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# Reverse a String using a Stack

Last Updated on December 14, 2022 by Prepbytes

### Problem Statement

You are given a string, and your task is to reverse the string using a stack data structure.

Input: String.
Output: Reversed string.

### Test cases:

Input: “Prepbytes”
Output: “setybpreP”

### Explanation:

Reading the input string “Prepbytes” from right to left we get “setybpreP”.
Approach – Using Stack
The idea is to traverse the given string from left to right and push each character in the stack. Then, pop the characters from the stack one by one and append them to a new string. The newly created will be the reverse of the input string because of the LIFO (Last-in-first-out) property of the stack data structure. The last character pushed into the stack will be the first to get popped.

### Algorithm

1. Initialize an empty stack and an array.
2. Traverse the given string from left to right
• Push the current character to the stack
3. Initialize an empty StringBuilder
4. Pop from the stack one by one.
• Append the current character at last.
5. Return the newly created StringBuilder as a string.

### Code Implementation:

```
import java.util.*;

public class Prepbytes
{
public static void main(String[] args) {
System.out.println(reverseString("Prepbytes"));
}

public static String reverseString(String str){

// Initialize an empty stack.
Stack<Character> st = new Stack<>();

// Traverse the given string from left to right
for(char ch : str.toCharArray()){

// Push the current character to the stack.
st.push(ch);
}

// Initialize an empty StringBuilder
StringBuilder reverseStr = new StringBuilder();

// Pop from the stack one by one.
while(!st.isEmpty()){

// Append the current character at the last.
reverseStr.append(st.pop());
}

return reverseStr.toString();
}
}
```

Output:
setybperP

Time complexity: O(n), where n is the length of the input string. Each character gets pushed and popped from the stack at most once. And it takes constant time to append a character at the end of a StringBuilder.

Space Complexity: O(n). We are using StringBuilder to store the reversed string and it will have n characters.

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