number<br>");

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

// Driver code
let bnum = "10010".split("");
document.write("Binary representation of next " +
               "greater number = " +
               nextGreaterWithSameDigits(bnum));

// This code is contributed by rag2127

</script>

Output

Binary representation of next greater number = 10100

Time Complexity : O(n) where n is number of bits in input.
Auxiliary Space: O(1)

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

Approach 2 :

Here’s the approach to find the next greater number with the same number of 1’s and 0’s in a binary string:

1. Find the rightmost non-trailing one (RT1) in the string. Let its index be i.
2. If there is no RT1, then the given binary string is already the largest possible binary string with the same number of 1’s and 0’s. Return “no greater number”.
3. Find the rightmost zero to the right of i (let its index be j), and swap it with the RT1.
4. Sort the substring to the right of j in ascending order.
5. Return the resulting string.

Here’s the corrected C++ and Java code for this approach:

C++

#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 = l - 1;

    // Find the rightmost non-trailing one
    while (i >= 0 && bnum[i] == '0') {
        i--;
    }
    if (i < 0) {
        return "no greater number";
    }

    // Find the rightmost zero to the right of i
    int j = i - 1;
    while (j >= 0 && bnum[j] == '1') {
        j--;
    }
    if (j < 0) {
        return "no greater number";
    }

    // Swap the RT1 with the rightmost zero to the right of i
    swap(bnum[i], bnum[j]);

    // Sort the substring to the right of j in ascending order
    sort(bnum.begin() + j + 1, bnum.end());

    // 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

import java.util.Arrays;

public class GFG {

    // Function to find the next greater number
    // with the same number of 1's and 0's
    public static String nextGreaterWithSameDigits(String bnum) {
        int l = bnum.length();
        int i = l - 1;

        // Find the rightmost non-trailing one
        while (i >= 0 && bnum.charAt(i) == '0') {
            i--;
        }
        if (i < 0) {
            return "no greater number";
        }

        // Find the rightmost zero to the right of i
        int j = i - 1;
        while (j >= 0 && bnum.charAt(j) == '1') {
            j--;
        }
        if (j < 0) {
            return "no greater number";
        }

        // Swap the RT1 with the rightmost zero to the right of i
        char[] bnumArray = bnum.toCharArray();
        char temp = bnumArray[i];
        bnumArray[i] = bnumArray[j];
        bnumArray[j] = temp;

        // Sort the substring to the right of j in ascending order
        Arrays.sort(bnumArray, j + 1, l);

        // Required next greater number
        return new String(bnumArray);
    }

    // Driver program to test above
    public static void main(String[] args) {
        String bnum = "10010";
        System.out.println("Binary representation of next greater number = " +
                nextGreaterWithSameDigits(bnum));
    }
}

C#

using System;

namespace NextGreaterNumberWithSameDigits
{
    class GFG
    {
        // Function to find the next greater number
        // with same number of 1's and 0's
        static string NextGreaterWithSameDigits(string bnum)
        {
            int l = bnum.Length;
            int i = l - 1;

            // Find the rightmost non-trailing one
            while (i >= 0 && bnum[i] == '0')
            {
                i--;
            }
            if (i < 0)
            {
                return "no greater number";
            }

            // Find the rightmost zero to the right of i
            int j = i - 1;
            while (j >= 0 && bnum[j] == '1')
            {
                j--;
            }
            if (j < 0)
            {
                return "no greater number";
            }

            // Swap the RT1 with the rightmost zero to the right of i
            char[] bnumArray = bnum.ToCharArray();
            char temp = bnumArray[i];
            bnumArray[i] = bnumArray[j];
            bnumArray[j] = temp;

            // Sort the substring to the right of j in ascending order
            Array.Sort(bnumArray, j + 1, l - j - 1);

            // Required next greater number
            return new string(bnumArray);
        }

        // Driver program to test above
        static void Main(string[] args)
        {
            string bnum = "10010";
            Console.WriteLine("Binary representation of next greater number = " + NextGreaterWithSameDigits(bnum));
        }
    }
}

Javascript

function nextGreaterWithSameDigits(bnum) {
    const l = bnum.length;
    let i = l - 1;

    // Find the rightmost non-trailing one
    while (i >= 0 && bnum[i] === '0') {
        i--;
    }
    if (i < 0) {
        return "no greater number";
    }

    // Find the rightmost zero to the right of i
    let j = i - 1;
    while (j >= 0 && bnum[j] === '1') {
        j--;
    }
    if (j < 0) {
        return "no greater number";
    }

    // Convert string to array for swapping
    bnum = bnum.split('');

    // Swap the RT1 with the rightmost zero to the right of i
    [bnum[i], bnum[j]] = [bnum[j], bnum[i]];

    // Sort the substring to the right of j in ascending order
    const sortedSubstring = bnum.slice(j + 1).sort().join('');

    // Required next greater number
    return bnum.slice(0, j + 1).join('') + sortedSubstring;
}

// Driver program to test above
function main() {
    const bnum = "10010";
    console.log("Binary representation of next greater number =", nextGreaterWithSameDigits(bnum));
}

main();

Output

Binary representation of next greater number = 10100

Time Complexity : O(n + m log m), where n is the length of the input string and m is the length of the substring to the right of the swapped characters.
Auxiliary Space : O(n)