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 Heren= (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++ 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; }

Output:

Binary representation of previous number = 10101

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

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