Check if given words are present in a string
Given a big string and an array of small strings, all of which are smaller in length than the big string. The task is to create an array of booleans, where each boolean represents whether the small string at that index in the array of small strings is contained in the big string.
Note : that you can’t use language-built-in string-matching methods.
Examples:
Input : bigString = “this is a big string”, smallStrings = [“this”, “yo”, “is”, “a”, “bigger”, “string”, “kappa”]
Output : [true, false, true, true, false, true, false]Input : bigString = “Mary goes to the shopping center every week.”, smallStrings = [“to”, “Mary”, “centers”, “shop”, “shopping”, “string”, “kappa”]
Output : [true, true, false, true, true, false, false]
Approach : Naive Approach
A simple way to solve this problem is to iterate through all of the small strings, checking if each of them is contained in the big string by iterating through the big string’s characters and comparing them to the given small string’s characters with a couple of loops.
Below is the implementation of the above approach:
C++
// Find the small string at that index in the array of // small strings is contained in the big string #include <bits/stdc++.h> using namespace std; bool isInBigString(string bigString, string smallString); bool isInBigStringHelper(string bigString, string smallString, int startIdx); // Function to the multiStringSearch vector<bool> multiStringSearch(string bigString, vector<string> smallStrings) { vector<bool> solution; // iterate in the smallString for (string smallString : smallStrings) { // calling the isInBigString Function solution.push_back(isInBigString(bigString, smallString)); } return solution; } // Function to the bigString bool isInBigString(string bigString, string smallString) { // iterate in the bigString for (int i = 0; i < bigString.length(); i++) { // Check if length of smallString + i is greater than // the length of bigString if (i + smallString.length() > bigString.length()) { break; } // call the function isInBigStringHelper if (isInBigStringHelper(bigString, smallString, i)) { return true; } } return false; } // Helper Function to the Finding bigString bool isInBigStringHelper(string bigString, string smallString, int startIdx) { // Initialize leftBigIdx and rightBigIdx and leftSmallIdx variables int leftBigIdx = startIdx; int rightBigIdx = startIdx + smallString.length() - 1; int leftSmallIdx = 0; int rightSmallIdx = smallString.length() - 1; // Iterate until leftBigIdx variable reaches less // than or equal to rightBigIdx while (leftBigIdx <= rightBigIdx) { // Check if bigString[leftBigIdx] is not equal // to smallString[leftSmallIdx] or Check if // bigString[rightBigIdx] is not equal to // smallString[rightSmallIdx] than return false // otherwise increment leftBigIdx and leftSmallIdx // decrement rightBigIdx and rightSmallIdx if (bigString[leftBigIdx] != smallString[leftSmallIdx] || bigString[rightBigIdx] != smallString[rightSmallIdx]) { return false; } leftBigIdx++; rightBigIdx--; leftSmallIdx++; rightSmallIdx--; } return true; } // Driver code int main(int argc, char* argv[]) { // initialize string string str = "this is a big string"; // initialize vector string vector<string> substr = { "this", "yo", "is", "a", "bigger", "string", "kappa" }; // Function call vector<bool> ans = multiStringSearch(str, substr); // Print answers for (int i = 0; i < ans.size(); i++) { // Check if ans[i] is equal to 1 // then Print true otherwise print false if (ans[i] == 1) { cout << "true" << " "; } else { cout << "false" << " "; } } return 0; } |
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Python3
# Find the small string at that index in the array of # small strings is contained in the big string # Helper Function to the Finding bigString def isInBigStringHelper(bigString,smallString,startIdx): # Initialize leftBigIdx and rightBigIdx and leftSmallIdx variables leftBigIdx = startIdx rightBigIdx = startIdx + len(smallString) - 1 leftSmallIdx = 0 rightSmallIdx = len(smallString) - 1 # Iterate until leftBigIdx variable reaches less # than or equal to rightBigIdx while (leftBigIdx <= rightBigIdx): # Check if bigString[leftBigIdx] is not equal # to smallString[leftSmallIdx] or Check if # bigString[rightBigIdx] is not equal to # smallString[rightSmallIdx] than return false # otherwise increment leftBigIdx and leftSmallIdx # decrement rightBigIdx and rightSmallIdx if (bigString[leftBigIdx] != smallString[leftSmallIdx] or bigString[rightBigIdx] != smallString[rightSmallIdx]): return False leftBigIdx += 1 rightBigIdx -= 1 leftSmallIdx += 1 rightSmallIdx -= 1 return True # Function to the bigString def isInBigString(bigString, smallString): # iterate in the bigString for i in range(len(bigString)): # Check if length of smallString + i is greater than # the length of bigString if (i + len(smallString) > len(bigString)): break # call the function isInBigStringHelper if (isInBigStringHelper(bigString, smallString, i)): return True return False # Function to the multiStringSearch def multiStringSearch(bigString, smallStrings): solution = [] # iterate in the smallString for smallString in smallStrings: # calling the isInBigString Function solution.append(isInBigString(bigString, smallString)) return solution # Driver code if __name__ == '__main__': # initialize string str1 = "this is a big string" # initialize vector string substr = ["this", "yo", "is", "a","bigger", "string", "kappa"] # Function call ans = multiStringSearch(str1, substr) # Print answers for i in range(len(ans)): # Check if ans[i] is equal to 1 # then Print true otherwise print false if (ans[i] == 1): print("true",end = " ") else: print("false",end = " ") # This code is contributed by Bhupendra_Singh |
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Output :
true false true true false true false
Time Complexity : O(b * n * s), where b is the length of the bigstring and n is the number of small strings and s is the length of longest small string.
Auxiliary Space : O(n)
Approach : Using Suffix Trie
Build a Suffix-trie data structure containing all of the big string’s suffixes. Then, iterate through all of the small strings and check if each of them is contained in the data structure you have created.
Below is the implementation of the above approach:
// Find the small string at that index in the array of // small strings is contained in the big string #include <bits/stdc++.h> using namespace std; // Blueprint of the TrieNode class TrieNode { public: // Declaring children to be of <char, TrieNode> type // key will be of char type and mapped value will // be of TrieNode type unordered_map<char, TrieNode*> children; }; // Blueprint of the ModifiedSuffixTrie class ModifiedSuffixTrie { public: TrieNode* root; ModifiedSuffixTrie(string str) { this->root = new TrieNode(); // Function call this->populateModifiedSuffixTrieFrom(str); } void populateModifiedSuffixTrieFrom(string str) { // iterate in the length of String for (int i = 0; i < str.length(); i++) { // Function call this->insertSubstringStartingAt(i, str); } } void insertSubstringStartingAt(int i, string str) { TrieNode* node = this->root; // iterate in the length of String for (int j = i; j < str.length(); j++) { // initialize char as a letter // put the value of str[j] in letter char letter = str[j]; // Check if letter is is equal to endnode or not if (node->children.find(letter) == node->children.end()) { TrieNode* newNode = new TrieNode(); node->children.insert({ letter, newNode }); } node = node->children[letter]; } } bool contains(string str) { TrieNode* node = this->root; // iterate in the String for (char letter : str) { // Check if letter is is equal to endnode or not if (node->children.find(letter) == node->children.end()) { return false; } node = node->children[letter]; } return true; } }; // Function to the multiStringSearch vector<bool> multiStringSearch(string bigString, vector<string> smallStrings) { ModifiedSuffixTrie modifiedSuffixTrie(bigString); vector<bool> solution; // iterate in the smallString for (string smallString : smallStrings) { solution.push_back(modifiedSuffixTrie.contains(smallString)); } return solution; } // Driver code int main(int argc, char* argv[]) { // initialize string string str = "this is a big string"; // initialize vector string vector<string> substr = { "this", "yo", "is", "a", "bigger", "string", "kappa" }; // Function call vector<bool> ans = multiStringSearch(str, substr); // Print answers for (int i = 0; i < ans.size(); i++) { // Check if ans[i] is equal to 1 // then Print true otherwise print false if (ans[i] == 1) { cout << "true" << " "; } else { cout << "false" << " "; } } return 0; } |
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Output :
true false true true false true false
Time Complexity : O(b*b + n * s), where b is the length of the bigstring and n is the number of small strings and s is the length of longest small string.
