Binary representation of previous number

• Difficulty Level : Easy
• Last Updated : 20 May, 2021

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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++

 // C++ implementation to find the binary// representation of previous number#include using namespace std; // function to find the required// binary representationstring 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 aboveint main(){    string num = "10110";    cout << "Binary representation of previous number = "         << previousNumber(num);    return 0;}

Java

 // Java implementation to find the binary// representation of previous numberclass 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 */

Python3

 # Python3 implementation to find the binary# representation of previous number # function to find the required# binary representationdef 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 codenum = "10110";print("Binary representation of previous number =",                              previousNumber(num));     # This code is contributed by mits

C#

 // C# implementation to find the binary// representation of previous numberusing 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

PHP

 = 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?>

Javascript



Output:

Binary representation of previous number = 10101

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