# Pattern Occurrences : Stack Implementation Java

Last Updated : 03 Apr, 2023

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. 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

Implementation:

## C++

 `// C++ program for the above approach` `#include ` `#include ` `#include ` `using` `namespace` `std;`   `// custom class for returning multiple values` `class` `Result {` `  ``public``:` `    ``vector<``int``> positions;` `    ``int` `counter;`   `    ``Result(vector<``int``> positions, ``int` `counter) {` `      ``this``->positions = positions;` `      ``this``->counter = counter;` `    ``}` `};`   `// function for finding the occurrences of a pattern in a text` `Result solution(string pattern, string text) {` `  ``vector<``int``> positions;` `  ``int` `counter = 0;` `  ``int` `lastOccurrence = -10;` `  ``stack s;`   `  ``// present index pointer searched for in` `  ``// the entire array of string characters` `  ``int` `p = 0;`   `  ``for` `(``int` `i = 0; i < text.length(); i++) {` `    ``char` `c = text[i];`   `    ``if` `(c == pattern[p]) {` `      ``if` `(c == pattern[pattern.length() - 1]) {` `        ``positions.push_back(i);` `        ``counter++;` `        ``lastOccurrence = i;` `        ``p = 0;` `      ``} ``else` `{` `        ``p++;` `      ``}` `    ``} ``else` `{` `      ``if` `(c == pattern[0]) {` `        ``string temp = pattern.substr(p);` `        ``s.push(temp);` `        ``p = 1;` `      ``} ``else` `if` `(lastOccurrence == i - 1) {` `        ``if` `(!s.empty()) {` `          ``string temp = s.top();` `          ``s.pop();` `          ``if` `(temp[0] == c) {` `            ``lastOccurrence = i;` `            ``if` `(temp[temp.length() - 1] == pattern[pattern.length() - 1]) {` `              ``positions.push_back(i);` `              ``counter++;` `            ``} ``else` `{` `              ``s.push(temp.substr(1));` `            ``}` `          ``} ``else` `{` `            ``s.push(temp);` `          ``}` `        ``} ``else` `{` `          ``p = 0;` `        ``}` `      ``} ``else` `{` `        ``while` `(!s.empty()) { ` `          ``s.pop(); ` `        ``}` `        `  `        ``p = 0;` `      ``}` `    ``}` `    `  `   ``}`   `   ``return` `Result(positions, counter);` `}`   `int` `main() {`   `   ``// the simple pattern to be matched` `   ``string pattern = ``"ABC"``;`   `   ``// the input string in which the number of` `   ``// occurrences can be found out after removing` `   ``// each occurrence.` `   ``string text = ``"ABABCABCC"``;` `   `  `   ``Result result = solution(pattern, text);` `   `  `   ``cout << result.counter << endl;`   `     ``if` `(result.counter > 0) { ` `       ``cout << ``"Occurrences found at:"` `<< endl; ` `       ``for``(``int` `position : result.positions){ ` `         ``cout << position << ``" "``; ` `       ``} ` `     ``}`   `}`

## Java

 `import` `java.util.ArrayList;` `import` `java.util.Stack;`   `public` `class` `StackImplementation {` `    ``// custom class for returning multiple values` `    ``class` `Data {` `        ``ArrayList present;` `        ``int` `count;`   `        ``public` `Data(ArrayList present, ``int` `count)` `        ``{` `            ``this``.present = present;` `            ``this``.count = count;` `        ``}` `    ``}` `    ``public` `Data Solution(``char` `pattern[], ``char` `text[])` `    ``{` `        ``// stores the indices for all occurrences` `        ``ArrayList list = ``new` `ArrayList<>();` `        ``Stack 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 list = data.present;` `        ``System.out.println(count);` `        ``if` `(count > ``0``) {` `            ``System.out.println(``"Occurrences found at:"``);` `            ``for` `(``int` `i : list)` `                ``System.out.print(i + ``" "``);` `        ``}` `    ``}` `}`

