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# Kth character after replacing each character of String by its frequency exactly X times

• Difficulty Level : Easy
• Last Updated : 18 Jan, 2023

Given a string S consisting of N digits from [1, 9] and a positive integer K and X. Every day each character of the string is replaced by its frequency and value. The task is to find the Kth character of the string after X days.

Examples:

Input: S = “1214”, K = 10, X = 3
Output: 4
Explanation:
1st day = “12214444”
2nd day = “1222214444444444444444”
3rd day = “122222222444444444444444444444444444444444444444444444444”
So, 10th character after 3rd day is 4.

Input: S =”123″, K = 6, X = 2
Output: 3

Naive Approach: The simplest approach to solve the problem is to create the string by appending the digits in the string digitX times and finding the Kth character of the string.

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the Kth character` `// after X days` `char` `FindKthChar(string str, ``long` `long` `K, ``int` `X)` `{` `    ``string s = str;`   `    ``for` `(``int` `i = 0; i < X; i++) {` `        ``string res = ``""``;` `        ``for` `(``int` `j = 0; j < s.size(); j++) {` `            ``int` `count = s[j] - ``'0'``;` `            ``while` `(count--) {` `                ``res += s[j];` `            ``}` `        ``}` `        ``s = res;` `    ``}`   `    ``return` `s[K - 1];` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Input` `    ``string str = ``"1214"``;` `    ``long` `long` `K = 10;` `    ``int` `X = 3;`   `    ``// Function Call` `    ``char` `ans = FindKthChar(str, K, X);` `    ``cout << ans << ``"\n"``;` `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `class` `Gfg {` `    ``// Function to find the Kth character` `    ``// after X days` `    ``static` `char` `findKthChar(String str, ``long` `K, ``int` `X)` `    ``{` `        ``String s = str;`   `        ``for` `(``int` `i = ``0``; i < X; i++) {` `            ``String res = ``""``;` `            ``for` `(``int` `j = ``0``; j < s.length(); j++) {` `                ``int` `count = s.charAt(j) - ``'0'``;` `                ``while` `(count-- > ``0``) {` `                    ``res += s.charAt(j);` `                ``}` `            ``}` `            ``s = res;` `        ``}`   `        ``return` `s.charAt((``int``)(K - ``1``));` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// Given Input` `        ``String str = ``"1214"``;` `        ``long` `K = ``10``;` `        ``int` `X = ``3``;`   `        ``// Function Call` `        ``char` `ans = findKthChar(str, K, X);` `        ``System.out.println(ans);` `    ``}` `}`

## C#

 `using` `System;`   `class` `Gfg` `{`   `  ``// Function to find the Kth character` `  ``// after X days` `  ``static` `char` `findKthChar(``string` `str, ``long` `K, ``int` `X)` `  ``{` `    ``string` `s = str;`   `    ``for` `(``int` `i = 0; i < X; i++) {` `      ``string` `res = ``""``;` `      ``for` `(``int` `j = 0; j < s.Length; j++) {` `        ``int` `count = s[j] - ``'0'``;` `        ``while` `(count-- > 0) {` `          ``res += s[j];` `        ``}` `      ``}` `      ``s = res;` `    ``}`   `    ``return` `s[(``int``)(K - 1)];` `  ``}`   `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``// Given Input` `    ``string` `str = ``"1214"``;` `    ``long` `K = 10;` `    ``int` `X = 3;`   `    ``// Function Call` `    ``char` `ans = findKthChar(str, K, X);` `    ``Console.WriteLine(ans);` `  ``}` `}`   `// This code is contributed by divya_p123.`

Output

`1`

Time Complexity: O(∑digit[i]X) for i in range [0, N-1]
Auxiliary Space: O(∑digit[i]X) for i in range [0, N-1]

Efficient Approach: The above approach can be optimized further by instead of creating a string add digitX to the sum and check if sum > K. Follow the steps below to solve the problem:

