Given a string array arr[], the task is to find the longest sub-sequence of the array with at least one character appearing in all the strings. Note that all the strings contain only lowercase English alphabets.
Examples:
Input: str = {“ab”, “bc”, “de”}
Output: 2
{“ab”, “bc”} is the required sub-sequence
with ‘b’ as the common character.Input: str = {“a”, “b”, “c”}
Output: 1
Approach: Create a count[] array such that count[0] will store the number of strings which contain ‘a’, count[1] will store the number of strings which contain ‘b’ and so on…
Now, it’s clear that the answer will be the maximum value from the count[] array. In order to update this array start traversing the string array and for every string, mark which characters are present in the current string in a hash[] array.
And after the traversal, for every character which is present in the current string update its count in the count[] array.
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 return the length of the longest // sub-sequence with at least one // common character in every string int largestSubSeq(string arr[], int n) { // count[0] will store the number of strings // which contain 'a', count[1] will store the // number of strings which contain 'b' and so on.. int count[MAX] = { 0 }; // For every string for (int i = 0; i < n; i++) { string str = arr[i]; // Hash array to set which character is // present in the current string bool hash[MAX] = { 0 }; for (int j = 0; j < str.length(); j++) { hash[str[j] - 'a'] = true; } for (int j = 0; j < MAX; j++) { // If current character appears in the // string then update its count if (hash[j]) count[j]++; } } return *(max_element(count, count + MAX)); } // Driver code int main() { string arr[] = { "ab", "bc", "de" }; int n = sizeof(arr) / sizeof(string); cout << largestSubSeq(arr, n); return 0; } |
Java
// Java implementation of the approach class GFG { static int MAX = 26; // Function to return the length of the longest // sub-sequence with at least one // common character in every string static int largestSubSeq(String arr[], int n) { // count[0] will store the number of strings // which contain 'a', count[1] will store the // number of strings which contain 'b' and so on.. int [] count = new int[MAX]; // For every string for (int i = 0; i < n; i++) { String str = arr[i]; // Hash array to set which character is // present in the current string boolean [] hash = new boolean[MAX]; for (int j = 0; j < str.length(); j++) { hash[str.charAt(j) - 'a'] = true; } for (int j = 0; j < MAX; j++) { // If current character appears in the // string then update its count if (hash[j]) count[j]++; } } int max = -1; for(int i=0;i< MAX; i++) { if(max < count[i]) max = count[i]; } return max; } // Driver code public static void main (String[] args) { String arr[] = { "ab", "bc", "de" }; int n = arr.length; System.out.println(largestSubSeq(arr, n)); } } // This code is contributed by ihritik |
Python3
# Python3 implementation of the approach MAX = 26 # Function to return the length of the longest # sub-sequence with at least one # common character in every string def largestSubSeq(arr, n): # count[0] will store the number of strings # which contain 'a', count[1] will store the # number of strings which contain 'b' and so on.. count = [0] * MAX # For every string for i in range(n): string = arr[i] # Hash array to set which character is # present in the current string _hash = [False] * MAX for j in range(len(string)): _hash[ord(string[j]) - ord('a')] = True for j in range(MAX): # If current character appears in the # string then update its count if _hash[j] == True: count[j] += 1 return max(count) # Driver code if __name__ == "__main__": arr = [ "ab", "bc", "de" ] n = len(arr) print(largestSubSeq(arr, n)) # This code is contributed by # sanjeev2552 |
C#
// C# implementation of the approach using System; class GFG { static int MAX = 26; // Function to return the length of the longest // sub-sequence with at least one // common character in every string static int largestSubSeq(string [] arr, int n) { // count[0] will store the number of strings // which contain 'a', count[1] will store the // number of strings which contain 'b' and so on.. int [] count = new int[MAX]; // For every string for (int i = 0; i < n; i++) { string str = arr[i]; // Hash array to set which character is // present in the current string bool [] hash = new bool[MAX]; for (int j = 0; j < str.Length; j++) { hash[str[j] - 'a'] = true; } for (int j = 0; j < MAX; j++) { // If current character appears in the // string then update its count if (hash[j]) count[j]++; } } int max = -1; for(int i=0;i< MAX; i++) { if(max < count[i]) max = count[i]; } return max; } // Driver code public static void Main () { string [] arr = { "ab", "bc", "de" }; int n = arr.Length; Console.WriteLine(largestSubSeq(arr, n)); } } // This code is contributed by ihritik |
2
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
