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.
Implementation:
C++
// C++ implementation to find the binary// representation of previous number#include <bits/stdc++.h>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 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 */ |
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 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 |
PHP
<?php// PHP implementation to find the binary// representation of previous number// function to find the required// binary representationfunction 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?> |
Javascript
<script>// Javascript implementation to find the binary// representation of previous number// function to find the required// binary representationfunction previousNumber(num){ var n = num.length; // if the number is '1' if (num == "1") return "0"; // examine bits from right to left var i; for (i = n - 1; 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 num.substring(1, n); // final binary representation // of the required number return num.join('');}// Driver program to test abovevar num = "10110".split('');document.write( "Binary representation of previous number = " + previousNumber(num));// This code is contributed by rrrtnx.</script> |
Output
Binary representation of previous number = 10101
Time Complexity: O(n) where n is number of bits in input.
Auxiliary Space: O(n)
If you like neveropen and would like to contribute, you can also write an article using write.neveropen.co.uk or mail your article to review-team@neveropen.co.uk. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!