Android Device Monitor

Android Device Monitor is a stand-alone tool that provides a graphical user interface for several Android application debugging and analysis tools. The Monitor tool does not require installation of a integrated development environment, such as Eclipse, and encapsulates the following tools:

  1. DDMS
  2. Tracer for OpenGL ES
  3. Hierarchy Viewer
  4. Traceview
  5. Pixel Perfect magnification viewer
Android ships with a debugging tool called the Dalvik Debug Monitor Server (DDMS), which provides port-forwarding services, screen capture on the device, thread and heap information on the device, logcat, process, and radio state information, incoming call and SMS spoofing, location data spoofing, and more.

Traceview is a graphical viewer for execution logs saved by your application. Traceview can help you debug your application and profile its performance.

Hierarchy Viewer allows you to debug and optimize your user interface. It provides a visual representation of the layout's View hierarchy (the Layout View) and a magnified inspector of the display (the Pixel Perfect View).

Traceview
Traceview is a graphical viewer for execution logs saved by your application. Traceview can help you debug your application and profile its performance.

Pixel Perfect magnification viewer

For Android Studio Tools Click Here

Demo Code

/******************************************************************************

*******************************************************************************/

public class Main
{
    void palindrome(String w)
{
    StringBuilder s2 = new StringBuilder(w);
        s2.reverse();
     
        String s = s2.toString();
        System.out.print("The String is "+s+" and is");
       
        if(w.equals(s))
          System.out.println(" a Palindrome");
         else
           System.out.println(" not a Palindrome");
}

int fibonacci (int i){
 
   if(i==0)
        return 0;
   else if(i==1 || i==2)
       return 1;
 
    return fibonacci(i-1)+fibonacci(i-2);
 
}

int factorial (int factorial_number){
 
    if(factorial_number==0)
        return 1;
    else
        return factorial_number*factorial(factorial_number-1);
}

void countFrequency(String str, char ch, int count){
 
    int occ = 0, i;

  // If given count is 0
  // print the given string and return
  if (count == 0) {
   System.out.println(str);
   return;
  }

  // Start traversing the string
  for (i = 0; i < str.length(); i++) {

   // Increment occ if current char is equal
   // to given character
   if (str.charAt(i) == ch)
    occ++;

   // Break the loop if given character has
   // been occurred given no. of times
   if (occ == count)
    break;
  }

  // Print the string after the occurrence
  // of given character given no. of times
  if (i < str.length() - 1)
   System.out.println(str.substring(i + 1));

  // Otherwise string is empty
  else
   System.out.println("Empty string");
 
}

public static void main(String[] args) {
 
    Main demo = new Main();
    demo.palindrome("atata");
 
    int number=5;
    System.out.println("Entered Number is "+number);
    System.out.println("The "+number+"th fibonacci number is: "+demo.fibonacci(number));
 
    System.out.print("Fibonacci Series: ");
    for (int i=0;i<number;i++)
            System.out.print(demo.fibonacci(i)+", ");
     
System.out.println("\nFactorial of "+number+" is "+demo.factorial(number));
 
        //Find Duplicate Items in Array List
    int [] arr = new int [] {1,2,3,4,3,5,6,6,8,9};
    System.out.print("Duplicate elements in given array: ");
    for(int i=0;i<arr.length;i++)
    {
        for(int j=i+1;j<arr.length;j++)
        {
            if(arr[i]==arr[j])
            System.out.print(arr[j]+" ");
        }
    }

    System.out.println(" ");
    demo.countFrequency("geeksforgeeks",'e',3);
 
}
}

Print the string after the specified character has occurred given no. of times

Given a string, a character, and a count, the task is to print the string after the specified character has occurred count number of times.Print “Empty string” in case of any unsatisfying conditions.(Given character is not present, or present but less than given count, or given count completes on last index). If given count is 0, then given character doesn’t matter, just print the whole string.

