Java Programming Challenges

Palindrome Check

Problem Statement

You are given a string. Your task is to check whether the given string is palindrome or not. A string is said to be a palindrome if the reverse of the string is the same as the given string.

Input

string = 'LeveL'
string = 'coding'

 

Output

Yes
No

Approach

This problem can be solved by various methods, but we will focus on solving it using two pointer approach. Here, one pointer will point at the beginning of the string and will move forward. At the same time, another pointer will point at the end of the string and will move backwards. We will compare each character while iterating and likewise return the result.

Code

class Solution {
	//Function to check palindrome strings
	public static boolean IsPalindrome(String s, int n) {
		int start = 0, end = n-1;
	
		while(start<=end) {
			//Comparing end characters
			if(s.charAt(start) != s.charAt(end)) {
				return false;
			}
			start++;
			end--;
		}
	
		return true;
	}
	public static void main(String[] args) {
		String s= "LeveL";
		int n = s.length();
		
		if(IsPalindrome(s, n) == true) {
			System.out.println("'"+ s + "' is palindrome");
		} 
		else {
			System.out.println("'"+ s+ "' is not palindrome");
		}
	}
}

Output

Explanation

In the above example, we created a function IsPalindrome() to check whether a string is a palindrome or not. We also created two variables which point at the beginning and at the end of the string. After that, we iterated over the string and compared the characters at opposite ends. If the characters do not matches, we will return true. Otherwise, we will iterate till the start variable becomes greater than the end variable.

Time Complexity: O(n)

Space Complexity: O(1)

Leaders in an Array

Problem Statement

Given a sequence of numbers. Find all leaders in the sequence. An element is called a leader if it is strictly greater than all elements to its right side.

Input

arr[] = {23, 22, 24, 9, 10}

 

Output

10, 24

Approach

We will start iterating from the last position element. The last index element is itself a leader as it has no element to its right, so we will print it and iterate all the way till we reach the first index element. While iterating, we will keep a variable to store the maximum value so far and print and update it whenever we encounter a bigger element.

Code

class Solution {
	public static void PrintLeaders(int arr[], int N) {

		//Storing the maximum element so far
		int Max_So_Far = arr[N - 1];

		// Printing the last element
		System.out.print(Max_So_Far + " ");

		//Iterating and checking for the next max value
		for (int i = N - 2; i >= 0; i--) {
			if (Max_So_Far < arr[i]) { 
				Max_So_Far = arr[i];
				System.out.print(Max_So_Far + " ");
			}
		}
	}
	public static void main(String[] args) {

		//Creating the array
		int arr[] = new int[] {23, 22, 24, 9, 10};
		int n = arr.length;

		//Calling the function
		PrintLeaders(arr, n);
	}
}

Output

Explanation

In the above example, we called the PrintLeader() function to print all the leaders in the given array. We created a variable Max_So_Far to store the maximum element till now and iterated the array backwards, and simultaneously updated the maximum element. Once we encounter a new maximum element, we update the max value and print it.

Time Complexity: O(n)

Space Complexity: O(1)

Stock Buy and Sell

Problem Statement

You are given an array of integers, prices where prices[i] is the price of a given stock on an ith day. You want to maximise the profit by choosing a single day to buy one stock and a different day to sell that stock. Return the maximum profit you can get from this transaction.

Input

prices = [7, 1, 5, 4, 3, 6]

Output

5

Explanation

Here, we will purchase the stock on 2nd day when the price is 1 and sell it on the sixth day wherein the price is 6. Thus, profit = 6 - 1 = 5.

Approach

We will iterate and store the minimum cost at which we can buy the stock on that particular day. Whenever we encounter a smaller cost, we will update the minimum cost and find the maximum profit so far.

