Remove duplicates from array

// simple java program to remove
// duplicates

class Main
{
// Function to remove duplicate elements
// This function returns new size of modified
// array.
static int removeDuplicates(int arr[], int n)
{
// Return, if array is empty
// or contains a single element
if (n==0 || n==1)
return n;

int[] temp = new int[n];

// Start traversing elements
int j = 0;
for (int i=0; i<n-1; i++)
// If current element is not equal
// to next element then store that
// current element
if (arr[i] != arr[i+1])
temp[j++] = arr[i];

// Store the last element as whether
// it is unique or repeated, it hasn't
// stored previously
temp[j++] = arr[n-1];

// Modify original array
for (int i=0; i<j; i++)
arr[i] = temp[i];

return j;
}

public static void main (String[] args)
{
int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5};
int n = arr.length;

n = removeDuplicates(arr, n);

// Print updated array
for (int i=0; i<n; i++)
System.out.print(arr[i]+" ");
}
}

Prime Number

A prime number is a whole number greater than 1 whose only factors are 1 and itself. A factor is a whole numbers that can be divided evenly into another number. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Numbersthat have more than two factors are called composite numbers.
int n = 15, f=0;

for (int i = 2; i < n; i++)
{
                if (n % i == 0)
                    f = 1;
}

            if (f == 0)
            {
                System.out.println( "Given Number is Prime Number");
            }else{
                System.out.println( "Given Number is Not Prime Number");
            }

Android Lint

Android Studio provides a code scanning tool called lint that can help you to identify and correct problems with the structural quality of your code without your having to execute the app or write test cases. Each problem detected by the tool is reported with a description message and a severity level, so that you can quickly prioritize the critical improvements that need to be made. Also, you can lower the severity level of a problem to ignore issues that are not relevant to your project, or raise the severity level to highlight specific problems.

The lint tool checks your Android project source files for potential bugs and optimization improvements for correctness, security, performance, usability, accessibility, and internationalization. When using Android Studio, configured lint and IDE inspections run whenever you build your app. However, you can manually run inspections or run lint from the command line.

Longest Palindrome Substring in a String

Longest Palindrome Substring in a String Algorithm

The key point here is that from the mid of any palindrome string if we go to the right and left by 1 place, it’s always the same character.

For example 12321, here mid is 3 and if we keep moving one position on both sides, we get 2 and then 1. We will use the same logic in our java program to find out the longest palindrome.

However, if the palindrome length is even, the mid-size is also even. So we need to make sure in our program that this is also checked. For example, 12333321, here mid is 33 and if we keep moving one position in both sides, we get 3, 2 and 1.

package com.journaldev.util;

public class LongestPalindromeFinder {

public static void main(String[] args) {

System.out.println(longestPalindromeString("1234"));

System.out.println(longestPalindromeString("12321"));

System.out.println(longestPalindromeString("9912321456"));

System.out.println(longestPalindromeString("9912333321456"));

System.out.println(longestPalindromeString("12145445499"));

System.out.println(longestPalindromeString("1223213"));

System.out.println(longestPalindromeString("abb"));

}

static public String intermediatePalindrome(String s, int left, int right) {

if (left > right) return null;

while (left >= 0 && right < s.length()

&& s.charAt(left) == s.charAt(right)) {

left--;

right++;

}

return s.substring(left + 1, right);

}

// O(n^2)

public static String longestPalindromeString(String s) {

if (s == null) return null;

String longest = s.substring(0, 1);

for (int i = 0; i < s.length() - 1; i++) {

//odd cases like 121

String palindrome = intermediatePalindrome(s, i, i);

if (palindrome.length() > longest.length()) {

longest = palindrome;

}

//even cases like 1221

palindrome = intermediatePalindrome(s, i, i + 1);

if (palindrome.length() > longest.length()) {

longest = palindrome;

}

}

return longest;

}

}