Given a string S containing letter and digit and an integer K where,
The new string S’ is formed from old string S by following steps:
1. If the character read is a letter, that letter is added at end of S’.
2. If the character read is a digit, then entire string S’ repeatedly written d-1 more times in total.
Note: The new string is guaranteed to have less than 2^63 letters.
Examples:
Input: S = “geeks2for2”, K = 15
Output: “e”
Explanation: The new string S’ = “geeksgeeksforgeeksgeeksfor”. The 15th letter is “e”.Input: S = “a2345”, K = 100
Output: “a”
Explanation: The new string S’=”a” repeated 120 times. The 100th letter is “a”.
Let us take a new string like S’ = “geeksgeeksgeeksgeeksgeeks” and an index K = 22, then the answer at K = 22 is the same if K = 2.
In general, when a string is equal to some word with size length repeated some number of times (such as geeks with size = 5 repeated 5 times), then the answer will be same for the index K as it is for the index K % size.
Using this insight and working backwards, we keep track of the size of the new string S’. Whenever the string S’ would equal some word repeated d times, we can reduce K to K % (lengthof(word)).
We first find the length of the new string S’. After, this we’ll work backwards, keeping track of size: the length of the new string after parsing symbols S[0], S[1], …, S[i].
If we see a digit S[i], it means the size of the new string after parsing S[0], S[1], …, S[i-1] will be (size / toInteger(S[i])). Otherwise, it will be size – 1.
Below is the implementation of above approach:
// CPP implementation of above approach#include <bits/stdc++.h>using namespace std;
// Function to return the K-th letter from new String.string K_thletter(string S, int K)
{ int N = S.size();
long size = 0;
// finding size = length of new string S'
for (int i = 0; i < N; ++i) {
if (isdigit(S[i]))
size = size * (S[i] - '0');
else
size += 1;
}
// get the K-th letter
for (int i = N - 1; i >= 0; --i) {
K %= size;
if (K == 0 && isalpha(S[i]))
return (string) "" + S[i];
if (isdigit(S[i]))
size = size / (S[i] - '0');
else
size -= 1;
}
return "";
}// Driver programint main()
{ string S = "geeks2for2";
int K = 15;
cout << K_thletter(S, K);
return 0;
}// This code is written by Sanjit_Prasad |
// Java implementation of above approach class GFG
{ // Function to return the K-th letter from new String.
static String K_thletter(String S, int K)
{
int N = S.length();
long size = 0;
// finding size = length of new string S'
for (int i = 0; i < N; ++i)
{
if (Character.isDigit(S.charAt(i)))
{
size = size * (S.charAt(i) - '0');
}
else
{
size += 1;
}
}
// get the K-th letter
for (int i = N - 1; i >= 0; --i)
{
K %= size;
if (K == 0 && Character.isAlphabetic(S.charAt(i)))
{
return (String) "" + S.charAt(i);
}
if (Character.isDigit(S.charAt(i)))
{
size = size / (S.charAt(i) - '0');
}
else
{
size -= 1;
}
}
return null;
}
// Driver program
public static void main(String[] args)
{
String S = "geeks2for2";
int K = 15;
System.out.println(K_thletter(S, K));
}
}// This code is contributed by Rajput-Ji |
# Python3 implementation of above approach # Function to return the K-th letter # from new String. def K_thletter(S, K):
N = len(S)
size = 0
# finding size = length of new string S'
for i in range(0, N):
if S[i].isdigit():
size = size * int(S[i])
else:
size += 1
# get the K-th letter
for i in range(N - 1, -1, -1):
K %= size
if K == 0 and S[i].isalpha():
return S[i]
if S[i].isdigit():
size = size // int(S[i])
else:
size -= 1
# Driver Codeif __name__ == "__main__":
S = "geeks2for2"
K = 15
print(K_thletter(S, K))
# This code is contributed # by Rituraj Jain |
// C# implementation of the above approachusing System;
class GFG
{ // Function to return the K-th letter from new String.
static String K_thletter(String S, int K)
{
int N = S.Length;
long size = 0;
// finding size = length of new string S'
for (int i = 0; i < N; ++i)
{
if (char.IsDigit(S[i]))
{
size = size * (S[i] - '0');
}
else
{
size += 1;
}
}
// get the K-th letter
for (int i = N - 1; i >= 0; --i)
{
K %= (int)size;
if (K == 0 && char.IsLetter(S[i]))
{
return (String) "" + S[i];
}
if (char.IsDigit(S[i]))
{
size = size / (S[i] - '0');
}
else
{
size -= 1;
}
}
return null;
}
// Driver code
public static void Main(String[] args)
{
String S = "geeks2for2";
int K = 15;
Console.WriteLine(K_thletter(S, K));
}
}// This code has been contributed by 29AjayKumar |
<script>// Javascript implementation of above approachfunction isAlphaNumeric(str) {
var code, i, len;
for (i = 0, len = str.length; i < len; i++) {
code = str.charCodeAt(i);
if (!(code > 47 && code < 58) && // numeric (0-9)
!(code > 64 && code < 91) && // upper alpha (A-Z)
!(code > 96 && code < 123)) { // lower alpha (a-z)
return false;
}
}
return true;
};// Function to return the K-th letter from new String.function K_thletter(S, K)
{ var N = S.length;
var size = 0;
// finding size = length of new string S'
for (var i = 0; i < N; ++i) {
if (Number.isInteger(parseInt(S[i])))
size = size * (S[i].charCodeAt(0) - '0'.charCodeAt(0));
else
size += 1;
}
// get the K-th letter
for (var i = N - 1; i >= 0; --i) {
K %= size;
if (K == 0 && isAlphaNumeric(S[i]))
return "" + S[i];
if (Number.isInteger(parseInt(S[i])))
size = size / (S[i].charCodeAt(0) - '0'.charCodeAt(0));
else
size -= 1;
}
}// Driver programvar S = "geeks2for2";
var K = 15;
document.write( K_thletter(S, K));// This code is contributed by imporrtantly.</script> |
Output:
e
Time Complexity: O(N), where N is the length of S.
Auxiliary Space: O(1) since constant space is being used