• Initialize a variable, say sum that stores the sum of the character of the string after X days.
• Initialize a variable, say ans that stores the Kth characters after X days.
• Iterate in the range [0, N-1] and perform the following steps:
• Initialize a variable say range as (S[i] – ‘0’)X and add it to the variable sum.
• If sum>=K, return S, modify ans as S[i], and terminate the loop.
• Print the value of ans as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the Kth character` `// after X days` `char` `FindKthChar(string str, ``long` `long` `K, ``int` `X)` `{`   `    ``// Variable to store the KthChar` `    ``char` `ans;` `    ``int` `sum = 0;`   `    ``// Traverse the string` `    ``for` `(``int` `i = 0; i < str.length(); i++) {`   `        ``// Convert char into int` `        ``int` `digit = str[i] - ``'0'``;`   `        ``// Calculate characters` `        ``int` `range = ``pow``(digit, X);` `        ``sum += range;`   `        ``// If K is less than sum` `        ``// than ans = str[i]` `        ``if` `(K <= sum) {` `            ``ans = str[i];` `            ``break``;` `        ``}` `    ``}` `    ``// Return answer` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Input` `    ``string str = ``"123"``;` `    ``long` `long` `K = 9;` `    ``int` `X = 3;`   `    ``// Function Call` `    ``char` `ans = FindKthChar(str, K, X);` `    ``cout << ans << ``"\n"``;` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `class` `GFG{`   `// Function to find the Kth character` `// after X days` `static` `char` `FindKthChar(String str, ``int` `K, ``int` `X)` `{` `    `  `    ``// Variable to store the KthChar` `    ``char` `ans = ``' '``;` `    ``int` `sum = ``0``;`   `    ``// Traverse the string` `    ``for``(``int` `i = ``0``; i < str.length(); i++) ` `    ``{` `        `  `        ``// Convert char into int` `        ``int` `digit = (``int``)str.charAt(i) - ``48``;`   `        ``// Calculate characters` `        ``int` `range = (``int``)Math.pow(digit, X);` `        ``sum += range;`   `        ``// If K is less than sum` `        ``// than ans = str[i]` `        ``if` `(K <= sum) ` `        ``{` `            ``ans = str.charAt(i);` `            ``break``;` `        ``}` `    ``}`   `    ``// Return answer` `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// Given Input` `    ``String str = ``"123"``;` `    ``int` `K = ``9``;` `    ``int` `X = ``3``;`   `    ``// Function Call` `    ``char` `ans = FindKthChar(str, K, X);` `    ``System.out.println(ans);` `}` `}`   `// This code is contributed by abhinavjain194`

## Python3

 `# Python3 program for the above approach` `import` `math`   `# Function to find the Kth character` `# after X days` `def` `FindKthChar(``Str``, K, X):` `     `  `    ``# Variable to store the KthChar` `    ``ans ``=` `' '` `    ``Sum` `=` `0` ` `  `    ``# Traverse the string` `    ``for` `i ``in` `range``(``len``(``Str``)):` `      `  `        ``# Convert char into int` `        ``digit ``=` `ord``(``Str``[i]) ``-` `48` ` `  `        ``# Calculate characters` `        ``Range` `=` `int``(math.``pow``(digit, X))` `        ``Sum` `+``=` `Range` ` `  `        ``# If K is less than sum` `        ``# than ans = str[i]` `        ``if` `(K <``=` `Sum``):` `            ``ans ``=` `Str``[i]` `            ``break` `     `  `    ``# Return answer` `    ``return` `ans`   `# Given Input` `Str` `=` `"123"` `K ``=` `9` `X ``=` `3`   `# Function Call` `ans ``=` `FindKthChar(``Str``, K, X)` `print``(ans)`   `# This code is contributed by divyeshrabadiya07.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to find the Kth character` `// after X days` `static` `char` `FindKthChar(``string` `str, ``int` `K, ``int` `X)` `{` `    `  `    ``// Variable to store the KthChar` `    ``char` `ans = ``' '``;` `    ``int` `sum = 0;`   `    ``// Traverse the string` `    ``for``(``int` `i = 0; i < str.Length; i++) ` `    ``{` `        `  `        ``// Convert char into int` `        ``int` `digit = (``int``)str[i] - 48;`   `        ``// Calculate characters` `        ``int` `range = (``int``)Math.Pow(digit, X);` `        ``sum += range;`   `        ``// If K is less than sum` `        ``// than ans = str[i]` `        ``if` `(K <= sum) ` `        ``{` `            ``ans = str[i];` `            ``break``;` `        ``}` `    ``}` `    `  `    ``// Return answer` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    `  `    ``// Given Input` `    ``string` `str = ``"123"``;` `    ``int` `K = 9;` `    ``int` `X = 3;`   `    ``// Function Call` `    ``char` `ans = FindKthChar(str, K, X);` `    ``Console.Write(ans);` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR`

## Javascript

 ``

Output

`2`

Time Complexity: O(N), since there is only one loop to carry out the operations the overall complexity turns out to be O(N)
Auxiliary Space: O(1), since there is no extra array or data structure used, it takes constant space.

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