09. Find the N-th character

The problem can be found at the following link: Question Link

My Approach

• Initialize a variable len to store the length of the input string s.

• Use a loop to iterate r times for the specified number of iterations.

• In each iteration :

  • Create a temporary string temp and copy the content of the original string s.

  • Use another loop to traverse each character in the original string s.

  • Update each character based on the following logic :

    • If the corresponding character in temp at position j/2 is '0', set s[j] to '0' + (j % 2).

    • Otherwise, set s[j] to '1' - (j % 2).

• After completing all iterations, the desired character is at index n in the final string s.

Time and Auxiliary Space Complexity

• Time Complexity : O(r*|s|), as there are two nested loops.

• Auxiliary Space Complexity : O(|s|), as the temporary string temp is of the same length as the input string.

Code (C++)

class Solution{
  public:
    char nthCharacter(string s, int r, int n)
    {
        int len=s.length();
        for (int i=0;i<r;i++)
        {
            string temp=s;
            for (int j=0;j<len;j++)
            {
                if (temp[j/2]=='0')
                    s[j]='0'+(j%2);
                else s[j]='1'-(j%2);
            }
        }
        return s[n];
    }
};

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