Pattern Occurrences : Stack Implementation Java

Suppose we have two Strings :- Pattern and Text 
pattern: consisting of unique characters 
text: consisting of any length
We need to find the number of patterns that can be obtained from text removing each and every occurrence of Pattern in the Text.
Example: 

Input : 
Pattern : ABC
Text : ABABCABCC
Output :
3
Occurrences found at: 
4 7 8
Explanation
Occurrences and their removal in the order
1. ABABCABCC
2. ABABCC
3. ABC

The idea is to use stack data structure. 
1. Initialize a pointer to beginning for matching the occurrences in the pattern with 0 and counter to 0. 
2. Check if pattern and text have same character at the present index. 
3. If the pointer is to the end of pattern that means all the previous characters have been found in an increasing subsequential order increment the counter by 1. 
4. If not, keep incrementing the pointer by 1 if characters are same. 
5. If the characters are different in both the strings, check if the character is same as the first character of the pattern (i.e. pointer = 0). 
6. If yes, add the remaining characters from the present pointer to length of the pattern to a stack and check if they are present in order that the pattern can be formed from the stack. Also, initialize the pointer now to 1 because we already had checked for pointer = 0 (in step 5). 
7. If matches, empty the stack to null. Else, remove the first character and keep adding the rest of the substring for checking for further of the steps. 
8. If any added String to the Stack matches the pattern increment counter by 1 and initialize pointer by 0. 
9. Repeat all these steps for all the indexes of the text length. 
10. Print the counter and occurrences. 
11. Basic task of Stack is handling the pending operations that might be possible occurrences.
Example Explanation according to above algorithm: 

TEXT: ABABCABCC
PATTERN: ABC
pointer = 0
counter = 0
A B A B C A B C C
0 1 2 3 4 5 6 7 8

at index = 0
pointer = 0
stack = []

at index = 1
pointer = 1
stack = []

at index = 2
pointer = 0
stack = ['C']

at index = 3
pointer = 1
stack = ['C']

at index = 4
pointer = 2
counter += 1
pointer = 0 
stack = ['C']

same for index 5,6,7 according to above method

at index = 8
pop from Stack
counter += 1
clear Stack 

Code in Java for the algorithm: 
Prerequisite : Stack class in Java

Java

import java.util.ArrayList;
import java.util.Stack;

class StackImplementation
{
  // custom class for returning multiple values
  class Data
  {
    ArrayList<Integer> present;
    int count;

    public Data(ArrayList<Integer> present, int count)
    {
      this.present = present;
      this.count = count;
    }
  }
  public Data Solution(char pattern[], char text[])
  {
    // stores the indices for all occurrences
    ArrayList<Integer> list = new ArrayList<>();
    Stack<String>  stack = new Stack<>();

    // present index pointer searched for in
    // the entire array of string characters
    int p = 0;

    //count of all the number of occurrences
    int counter = 0 ;

    // any random number less than 0 to mark
    // the previous index where the occurrence
    // was found
    int lastOccurrence = -10;

    // traversing all the indexes of the text
    // searching for possible pattern
    for (int i = 0; i < text.length; i ++)
    {
      // if the present index and the pointer in
      // the pattern is at same character
      if(text[i] == pattern[p])
      {
        // and if that character is the end of
        // the pattern to be found
        if(text[i] == pattern[pattern.length - 1])
        {
          //index at which pattern is found
          list.add(i);

          // incrementing total occurrences by 1
          counter ++;

          // last found index to be initialized
          // to present index
          lastOccurrence = i;

          // begin the search for the next pointer
          // again from 0th index of the pattern
          p = 0;
        }
        else
        {
          // if present character at pattern and index
          // is same but still not the end of pattern
          p ++;
        }
      }

      // if characters are not same
      else
      {
        // if the present character is same as the 1st
        // character of the pattern
        // here 0 = pointer in the pattern fixed to 0
        if(text[i] == pattern[0])
        {
          // assume a temporary string
          String temp = "";

          // and add all characters to it to the pattern
          // length from the present pointer to the end
          for (int i1 = p; i1 < pattern.length; i1 ++)
            temp += pattern[i1];

          // push the present pattern length into the stack
          // for checking if pattern is same as subsequence
          // of the text
          stack.push(temp);

          //pattern at pointer = 0 already checked so we
          // start from 1 for the next step
          p = 1;
        }
        else
        {
          // if the previous occurrence was just before
          // the present index
          if (lastOccurrence == i - 1)
          {
            // if the stack is empty place the pointer = 0
            if (stack.isEmpty())
              p = 0;
            else
            {
              // pick up the present possible pattern
              String temp = stack.pop();

              // check if it's character has the matching
              // occurrence
              if (temp.charAt(0) == text[i])
              {
                //increment last index by the present index
                // so that net index is checked
                lastOccurrence = i;

                // check if stack character is last character
                // in the pattern
                if (temp.charAt(0) == pattern[pattern.length - 1])
                {
                  // index found
                  list.add(i);

                  // increment occurrences by 1
                  counter ++;
                }
                else
                {
                  // if present index character doesn't
                  // match the last character in the pattern
                  // remove the first character which was same
                  // and check for further occurrences of the
                  // remaining letters in the stack string
                  temp = temp.substring(1, temp.length());

                  // add the remaining string back to stack
                  // for further review
                  stack.push(temp);
                }
              }
              // if first string character in the stack doesn't
              // match the present character in the text
              else
              {
                // if stack is not empty empty it.
                if (!stack.isEmpty())
                  stack.clear();

                // reinitialize the pointer back to 0 for
                // checking pattern from beginning
                p = 0;
              }
            }
          }
          else
          {
            // empty the stack under any other circumstances
            if (!stack.isEmpty())
              stack.clear();

            // reinitialize the pointer back to 0 for
            // checking pattern from beginning
            p = 0;
          }
        }
      }
    }
    // return the result
    return new Data(list, counter);
  }

  public static void main(String args[])
  {
    // the simple pattern to be matched
    char[] pattern = "ABC".toCharArray();

    // the input string in which the number of
    // occurrences can be found out after removing
    // each occurrence.
    char[] text = "ABABCABCC".toCharArray();

    StackImplementation obj = new StackImplementation();
    Data data = obj.Solution(pattern, text);

    int count = data.count;
    ArrayList<Integer> list = data.present;
    System.out.println(count);
    if (count > 0)
    {
      System.out.println("Occurrences found at:");
      for (int i : list)
        System.out.print(i + " ");
    }
  }
}

Output: 

3
Occurrences found at:
4 7 8

References: 
1. Stack Java Documentation.
2. Custom ArrayList and Class in Java.
3. More on Stack Operations.
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