Given a decimal number m, convert it into a binary string and apply n iterations. In each iteration, 0 becomes “01” and 1 becomes “10”. Find the (based on indexing) index character in the string after the nth iteration.
Examples:
Input : m = 5, n = 2, i = 3
Output : 1
Input : m = 3, n = 3, i = 6
Output : 1
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
- Change a decimal number into a binary and store it in string s.
- Run loop n times in each iteration. Run another loop of string length s to convert 0 to “01” and 1 to “10” and store in another string s1. After completion of each iteration, assign string s1 to s.
- Finally, return the value of the ith index in string s.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void binary_conversion(string &s, int m) {
while (m) {
int tmp = m % 2;
s += tmp + '0';
m = m / 2;
}
reverse(s.begin(), s.end());
}
int find_character(int n, int m, int i) {
string s;
binary_conversion(s, m);
string s1 = "";
for (int x = 0; x < n; x++) {
for (int y = 0; y < s.length(); y++) {
if (s[y] == '1')
s1 += "10";
else
s1 += "01";
}
s = s1;
s1 = "";
}
return s[i] - '0';
}
int main() {
int m = 5, n = 2, i = 8;
cout << find_character(n, m, i);
return 0;
}
|
Java
import java.io.*;
import java.util.Arrays;
class GFG
{
static String s = "";
static String ReverseString(String s)
{
char[] arr = s.toCharArray();
for(int i = 0;
i < arr.length / 2; i++)
{
char temp = arr[i];
arr[i] = arr[arr.length - i -1];
arr[arr.length - i - 1] = temp;
}
return new String(arr);
}
static void binary_conversion(int m)
{
while (m != 0)
{
int tmp = m % 2;
s += Integer.toString(tmp);
m = (int)(m / 2);
}
s = ReverseString(s);
}
static int find_character(int n,
int m,
int i)
{
binary_conversion(m);
String s1 = "";
for (int x = 0; x < n; x++)
{
for (int y = 0;
y < s.length(); y++)
{
if (s.charAt(y) == '1')
s1 += "10";
else
s1 += "01";
}
s = s1;
s1 = "";
}
return s.charAt(i) - '0';
}
public static void main(String args[])
{
int m = 5, n = 2, i = 8;
System.out.print(
find_character(n, m, i));
}
}
|
Python3
def binary_conversion(s, m):
while(m):
temp = m % 2
s += str(temp)
m = m // 2
return s[::-1]
def find_character(n, m, i):
s = ""
s = binary_conversion(s, m)
s1 = ""
for x in range(n):
for j in range(len(s)):
if s[j] == "1":
s1 += "10"
else:
s1 += "01"
s = s1
s1 = ""
e = ord(s[i])
r = ord('0')
return e-r
m, n, i = 5, 2, 8
print(find_character(n,m,i))
|
C#
using System;
class GFG
{
static string ReverseString(string s)
{
char[] arr = s.ToCharArray();
Array.Reverse(arr);
return new string(arr);
}
static void binary_conversion(ref string s,
int m)
{
while (m != 0)
{
int tmp = m % 2;
s += tmp.ToString();
m = (int)(m / 2);
}
s = ReverseString(s);
}
static int find_character(int n,
int m, int i)
{
string s = "";
binary_conversion(ref s, m);
string s1 = "";
for (int x = 0; x < n; x++)
{
for (int y = 0; y < s.Length; y++)
{
if (s[y] == '1')
s1 += "10";
else
s1 += "01";
}
s = s1;
s1 = "";
}
return s[i] - '0';
}
static void Main()
{
int m = 5, n = 2, i = 8;
Console.Write(find_character(n, m, i));
}
}
|
Javascript
<script>
let s = "";
function ReverseString(s)
{
let arr = s.split("");
for(let i = 0;
i < arr.length / 2; i++)
{
let temp = arr[i];
arr[i] = arr[arr.length - i -1];
arr[arr.length - i - 1] = temp;
}
return arr.join("");
}
function binary_conversion(m)
{
while (m != 0)
{
let tmp = m % 2;
s += tmp.toString();
m = Math.floor(m / 2);
}
s = ReverseString(s);
}
function find_character(n,m,i)
{
binary_conversion(m);
let s1 = "";
for (let x = 0; x < n; x++)
{
for (let y = 0;
y < s.length; y++)
{
if (s[y] == '1')
s1 += "10";
else
s1 += "01";
}
s = s1;
s1 = "";
}
return s[i] - '0';
}
let m = 5, n = 2, i = 8;
document.write(find_character(n, m, i));
</script>
|
Time complexity : O(n^2)
Auxiliary Space : O(n)
Refer Set-2 for an optimized solution.
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