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Check if given words are present in a string

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  • Difficulty Level : Medium
  • Last Updated : 04 Jun, 2022
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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;
}

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

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: 

CPP




// 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;
}

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: 

CPP




// 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;
}

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)

Python3 approach using find() method

Python3




# python code to check whether a word is present in a string or not
bigString = "this is a big string"
smallStrings = ["this", "yo", "is", "a", "bigger", "string", "kappa"]
x = []
# find() - returns position of specified value or else returns -1
for i in smallStrings:
  if(bigString.find(i) != -1):
    x.append(True)
  else:
    x.append(False)
print(x)

Output

[True, False, True, True, False, True, False]

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