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.
- Initialize a pointer to beginning for matching the occurrences in the pattern with 0 and counter to 0.
- Check if pattern and text have same character at the present index.
- 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.
- If not, keep incrementing the pointer by 1 if characters are same.
- 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).
- 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.
- If any added String to the Stack matches the pattern increment counter by 1 and initialize pointer by 0.
- Repeat all these steps for all the indexes of the text length.
- Print the counter and occurrences.
- 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 <iostream>#include <vector>#include <stack>using namespace std;// custom class for returning multiple valuesclass 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 textResult solution(string pattern, string text) { vector<int> positions; int counter = 0; int lastOccurrence = -10; stack<string> 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<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 + " "); } }} |
Python3
# Python program for the above approachfrom typing import List, Tuplefrom collections import dequeclass 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, counterif __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 valuesclass 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 valuesclass Result { constructor(positions, counter) { this.positions = positions; this.counter = counter; }}// function for finding the occurrences of a pattern in a textfunction 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).
References:
1. Stack Java Documentation.
2. Custom ArrayList and Class in Java.
3. More on Stack Operations.
This article is contributed by Shubham Saxena. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