// Java program for above implementation 

public class GFG 
// Method to print the string 
static void printString(String str, char ch, int count) 
int occ = 0, i; 
// If given count is 0 
// print the given string and return 
if (count == 0) { 
System.out.println(str); 
return; 
// Start traversing the string 
for (i = 0; i < str.length(); i++) { 
// Increment occ if current char is equal 
// to given character 
if (str.charAt(i) == ch) 
occ++; 
// Break the loop if given character has 
// been occurred given no. of times 
if (occ == count) 
break; 
// Print the string after the occurrence 
// of given character given no. of times 
if (i < str.length() - 1) 
System.out.println(str.substring(i + 1)); 
// Otherwise string is empty 
else
System.out.println("Empty string"); 
// Driver Method 
public static void main(String[] args) 
String str = "geeks for geeks"; 
printString(str, 'e', 2); 


Examples:

Input  :  str = "This is demo string" 
          char = i,    
          count = 3
Output :  ng

Input :  str = "geeksforgeeks"
         char = e, 
         count = 2
Output : ksforgeeks

Count frequency of characters in a string

Use a java Map and map a char to an int. You can then iterate over the characters in the string and check if they have been added to the map, if they have, you can then increment its value.

HashMap<Character, Integer> map = new HashMap<Character, Integer>();
String s = "aasjjikkk"; 

for (int i = 0; i < s.length(); i++)
 {
        char c = s.charAt(i);
        Integer val = map.get(c);
       
        if (val != null) { 
                                  map.put(c, new Integer(val + 1));
                                } else { 
                                              map.put(c, 1);
                                            }
                                }
}

Reverse a String

There are many ways of reversing a String in Java for whatever reason you may have. Today, we will look at a few simple ways of reversing a String in Java.

Method 1:
import java.util.Scanner;

public class ReverseString
{
public static void main(String[] args)
{
System.out.println("Enter string to reverse:");

Scanner read = new Scanner(System.in);
String str = read.nextLine();
String reverse = "";

for(int i = str.length() - 1; i >= 0; i--)
{
reverse = reverse + str.charAt(i);
}

System.out.println("Reversed string is:");
System.out.println(reverse);
}
}

Method 2:
import java.util.Scanner;

public class ReverseString
{
public static void main(String[] args)
{
System.out.println("Enter string to reverse:");

Scanner read = new Scanner(System.in);
String str = read.nextLine();

StringBuilder sb = new StringBuilder();

for(int i = str.length() - 1; i >= 0; i--)
{
sb.append(str.charAt(i));
}

System.out.println("Reversed string is:");
System.out.println(sb.toString());
}
}

Method 3:
import java.util.Scanner;

public class ReverseString
{
public static void main(String[] args)
{
System.out.println("Enter string to reverse:");

Scanner read = new Scanner(System.in);
String str = read.nextLine();

StringBuilder sb = new StringBuilder(str);

System.out.println("Reversed string is:");
System.out.println(sb.reverse().toString());
}
}

Print the duplicate elements of an array

In this program, we need to print the duplicate elements present in the array. This can be done through two loops. The first loop will select an element and the second loop will iteration through the array by comparing the selected element with other elements. If a match is found, print the duplicate element.



public class DuplicateElement {
public static void main(String[] args) {

//Initialize example array
int [] arr = new int [] {1, 2, 3, 4, 2, 7, 8, 8, 3};

System.out.println("Duplicate elements in given array: ");

//Searches for duplicate element
for(int i = 0; i < arr.length; i++) {
    for(int j = i + 1; j < arr.length; j++) {
                            if(arr[i] == arr[j])
                             System.out.println(arr[j]);
}}}}

OR

int[] array = {1,1,2,3,4,5,6,7,8,8};

Set<Integer> set = new HashSet<Integer>();

for(int i = 0; i < array.length ; i++)
{
//If same integer is already present then add method will return FALSE
if(set.add(array[i]) == false)
{
          System.out.println("Duplicate element found : " + array[i]);
}

}