## Python3

 `# Python program for the above approach`   `from` `typing ``import` `List``, ``Tuple` `from` `collections ``import` `deque`   `class` `StackImplementation:` `    ``# custom class for returning multiple values` `    ``def` `solution(``self``, pattern: ``List``[``str``], text: ``List``[``str``]) ``-``> ``Tuple``[``List``[``int``], ``int``]:` `        ``positions ``=` `[]` `        ``counter ``=` `0` `        ``last_occurrence ``=` `-``10` `        ``stack ``=` `deque()` `        ``# present index pointer searched for in` `        ``# the entire array of string characters` `        ``p ``=` `0` `        ``for` `i, c ``in` `enumerate``(text):` `            ``if` `c ``=``=` `pattern[p]:` `                ``if` `c ``=``=` `pattern[``-``1``]:` `                    ``positions.append(i)` `                    ``counter ``+``=` `1` `                    ``last_occurrence ``=` `i` `                    ``p ``=` `0` `                ``else``:` `                    ``p ``+``=` `1` `            ``else``:` `                ``if` `c ``=``=` `pattern[``0``]:` `                    ``temp ``=` `''.join(pattern[p:])` `                    ``stack.append(temp)` `                    ``p ``=` `1` `                ``elif` `last_occurrence ``=``=` `i ``-` `1``:` `                    ``if` `stack:` `                        ``temp ``=` `stack.pop()` `                        ``if` `temp[``0``] ``=``=` `c:` `                            ``last_occurrence ``=` `i` `                            ``if` `temp[``-``1``] ``=``=` `pattern[``-``1``]:` `                                ``positions.append(i)` `                                ``counter ``+``=` `1` `                            ``else``:` `                                ``stack.append(temp[``1``:])` `                    ``else``:` `                        ``# reinitialize the pointer` `                        ``# back to 0 for checking` `                        ``# pattern from beginning` `                        ``p ``=` `0` `                ``else``:` `                    ``# empty the stack under any other` `                    ``# circumstances` `                    ``stack.clear()` `                    ``# reinitialize the pointer back to` `                    ``# 0 for checking pattern from` `                    ``# beginning` `                    ``p ``=` `0` `        ``# return the result` `        ``return` `positions, counter`   `if` `__name__ ``=``=` `"__main__"``:` `    ``# the simple pattern to be matched` `    ``pattern ``=` `list``(``"ABC"``)` `    `  `    ``# the input string in which the number of` `    ``# occurrences can be found out after removing` `    ``# each occurrence.` `    ``text ``=` `list``(``"ABABCABCC"``)` `    ``obj ``=` `StackImplementation()` `    ``positions, counter ``=` `obj.solution(pattern, text)` `    ``print``(counter)` `    ``if` `counter > ``0``:` `        ``print``(``"Occurrences found at:"``)` `        ``print``(``*``positions, sep``=``" "``)` `        `  `        `  `# This code is contributed by sdeadityasharma`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `// custom class for returning multiple values` `class` `Result` `{` `    ``public` `List<``int``> positions;` `    ``public` `int` `counter;`   `    ``public` `Result(List<``int``> positions, ``int` `counter)` `    ``{` `        ``this``.positions = positions;` `        ``this``.counter = counter;` `    ``}` `}`   `class` `Program` `{` `    ``// function for finding the occurrences of a pattern in a text` `    ``static` `Result solution(``string` `pattern, ``string` `text)` `    ``{` `        ``List<``int``> positions = ``new` `List<``int``>();` `        ``int` `counter = 0;` `        ``int` `lastOccurrence = -10;` `        ``Stack<``string``> s = ``new` `Stack<``string``>();`   `        ``// present index pointer searched for in` `        ``// the entire array of string characters` `        ``int` `p = 0;`   `        ``for` `(``int` `i = 0; i < text.Length; i++)` `        ``{` `            ``char` `c = text[i];`   `            ``if` `(c == pattern[p])` `            ``{` `                ``if` `(c == pattern[pattern.Length - 1])` `                ``{` `                    ``positions.Add(i);` `                    ``counter++;` `                    ``lastOccurrence = i;` `                    ``p = 0;` `                ``}` `                ``else` `                ``{` `                    ``p++;` `                ``}` `            ``}` `            ``else` `            ``{` `                ``if` `(c == pattern[0])` `                ``{` `                    ``string` `temp = pattern.Substring(p);` `                    ``s.Push(temp);` `                    ``p = 1;` `                ``}` `                ``else` `if` `(lastOccurrence == i - 1)` `                ``{` `                    ``if` `(s.Count > 0)` `                    ``{` `                        ``string` `temp = s.Pop();` `                        ``if` `(temp[0] == c)` `                        ``{` `                            ``lastOccurrence = i;` `                            ``if` `(temp[temp.Length - 1] == pattern[pattern.Length - 1])` `                            ``{` `                                ``positions.Add(i);` `                                ``counter++;` `                            ``}` `                            ``else` `                            ``{` `                                ``s.Push(temp.Substring(1));` `                            ``}` `                        ``}` `                        ``else` `                        ``{` `                            ``s.Push(temp);` `                        ``}` `                    ``}` `                    ``else` `                    ``{` `                        ``p = 0;` `                    ``}` `                ``}` `                ``else` `                ``{` `                    ``while` `(s.Count > 0)` `                    ``{` `                        ``s.Pop();` `                    ``}`   `                    ``p = 0;` `                ``}` `            ``}` `        ``}`   `        ``return` `new` `Result(positions, counter);` `    ``}`   `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``// the simple pattern to be matched` `        ``string` `pattern = ``"ABC"``;`   `        ``// the input string in which the number of` `        ``// occurrences can be found out after removing` `        ``// each occurrence.` `        ``string` `text = ``"ABABCABCC"``;`   `        ``Result result = solution(pattern, text);`   `        ``Console.WriteLine(result.counter);`   `        ``if` `(result.counter > 0)` `        ``{` `            ``Console.WriteLine(``"Occurrences found at:"``);` `            ``foreach` `(``int` `position ``in` `result.positions)` `            ``{` `                ``Console.Write(position + ``" "``);` `            ``}` `        ``}` `    ``}` `}`

