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 Heren= (18)_{10}= (10010)_{2}next greater= (20)_{10}= (10100)_{2}Binary representation of20contains same number of 1's and 0's as in18. 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.

- Traverse the binary string
**bstr**from the right. - While traversing find the first index
**i**such that bstr[i] = ‘0’ and bstr[i+1] = ‘1’. - Exchange character of at index ‘i’ and ‘i+1’.
- 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 C++ implementation of above steps.

// 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; }

Output:

Binary representation of next greater number = 10100

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

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