# Find Array formed by reversing Prefix each time given character is found

• Difficulty Level : Expert
• Last Updated : 27 Oct, 2022

Given an array arr[] of length N consisting of uppercase English letters only and a letter ch. the task is to find the final array that will form by reversing the prefix each time the letter ch is found in the array.

Examples:

Input: arr[] = {‘A’, ‘B’, ‘X’, ‘C’, ‘D’, ‘X’, ‘F’}, ch= ‘X’
Output: D C X A B X F
Explanation
First encounter of ‘X’ at index 2, Initial subarray = A, B, Final subarray = B, A, X.
Second encounter of ‘X’ at index 5, Initial subarray = B, A, X, C, D
Final subarray = D, C, X, A, B, X(added).
Final subarray after traversing, = D, C, X, A, B, X, F

Input: arr[] = {‘A’, ‘B’, ‘C’, ‘D’, ‘E’}, ch = ‘F’
Output: A B C D E

Approach: The idea to solve the problem is as follows:

If each portion between two occurrences of ch (or the ends of the array) is considered a segment, then the prefix reversals of the string can be visualised as appending the characters of a segment alternatively at the starting and the ending of string and keep expanding outwards.

• So idea is to push the element of the array into back of a list till ch occurs for first time.
• When first ch occurs, push the elements, including ch, to the front of the list till the next ch occurs. Again if ch occurs push the elements to the back of the list, including ch
• So, the conclusion is that each time ch occurs, you have to change the direction of pushing the elements.

Note:  If there is odd number of K in the array, you need to reverse the list as we start pushing element from back.

•  Create a list named li to store the elements
•  Create a variable named found to check which side you have to add the elements from
•  If ch occurs flip the value of found from 1 to 0 or 0 to 1.
• If found is equal to 1 add the elements to the front
• Else add the elements to the back
•  If ch occurs odd number of times reverse the list and print, else print it simply

Below is the implementation of the above approach:

## C++

 // C++ code to implement the above approach   #include using namespace std;   // Final string after performing all the // operations   list findFinalString(char* arr, int n, char ch) {     list li;     bool found = 0;       for (int i = 0; i < n; i++) {           // ch Found         if (arr[i] == ch)             found = !found;           // Push character at front of list         if (found)             li.push_front(arr[i]);           // Push character at back of list         else             li.push_back(arr[i]);     }       // If there is odd number of ch     if (found == 1)         li.reverse();       // Return the list     return li; }   // Driver Code int main() {     char arr[] = { 'A', 'B', 'X', 'C', 'D', 'X', 'F' };     char ch = 'X';     int N = sizeof(arr) / sizeof(arr[0]);       // Function call     list ans = findFinalString(arr, N, ch);     for (auto itr = ans.begin();          itr != ans.end(); itr++)         cout << *itr << " ";       return 0; }

## Java

 // Java code to implement the above approach import java.util.ArrayList; import java.util.Collections;   class GFG {     // Final string after performing all the   // operations   static ArrayList findFinalString(char[] arr, int n, char ch) {     ArrayList li = new ArrayList();     Boolean found = false;       for (int i = 0; i < n; i++) {         // ch Found       if (arr[i] == ch)         found = !found;         // Push character at front of list       if (found)         li.add(0, arr[i]);         // Push character at back of list       else         li.add(arr[i]);     }       // If there is odd number of ch     if (found == true)       Collections.reverse(li);       // Return the list     return li;   }     // Driver Code   public static void main(String args[]) {     char arr[] = { 'A', 'B', 'X', 'C', 'D', 'X', 'F' };     char ch = 'X';     int N = arr.length;       // Function call     ArrayList ans = findFinalString(arr, N, ch);     for (char itr : ans)       System.out.print(itr + " ");     } }   // This code is contributed by saurabh_jaiswal.

## Python3

 # Python code to implement the above approach   # Final string after performing all the # operations def findFinalString(arr, n, ch):     li = []     found = 0     for i in range(n):                 # ch found         if arr[i] == ch:             found = 1-found           # Push character at front of list         if found:             li.insert(0, arr[i])         # Push character at back of list         else:             li.append(arr[i])            #  If there is odd number of ch     if found == 1:         li = li[::-1]           # Return the list     return li   #  Driver Code if __name__ == "__main__":     arr = ['A', 'B', 'X', 'C', 'D', 'X', 'F']     ch = 'X'     N = len(arr)       # Function call     ans = findFinalString(arr, N, ch)     for val in ans:         print(val, end=" ")           # This Code is Contributed By Vivek Maddeshiya

## C#

 using System; using System.Collections.Generic;   // C# code to implement the above approach public class GFG {     // Final string after performing all the   // operations   public static List findFinalString(char[] arr,                                            int n, char ch)   {     List li = new List();     bool found = false;       for (int i = 0; i < n; i++) {         // ch Found       if (arr[i] == ch)         found = !found;         // Push character at front of list       if (found == true)         li.Insert(0, arr[i]);         // Push character at back of list       else         li.Add(arr[i]);     }       // If there is odd number of ch     if (found == true)       li.Reverse();       // Return the list     return li;   }     // Driver Code   static public void Main()   {       char[] arr = { 'A', 'B', 'X', 'C', 'D', 'X', 'F' };     char ch = 'X';     int N = arr.Length;       // Function call     List ans = new List();     ans = findFinalString(arr, N, ch);     for (int i = 0; i < ans.Count; i++) {       Console.Write(" ");       Console.Write(ans[i]);     }   } }   // This code is contributed by akashish__

## Javascript

 // JavaScript code to implement the above approach   // Final string after performing all the // operations function findFinalString(arr, n, ch) {   let li = [];   let found = 0;     for (let i = 0; i < n; i++) {     // ch Found     if (arr[i] == ch) found = !found;       // Push character at front of list     if (found) li.unshift(arr[i]);     // Push character at back of list     else li.push(arr[i]);   }     // If there is odd number of ch   if (found == 1) li.reverse();     // Return the list   return li; }   // Driver Code   let arr = ["A", "B", "X", "C", "D", "X", "F"]; let ch = "X"; let N = 7;   // Function call let ans = findFinalString(arr, N, ch); console.log(ans);   // This code is contributed by ishankhandelwals.

Output

D C X A B X F

Time Complexity: O(N)
Auxiliary Space: O(N)

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