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.

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();