Binary representation of previous number

Given a binary input that represents binary representation of positive number n, find binary representation of n-1. It may be assumed that input binary number is greater than 0.

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

Examples:

Input : 10110
Output : 10101
Here n  = (22)10 = (10110)2
Previous number = (21)10 = (10101)2

Input : 11000011111000000
Output : 11000011110111111



We store input as string so that large numbers can be handled. We traverse the string from rightmost character and convert all 0’s to 1’s until we find a 1. Finally convert the found 1 to 0. The number so formed after this process is the required number. If input is “1”, then previous number will be “0”. If only the first character in the entire string is ‘1’, then we discard this character and change all the 0’s to 1’s.

C++

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// C++ implementation to find the binary
// representation of previous number
#include <bits/stdc++.h>
using namespace std;
  
// function to find the required
// binary representation
string previousNumber(string num)
{
    int n = num.size();
  
    // if the number is '1'
    if (num.compare("1") == 0)
        return "0";
      
    // examine bits from right to left
    int i;
    for (i = n - 1; i >= 0; i--) {
  
        // if '1' is encountered, convert
        // it to '0' and then break
        if (num.at(i) == '1') {
            num.at(i) = '0';
            break;
        }
  
        // else convert '0' to '1'
        else
            num.at(i) = '1';
    }
  
    // if only the 1st bit in the
    // binary representation was '1'
    if (i == 0)
        return num.substr(1, n - 1);
  
    // final binary representation
    // of the required number
    return num;
}
  
// Driver program to test above
int main()
{
    string num = "10110";
    cout << "Binary representation of previous number = "
         << previousNumber(num);
    return 0;
}

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Java

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// Java implementation to find the binary 
// representation of previous number 
class GFG 
{
  
    // function to find the required 
    // binary representation 
    static String previousNumber(String num) 
    {
        int n = num.length();
  
        // if the number is '1' 
        if (num.compareTo("1") == 0
        {
            return "0";
        }
  
        // examine bits from right to left 
        int i;
        for (i = n - 1; i >= 0; i--)
        {
  
            // if '1' is encountered, convert 
            // it to '0' and then break 
            if (num.charAt(i) == '1'
            {
                num = num.substring(0, i) + '0'
                            num.substring(i + 1);
                  
                // num.charAt(i) = '0'; 
                break;
            
              
            // else convert '0' to '1' 
            else 
            {
                num = num.substring(0, i) + '1'
                            num.substring(i + 1);
            }
            //num.at(i) = '1'; 
        }
  
        // if only the 1st bit in the 
        // binary representation was '1' 
        if (i == 0
        {
            return num.substring(1, n - 1);
        }
  
        // final binary representation 
        // of the required number 
        return num;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String num = "10110";
        System.out.print("Binary representation of previous number = "
                + previousNumber(num));
    }
}
  
/* This code contributed by PrinciRaj1992 */

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Python3

# Python3 implementation to find the binary
# representation of previous number

# function to find the required
# binary representation
def previousNumber(num1):
n = len(num1);
num = list(num1);

# if the number is ‘1’
if (num1 == “1”):
return “0”;
i = n – 1;

# examine bits from right to left
while (i >= 0):

# if ‘1’ is encountered, convert
# it to ‘0’ and then break
if (num[i] == ‘1’):
num[i] = ‘0’;
break;

# else convert ‘0’ to ‘1’
else:
num[i] = ‘1’;
i -= 1;

# if only the 1st bit in the
# binary representation was ‘1’
if (i == 0):
return num[1:n];

# final binary representation
# of the required number
return ” . join(num);

# Driver code
num = “10110”;
print(“Binary representation of previous number =”,
previousNumber(num));

# This code is contributed by mits

C#

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// C# implementation to find the binary 
// representation of previous number 
using System;
  
class GFG 
{
  
    // function to find the required 
    // binary representation 
    static String previousNumber(String num) 
    {
        int n = num.Length;
  
        // if the number is '1' 
        if (num.CompareTo("1") == 0) 
        {
            return "0";
        }
  
        // examine bits from right to left 
        int i;
        for (i = n - 1; i >= 0; i--)
        {
  
            // if '1' is encountered, convert 
            // it to '0' and then break 
            if (num[i] == '1'
            {
                num = num.Substring(0, i) + '0'
                            num.Substring(i + 1);
                  
                // num.charAt(i) = '0'; 
                break;
            
              
            // else convert '0' to '1' 
            else
            {
                num = num.Substring(0, i) + '1'
                            num.Substring(i + 1);
            }
            //num.at(i) = '1'; 
        }
  
        // if only the 1st bit in the 
        // binary representation was '1' 
        if (i == 0) 
        {
            return num.Substring(1, n - 1);
        }
  
        // final binary representation 
        // of the required number 
        return num;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        String num = "10110";
        Console.Write("Binary representation of previous number = "
                + previousNumber(num));
    }
}
  
// This code contributed by Rajput-Ji

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PHP

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<?php
// PHP implementation to find the binary
// representation of previous number
  
// function to find the required
// binary representation
function previousNumber($num)
{
    $n = strlen($num);
  
    // if the number is '1'
    if ($num == "1")
        return "0";
    $i = $n - 1;
      
    // examine bits from right to left
    for (; $i >= 0; $i--)
    {
  
        // if '1' is encountered, convert
        // it to '0' and then break
        if ($num[$i] == '1'
        {
            $num[$i] = '0';
            break;
        }
  
        // else convert '0' to '1'
        else
            $num[$i] = '1';
    }
  
    // if only the 1st bit in the
    // binary representation was '1'
    if ($i == 0)
        return substr($num,1, $n - 1);
  
    // final binary representation
    // of the required number
    return $num;
}
  
    // Driver code
    $num = "10110";
    echo "Binary representation of previous number = ".previousNumber($num);
      
// This code is contributed by mits
?>

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Output:

Binary representation of previous number = 10101

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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