# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

```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

// incrementing total occurrences by 1
counter ++;

// last found index to be initizalized
// 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

// 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
```

This article is contributed by Shubham Saxena. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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