Find Palindrome Strings in an Array Using Java

In this article, we will see how to find all palindrome strings from an array. This is a very frequently asked questions in interview.  Also there is a question to find all palindrome number from an array.  Find few more collection interview question.  

PalindromeStrings.java

package com.techbyteslearn.lab.basic;

/**
 * This program illustrates how to find palindrome strings from an array.
 */
public class PalindromeStrings {

    public static void main(String[] args) {
        String[] stringArray = {
            "eye",
            "jdg",
            "javadevelopersguide",
            "aabaa",
            "hello",
            "pip"
        };

        for (int i = 0; i < stringArray.length; i++) {
            printOnlyPalindrome(stringArray[i]);
        }
    }

    private static void printOnlyPalindrome(String str) {
        String oldString = str;
        StringBuilder builder = new StringBuilder(str);

        if (builder.reverse().toString().equals(oldString)) {
            System.out.println(oldString + " is a Palindrome String.");
        }
    }
}

Output:

eye is a Palindrome String.
aabaa is a Palindrome String.
pip is a Palindrome String.

Happy Learning.

Find All Palindrome Numbers from an Array in Java

In this article, we will see how to find all palindrome from an array. This is a very basic questions in interview, the interviewer will ask the same questions in different way. So, its good to know all possible questions from palindrome. Also there is a question to find all palindrome number from a list.  Find few more collection interview question.  Today we will see how to check a number is palindrome or not. 

FindAllPalindrome.java

package com.techbyteslearn.lab.basic;

public class FindAllPalindrome {
public static void main(String[] args) {
int numberArray[] = { 120, 990, 121, 777, 808, 1280 };
for (int i = 0; i < numberArray.length; i++) {
printOnlyPalindrom(numberArray[i]);
}
}
private static void printOnlyPalindrom(int number) {
int finalNumber = 0;
int oldNumber = number;
// Repeat the loop until the number became zero.
while (number != 0) {
// Get the First Digit (i.e. 1)
int firstDigit = number % 10;
// Get the Result number.
finalNumber = (finalNumber * 10) + firstDigit;
// Now get the remaining digits , after finding the first digit
number = number / 10;
}
// Now compare the finalNumber and oldNumber both are same or not.
if (finalNumber == oldNumber)
System.out.println(finalNumber + " is a Palindrome.");
}
}

Output -

121 is a Palindrome.
777 is a Palindrome.
808 is a Palindrome.

How to Check Whether a Number Is a Palindrome in Java

In this article, we will see how to check if a number is palindrome or not. This is a very basic questions in interview. But, you never know about what kind of question the interviewer will ask. So, better you prepare for every certain questions. Also there is a question to find all palindrome number from a list.  Find few more collection interview question.  Today we will see how to check a number is palindrome or not. 

Palindrome.java

package com.techbyteslearn.lab.basic;

public class Palindrome {

    public static void main(String[] args) {
        int number = 121;
        int temp = number;
        int finalNumber = 0;

        // Repeat the loop until the number becomes zero.
        while (number != 0) {

            // Get the last digit.
            int lastDigit = number % 10;

            // Build the reversed number.
            finalNumber = (finalNumber * 10) + lastDigit;

            // Remove the last digit from the number.
            number = number / 10;
        }

        // Compare the reversed number with the original number.
        if (finalNumber == temp) {
            System.out.println("This number is a Palindrome.");
        } else {
            System.out.println("This number is not a Palindrome.");
        }
    }
}

Output:

This number is a Palindrome.

One important correction from the original comments: % 10 gets the last digit, not the first digit. Also, number has already become 0 by the time of the final comparison, so comparing with temp is the correct approach.

Happy Learning.

Find Duplicate Values in a List Using Java

In this article, we will see how to find the duplicate values from an array or list using java.  This is one of important programming questions in technical interview. Each interviewer has different approach to access the candidate. But, the logic and the approach by candidate is really matter. In this program we have used Map and List both, so its a kind of collections interview questions. You can find few more collection interview question.  Today we will see how to find the duplicate values from array. 

The logic is very simple here, see the below.

• At first we need we need to create a Map to hold the key-value pair. Where key is the array element and value is the counter for number of time the array element repeats.
• Then we will iterate the array and put into the map as per the above step. If the map contains the element earlier, then we will update the value +1.
• Finally we will have the map , which holds the array elements with the counter for repentance. 
• Now, we will iterate the Map , by checking the condition where the counter is more than 1 (i.e. its duplicated or repeated).

DuplicateFinder.java

package com.techbyteslearn.lab.basic;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

public class DuplicateFinder {

    public static void main(String[] args) {

        ArrayList<Integer> list = new ArrayList<>(
                Arrays.asList(4, 3, 5, 25, 25, 25, 13, 5, 22, 4, 90));

        System.out.println("Input List Data = " + list);

        Map<Integer, Integer> map = new HashMap<Integer, Integer>();

        for (int i = 0; i < list.size(); i++) {

            if (map.isEmpty()) {
                map.put(list.get(i), 1);
            } else if (map.containsKey(list.get(i))) {
                map.put(list.get(i), map.get(list.get(i)) + 1);
            } else {
                map.put(list.get(i), 1);
            }
        }

        System.out.println("\nDuplicate values are: ");

        // Iterate the Map and display the duplicate values.
        for (Entry<Integer, Integer> entry : map.entrySet()) {

            if (entry.getValue() > 1) {
                System.out.println(entry.getKey());

                // TODO: We can now put these values into any list.
            }
        }
    }
}

Output:

Input List Data = [4, 3, 5, 25, 25, 25, 13, 5, 22, 4, 90]

Duplicate values are:
4
5
25

Note- One small point: because HashMap does not guarantee iteration order, the order of 4, 5, and 25 in the output can vary.