## Javascript

 `// custom class for returning multiple values` `class Result {` `    ``constructor(positions, counter) {` `        ``this``.positions = positions;` `        ``this``.counter = counter;` `    ``}` `}`   `// function for finding the occurrences of a pattern in a text` `function` `solution(pattern, text) {` `    ``let positions = [];` `    ``let counter = 0;` `    ``let lastOccurrence = -10;` `    ``let s = [];` `    ``let p = 0;` `    ``// present index pointer searched for in` `    ``// the entire array of string characters`   `    ``for` `(let i = 0; i < text.length; i++) {` `        ``let c = text[i];`   `        ``if` `(c == pattern[p]) {` `            ``if` `(c == pattern[pattern.length - 1]) {` `                ``positions.push(i);` `                ``counter++;` `                ``lastOccurrence = i;` `                ``p = 0;` `            ``} ``else` `{` `                ``p++;` `            ``}` `        ``} ``else` `{` `            ``if` `(c == pattern[0]) {` `                ``let temp = pattern.substr(p);` `                ``s.push(temp);` `                ``p = 1;` `            ``} ``else` `if` `(lastOccurrence == i - 1) {` `                ``if` `(s.length != 0) {` `                    ``let temp = s.pop();` `                    ``if` `(temp[0] == c) {` `                        ``lastOccurrence = i;` `                        ``if` `(temp[temp.length - 1] == pattern[pattern.length - 1]) {` `                            ``positions.push(i);` `                            ``counter++;` `                        ``} ``else` `{` `                            ``s.push(temp.substr(1));` `                        ``}` `                    ``} ``else` `{` `                        ``s.push(temp);` `                    ``}` `                ``} ``else` `{` `                    ``p = 0;` `                ``}` `            ``} ``else` `{` `                ``while` `(s.length != 0) { ` `                    ``s.pop(); ` `                ``}`   `                ``p = 0;` `            ``}` `        ``}` `    ``}`   `    ``return` `new` `Result(positions, counter);` `}`   `let pattern = ``"ABC"``;` `let text = ``"ABABCABCC"``;` `// the input string in which the number of` `// occurrences can be found out after removing` `// each occurrence.` `let result = solution(pattern, text);`   `console.log(result.counter);`   `if` `(result.counter > 0) { ` `    ``console.log(``"Occurrences found at:"``); ` `    ``for``(let position of result.positions){ ` `        ``console.log(position); ` `    ``} ` `}`

Output

```3
Occurrences found at:
4 7 8 ```

Time Complexity: O(N*M),  N is the length of text and M is length of pattern.
Auxiliary Space: O(N).