C++ Program for Check if given string can be formed by two other strings or their permutations
Given a string str and an array of strings arr[], the task is to check if the given string can be formed by any of the string pair from the array or their permutations.
Examples:
Input: str = “amazon”, arr[] = {“loa”, “azo”, “ft”, “amn”, “lka”}
Output: Yes
The chosen strings are “amn” and “azo”
which can be rearranged as “amazon”.Input: str = “geeksforgeeks”, arr[] = {“geeks”, “geek”, “for”}
Output: No
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if str can be // generated from any permutation of the // two strings selected from the given vector bool isPossible(vector<string> v, string str) { // Sort the given string sort(str.begin(), str.end()); // Select two strings at a time from given vector for ( int i = 0; i < v.size() - 1; i++) { for ( int j = i + 1; j < v.size(); j++) { // Get the concatenated string string temp = v[i] + v[j]; // Sort the resultant string sort(temp.begin(), temp.end()); // If the resultant string is equal // to the given string str if (temp.compare(str) == 0) { return true ; } } } // No valid pair found return false ; } // Driver code int main() { string str = "amazon" ; vector<string> v{ "fds" , "oxq" , "zoa" , "epw" , "amn" }; if (isPossible(v, str)) cout << "Yes" ; else cout << "No" ; return 0; } |
Method 2: Counting sort can be used to reduce the running time of the above approach. Counting sort uses a table to store the count of each character. We have 26 alphabets hence we make an array of size 26 to store counts of each character in the string. Then take the characters in increasing order to get the sorted string.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAX 26 // Function to sort the given string // using counting sort void countingsort(string& s) { // Array to store the count of each character int count[MAX] = { 0 }; for ( int i = 0; i < s.length(); i++) { count[s[i] - 'a' ]++; } int index = 0; // Insert characters in the string // in increasing order for ( int i = 0; i < MAX; i++) { int j = 0; while (j < count[i]) { s[index++] = i + 'a' ; j++; } } } // Function that returns true if str can be // generated from any permutation of the // two strings selected from the given vector bool isPossible(vector<string> v, string str) { // Sort the given string countingsort(str); // Select two strings at a time from given vector for ( int i = 0; i < v.size() - 1; i++) { for ( int j = i + 1; j < v.size(); j++) { // Get the concatenated string string temp = v[i] + v[j]; // Sort the resultant string countingsort(temp); // If the resultant string is equal // to the given string str if (temp.compare(str) == 0) { return true ; } } } // No valid pair found return false ; } // Driver code int main() { string str = "amazon" ; vector<string> v{ "fds" , "oxq" , "zoa" , "epw" , "amn" }; if (isPossible(v, str)) cout << "Yes" ; else cout << "No" ; return 0; } |
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