Happy Learning.

Find the Largest Number in an Array Using Java

In this article, we will see how to find the largest number from an array using java.  This is one of basic questions in technical interviews. Earlier post we had seen how to find the smallest element from array. Now we will see how to find the largest number from integer array using java. 

The logic is very simple here, see the below.

• At first we need to assume any element as largest value. Example - 0th location.
• Then iterate over the array and compare with each element , whether its larger than the assumed larger value or not. If array element is larger then assign the array element value to assumed variable. Repeat the entire until end. 

FindLargestNumberInArray.java

package com.techbyteslearn.lab.basic;

public class FindLargestNumberInArray {

    // Find the largest value from an array.
    public static void main(String[] args) {
        int[] arr = {200, 3, 4, 24, 33, 24, 22, 55, 90, 103, 150};

        // Assume the largest value is at the 0th index.
        int largest = arr[0];

        for (int i = 0; i < arr.length; i++) {
            if (arr[i] >= largest) {
                largest = arr[i];
            }
        }

        System.out.println("Largest Number is ::" + largest);
    }
}

Output:

Largest Number is ::200

The original i < arr.length - 1 skips the last element. Using i < arr.length checks the complete array.

Using Java 8

int largest = IntStream.of(arr).boxed().max(Comparator.naturalOrder()).get().intValue() ;

 Happy Learning.

Find the Smallest Number in an Array Using Java

In this article, we will see how to find the smallest number from an array using java.  This is one of basic questions in technical interviews. Earlier post we had seen how to use stream for finding the missing number. Now we will see how to find the smallest number from integer array using java. 


The logic is very simple here, see the below.

• At first we need to assume the first smallest element.
• Then iterate over the array and compare with each element , whether its smaller than the assumed value or not. If array element is smaller then assign the array element value to assumed variable. Repeat the entire until end. 

FindSmallestNumberInArray.java

package com.techbyteslearn.lab.basic;

public class FindSmallestNumberInArray {

    // Find the smallest value from an array.
    public static void main(String[] args) {
        int[] arr = {200, 3, 4, 24, 33, 24, 22, 55, 90, 103, 150};

        // Assign the 0th index as the first smallest number.
        int smallest = arr[0];

        for (int i = 0; i < arr.length; i++) {
            if (arr[i] <= smallest) {
                smallest = arr[i];
            }
        }

        System.out.println("Smallest Element is - " + smallest);
    }
}

Output:

Smallest Element is - 3

The original i < arr.length - 1 skips the last element. Using i < arr.length is the correct condition.

Using Java 8

IntStream.of(arr).boxed().min(Comparator.naturalOrder()).get().intValue();

Find One Missing Number from a List Using Java

In this article, we will see how to find the missing number from a list using java.  This is one of important common interview question asked in interview. You can see, how to find all missing numbers from a list. In this program we will use core java or the traditional way using for loop for finding the miss number from a list. Earlier post we had seen how to use stream for finding the missing number. Now we will see how to find one missing number using traditional core java style. 


The logic is very simple here, see the below.

• At first we need to find the MAX number from the list. We need this MAX number because , we need to calculate the SUM of all natural number up to that max number. 
• Then , we need to calculate the sum of all those natural number.
• Then we will subtract each element from the given list from sumOfNaturalNumbers.
• Now, the at the last  the value inside sumOfNaturalNumbers is the missing number.

FindOneMissingNumber.java

package com.techbyteslearn.lab.basic;

import java.util.ArrayList;
import java.util.Arrays;

public class FindOneMissingNumber {

    public static void main(String[] args) {
        ArrayList<Integer> numberList = new ArrayList<>(
                Arrays.asList(10, 3, 2, 4, 5, 6, 7, 9, 8, 14, 1, 11, 13));

        int sumOfNaturalNumbers = getSumUptoMax(findMax(numberList));

        for (int i = 0; i < numberList.size(); i++) {
            /*
             * Subtract each element from the list from sumOfNaturalNumbers.
             * The final value will be the missing number.
             */
            sumOfNaturalNumbers = sumOfNaturalNumbers - numberList.get(i);
        }

        int missingNumber = sumOfNaturalNumbers;
        System.out.println("Missing Number is :: " + missingNumber);
    }

    // Find the sum of all natural numbers up to limitNumber.
    private static int getSumUptoMax(int limitNumber) {
        int sum = 0;

        for (int i = 1; i <= limitNumber; i++) {
            sum = sum + i;
        }

        return sum;
    }

    // Find the greatest value from the list.
    private static int findMax(ArrayList<Integer> numberList) {
        int largest = numberList.get(0);

        for (int i = 1; i < numberList.size(); i++) {
            if (numberList.get(i) > largest) {
                largest = numberList.get(i);
            }
        }

        return largest;
    }
}

Output:

Missing Number is :: 12

 

This approach assumes the list contains the numbers from 1 through the maximum value, with exactly one number missing.

Happy Learning.