How can I reverse a single String in Java 8 using Lambda and Streams? - Stack Overflow

import javax.swing.*;  
public class FirstSwingExample {  
public static void main(String[] args) {  
JFrame f=new JFrame();//creating instance of JFrame  
          
JButton b=new JButton("click");//creating instance of JButton  
b.setBounds(130,100,100, 40);//x axis, y axis, width, height  
          
f.add(b);//adding button in JFrame  
          
f.setSize(400,500);//400 width and 500 height  
f.setLayout(null);//using no layout managers  
f.setVisible(true);//making the frame visible  
}  
}  

List<String> songList = new ArrayList<>();
songList.add("Some song"); //Repeat until satisfied

System.out.println("\n\tWelcome! Please choose a song!");
String songChoice = scan.nextLine();
while (!songList.contains(songChoice)) {
    //Do stuff when input is not a recognised song
}

String[][] deepArray = new String[][] {{"John", "Mary"}, {"Alice", "Bob"}};
System.out.println(Arrays.toString(deepArray));
//output: [[Ljava.lang.String;@106d69c, [Ljava.lang.String;@52e922]
System.out.println(Arrays.deepToString(deepArray));

Map<String, String> map = ...
for (Map.Entry<String, String> entry : map.entrySet()) {
    System.out.println(entry.getKey() + "/" + entry.getValue());
}

import java.util.Stack;

/**
 * Java Program to implement a binary search tree. A binary search tree is a
 * sorted binary tree, where value of a node is greater than or equal to its
 * left the child and less than or equal to its right child.
 * 
 * @author WINDOWS 8
 *
 */
public class BST {

    private static class Node {
        private int data;
        private Node left, right;

        public Node(int value) {
            data = value;
            left = right = null;
        }
    }

    private Node root;

    public BST() {
        root = null;
    }

    public Node getRoot() {
        return root;
    }

    /**
     * Java function to check if binary tree is empty or not
     * Time Complexity of this solution is constant O(1) for
     * best, average and worst case. 
     * 
     * @return true if binary search tree is empty
     */
    public boolean isEmpty() {
        return null == root;
    }

    
    /**
     * Java function to return number of nodes in this binary search tree.
     * Time complexity of this method is O(n)
     * @return size of this binary search tree
     */
    public int size() {
        Node current = root;
        int size = 0;
        Stack<Node> stack = new Stack<Node>();
        while (!stack.isEmpty() || current != null) {
            if (current != null) {
                stack.push(current);
                current = current.left;
            } else {
                size++;
                current = stack.pop();
                current = current.right;
            }
        }
        return size;
    }


    /**
     * Java function to clear the binary search tree.
     * Time complexity of this method is O(1)
     */
    public void clear() {
        root = null;
    }

}

private boolean oneQueenPerRow() {
    int foundQueens;
    for (int i = 0; i < board.length; i++) {
        foundQueens = 0;//each loop is a checked row
        for (int j = 0; j < board.length; j++) {
            if (board[i][j] == QUEEN)
                foundQueens++;
        }
        if (foundQueens > 1) return false;
    }
    return true;
}

public <T> List<Class<? extends T>> findAllMatchingTypes(Class<T> toFind) {
    foundClasses = new ArrayList<Class<?>>();
    List<Class<? extends T>> returnedClasses = new ArrayList<Class<? extends T>>();
    this.toFind = toFind;
    walkClassPath();
    for (Class<?> clazz : foundClasses) {
        returnedClasses.add((Class<? extends T>) clazz);
    }
    return returnedClasses;
}

import java.io.*;
import java.util.*;

class GFG 
{
	static int maxCuts(int n, int a, int b, int c)
	{
		if(n == 0) return 0;
		if(n < 0)  return -1;

		int res = Math.max(maxCuts(n-a, a, b, c), 
		          Math.max(maxCuts(n-b, a, b, c), 
		          maxCuts(n-c, a, b, c)));

		if(res == -1)
			return -1;

		return res + 1; 
	}
  
    public static void main(String [] args) 
    {
    	int n = 5, a = 2, b = 1, c = 5;
    	
    	System.out.println(maxCuts(n, a, b, c));
    }
}

class Solution{
    //Function to count the frequency of all elements from 1 to N in the array.
    public static void frequencyCount(int arr[], int N, int P)
    {
        //Decreasing all elements by 1 so that the elements
        //become in range from 0 to n-1.
        int maxi = Math.max(P,N);
        int count[] = new int[maxi+1];
        Arrays.fill(count, 0);
        for(int i=0;i<N;i++){
            count[arr[i]]++; 
        }
        
        for(int i=0;i<N;i++){
            arr[i] = count[i+1];
        }
    }
}

<?xml version="1.0" encoding="utf-8"?>
<shape xmlns:android="http://schemas.android.com/apk/res/android">
    <gradient
        android:angle="45"
        android:startColor="#ff00ff"
        android:endColor="#000000"/>
</shape>

for(int i = 0; i < 108; i++) {
System.out.println('Jai Shri Ram!');
}

// IntendedActivity - the file you wish to open
// CurrentActivity - the file where this code is placed  

Intent intent = new Intent (this, IntendedActivity.class);
    CurrentActivity.this.startActivity(intent);

 public void sort(int array[]) { 
     for (int i = 1; i < array.length; i++) { 
          int key = array[i]; 
          int j = i - 1; 
          while (j >= 0 && array[j] > key) { 
              array[j + 1] = array[j]; 
              j = j - 1; 
          } 
          array[j + 1] = key; 
      } 
 }

public class LocalDatabase extends SQLiteOpenHelper {
         
    private static final String mDatabaseName = "LocalDatabase";
    private static final int mDatabaseVersion = 1;

public LocalDatabase(Context context) {
        super(context, mDatabaseName, null, mDatabaseVersion);
        SQLiteDatabase db = this.getWritableDatabase();
    }      

}

    AlarmManager alarmManager;