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Find i’th index character in a binary string obtained after n iterations in C++
Suppose we have a binary string bin. Then apply n iterations on it, and in each iteration 0 becomes 01 and 1 becomes 10, after that ith index character in the string after nth iteration. So if the binary string is 101, and n = 2, and i = 3, so after first iteration it will be 100110, in the next iteration, it will be 100101101001, so ith index is holding 1.
To solve this, we have to follow these steps −
- Run loop n times, and in each iteration run another loop on the string
- Convert each character of binary string, and if that is 0, then store 01 or if that is 1, then store 10 into another temporary string
- After completion of inner loop, store temporary string to binary string.
- Then return ith index.
Example
#include<iostream>
using namespace std;
char getCharacter(string bin_str, int n, int i) {
string temp = "";
for (int x = 0; x < n; x++) {
for (int y = 0; y < bin_str.length(); y++) {
if (bin_str[y] == '1')
temp += "10";
else
temp += "01";
}
bin_str = temp;
temp = "";
}
return bin_str[i];
}
int main() {
int n = 2;
string bin = "101";
cout << 3 << "rd character is: "<< getCharacter(bin, n, 3)<<endl;
cout << 9 << "th character is: "<< getCharacter(bin, n, 9);
}Output
3rd character is: 1 9th character is: 0
Updated on: 17-Dec-2019
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