Code

public class Solution{

	//Function to calculate maximum profit
	public static int findMaxProfit(int[] prices) {
		
		//Initialising variables
		int minCost = Integer.MAX_VALUE;
		int profit = 0;

		for(int i = 0; i < prices.length; i++) {
			minCost = Math.min(minCost, prices[i]);
			profit = Math.max(profit, prices[i] - minCost);
		}
		return profit;
	}

	//Main starts here
	public static void main(String[] args) {

		System.out.print("Maximum profit is: ");
		//Prices array
		int[] prices = {7, 1, 5, 4, 3, 6};

		//Printing result
		System.out.println(findMaxProfit(prices));
	}
}

Output

Explanation

In the above example, we initialised a variable minCost with the maximum integer value to store the stock's minimum cost. After that, we iterated over the prices, and whenever we found any lower-price stock, we updated the variable minCost and calculated the profit so far. We returned the maximum profit possible after iterating over the complete array.

Time Complexity: O(n)

Space Complexity: O(1)

Assign Mice to Holes

Problem Statement

There are n mice and n holes in a straight line. Each hole can only accommodate only a single mouse. A mouse can stay at his position (x), move one step to the right (x+1), or move a step to the left(x-1). Any of these moves will take 1 minute. Your task is to assign mice to holes such that the time when the last mouse gets inside the hole is minimised.

Input

Position of mice = [3, -3, 2]
Position of holes = [3, 0, 4]

 

Output

3

 

Explanation

The answer would be 3 because:

Assign the mouse at x=-3 to hole at position x = 0: Time = 3

Assign the mouse at x = 2 to hole at position x= 4: Time = 2

Assign the mouse at x = 3 to hole at position x = 3: time = 0  

Thus, after 3 minutes, all the mice will be in holes.

Approach

We can solve this problem using a greedy strategy. We will sort the positions of mice and holes and put every mouse to its corresponding nearest hole to minimise the time. After that, we will find the maximum value of the absolute difference between the ith hole and mouse positions, which would be our final output.

Code

import java.util.*;
class Solution {
	//Function to calculate minimum time taken
	public static int MaxTimeTaken(ArrayList<Integer> mice, ArrayList<Integer> holes) {
		//Checking the size of the list
		if (mice.size( ) != holes.size( )) {
			return -1 ;
		}
          
		// Sort the positions 
		Collections.sort(mice);
		Collections.sort(holes);
 
		//finding the max difference between ith mice and hole 
		int max_time = 0;

		//Iterating over the list
		for (int i=0; i<mice.size(); i++) {
			if (max_time < Math.abs(mice.get(i)-holes.get(i))) {
				max_time = Math.abs(mice.get(i)-holes.get(i));
			}
 		}
		//Returning the max value of time
		return Math.abs(max_time);
	}

	//Main function start here
	public static void main(String[] args) {
	
		ArrayList<Integer> mice = new ArrayList<Integer>();
		mice.add(3);
		mice.add(-3);
		mice.add(2);

		ArrayList<Integer> holes= new ArrayList<Integer>();
		holes.add(3);
		holes.add(0);
		holes.add(4);
 
		System.out.println("The minimum time in which every mouse is in the hole is: "+ MaxTimeTaken(mice, holes));
	}
}

Output

Explanation

In the above example, we create and pass the argument to the function MaxTimeTaken, which would calculate the minimum time. We sorted both the list using the Collections.sort() method. After that, we iterated over both lists and calculated the maximum absolute difference between the corresponding hole position and mouse position. Once found, we returned the maximum value, which is the minimum time taken to place all the mice.

Time Complexity: O(nlog(n))

Space Complexity: O(1)

Try this code on Online Java Compiler.

Frequently Asked Questions

What are programming challenges?

The programming challenges are the coding problems given by the interviewer to test the problem-solving skills of the candidate.

How to solve a programming challenge?

To solve a challenge, you need to understand the problem first. Next, you need to come up with the best approach for the problem and then code it.

What is problem-solving?

Problem-solving is the process of understanding the problem, finding various approaches to solve it and choosing the best possible approach.

Is Java more hard compared to Python?

Yes, Java is more difficult to master than Python, as it uses explicit typing and more concise syntax.

Conclusion

This article discusses some most asked Java programming challenges. We discussed various coding questions with their best approach. We hope this blog has helped you enhance your knowledge of Java programming challenges. If you want to learn more, then check out our articles.

• Problems in Java
• Prime Number Program in Java
• Menu Driven Program in Java

 

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