Given a string S and an array arr[] of words, the task is to return the number of words from the array which is a subsequence of S.
Examples:
Input: S = “programming”, arr[] = {“prom”, “amin”, “proj”}
Output: 2
Explanation: “prom” and “amin” are subsequence of S while “proj” is not)Input: S = “geeksforgeeks”, arr[] = {“gfg”, “geek”, “geekofgeeks”, “for”}
Output: 3
Explanation:” gfg”, “geek” and “for” are subsequences of S while “geekofgeeks” is not.
Naive Approach: The basic way to solve the problem is as follows:
The idea is to check all strings in the words array arr[] which are subsequences of S by recursion.
Below is the implementation of the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
bool isSubsequence(string& str1, int m, string& str2, int n)
{
if (m == 0)
return true;
if (n == 0)
return false;
// If last characters of two strings
// are matching
if (str1[m - 1] == str2[n - 1])
return isSubsequence(str1, m - 1, str2, n - 1);
// If last characters are not matching
return isSubsequence(str1, m, str2, n - 1);
}
// Function to count number of words that
// are subsequence of given string S
int countSubsequenceWords(string s, vector<string>& arr)
{
int n = arr.size();
int m = s.length();
int res = 0;
for (int i = 0; i < n; i++) {
if (isSubsequence(arr[i], arr[i].size(), s, m)) {
res++;
}
}
return res;
}
// Driver Code
int main()
{
string S = "geeksforgeeks";
vector<string> arr
= { "geek", "for", "geekofgeeks", "gfg" };
// Funtion call
cout << countSubsequenceWords(S, arr) << "\n";
return 0;
}
Time Complexity: O(m*n)
Auxiliary Space: O(m) for recursion stack space
Efficient Approach: The above approach can be optimized based on the following idea:
- Map the index of characters of the given string to the respective characters array.
- Initialize the ans with the size of arr.
- Iterate over all the words in arr one by one.
- Iterate over each character.
- Find strictly greater index than prevIndex in dict.
- If the strictly greater element is not found, it means the current word is not a subsequence of the given string, so decrease res by 1.
- Else update prevIndex.
- After iterating over all the words, return ans.
Below is the implementation of the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
// Function to count number of words that
// are subsequence of given string S
int countSubsequenceWords(string s, vector<string>& arr)
{
unordered_map<char, vector<int> > dict;
// Mapping index of characters of given
// string to respective characters
for (int i = 0; i < s.length(); i++) {
dict[s[i]].push_back(i);
}
// Initializing res with size of arr
int res = arr.size();
for (auto word : arr) {
// Index where last character
// is found
int prevIndex = -1;
for (int j = 0; j < word.size(); j++) {
// Searching for strictly
// greater element than prev
// using binary search
auto x = upper_bound(dict[word[j]].begin(),
dict[word[j]].end(),
prevIndex);
// If strictly greater index
// not found, the word cannot
// be subsequence of string s
if (x == dict[word[j]].end()) {
res--;
break;
}
// Else, update the prevIndex
else {
prevIndex = *x;
}
}
}
return res;
}
// Driver Code
int main()
{
string S = "geeksforgeeks";
vector<string> arr
= { "geek", "for", "geekofgeeks", "gfg" };
// Function call
cout << countSubsequenceWords(S, arr) << "\n";
return 0;
}
Time Complexity: O( m * s * log(n) ), where m is the length of the given string, s is the max length of the word of arr and n is the length of arr
Auxiliary Space: O(n)