Auxiliary Space : O(b*2 + n)
Approach : Using Trie
Try building a trie containing all of the small strings. Then, iterate through the big string’s characters and check if any part of the big string is a string contained in the trie you have created.
Below is the implementation of the above approach:
// Find the small string at that index in the array of // small strings is contained in the big string #include <bits/stdc++.h> using namespace std; // Blueprint of the TrieNode class TrieNode { public: // Declaring children to be of <char, TrieNode> type // key will be of char type and mapped value will // be of TrieNode type unordered_map<char, TrieNode*> children; string word; }; // Blueprint of the Trie class Trie { public: TrieNode* root; char endSymbol; Trie() { this->root = new TrieNode(); this->endSymbol = '*'; } // function to insert string void insert(string str) { TrieNode* current = this->root; // iterate in the length of String for (int i = 0; i < str.length(); i++) { // initialize char as a letter // put the value of str[i] in letter char letter = str[i]; // Check if letter is is equal to endnode or not if (current->children.find(letter) == current->children.end()) { TrieNode* newNode = new TrieNode(); current->children.insert({ letter, newNode }); } current = current->children[letter]; } current->children.insert({ this->endSymbol, NULL }); current->word = str; } }; // define a findSmallStringsIn function void findSmallStringsIn(string str, int startIdx, Trie* trie, unordered_map<string, bool>* containedStrings); // Function to the multiStringSearch vector<bool> multiStringSearch(string bigString, vector<string> smallStrings) { Trie* trie = new Trie(); // iterate in the smallString for (string smallString : smallStrings) { trie->insert(smallString); } // Declaring containedStrings to be of <string, bool> type // key will be of string type and mapped value will // be of boolean type unordered_map<string, bool> containedStrings; // iterate in the bigString for (int i = 0; i < bigString.length(); i++) { findSmallStringsIn(bigString, i, trie, &containedStrings); } vector<bool> solution; // iterate in the smallString for (string smallString : smallStrings) { solution.push_back(containedStrings.find(smallString) != containedStrings.end()); } return solution; } // Function to findSmallStringsIn void findSmallStringsIn(string str, int startIdx, Trie* trie, unordered_map<string, bool>* containedStrings) { TrieNode* currentNode = trie->root; // iterate the length of the string for (int i = startIdx; i < str.length(); i++) { // Check if letter is is equal to endnode or not if (currentNode->children.find(str[i]) == currentNode->children.end()) { break; } currentNode = currentNode->children[str[i]]; if (currentNode->children.find(trie->endSymbol) != currentNode->children.end()) { containedStrings->insert({ currentNode->word, true }); } } } // Driver code int main(int argc, char* argv[]) { // initialize string string str = "this is a big string"; // initialize vector string vector<string> substr = { "this", "yo", "is", "a", "bigger", "string", "kappa" }; // Function call vector<bool> ans = multiStringSearch(str, substr); // Print answers for (int i = 0; i < ans.size(); i++) { // Check if ans[i] is equal to 1 // then Print true otherwise print false if (ans[i] == 1) { cout << "true" << " "; } else { cout << "false" << " "; } } return 0; } |
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Output :
true false true true false true false
Time Complexity : O(n*s + b * s), where b is the length of the bigstring and n is the number of small strings and s is the length of longest small string.
Auxiliary Space : O(ns)
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