Longest subsequence with at least one character appearing in every string

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

Javascript

<script>

// Javascript implementation of the approach

var MAX = 26;

// Function to return the length of the longest
// sub-sequence with at least one
// common character in every string
function 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..
    var count = Array(MAX).fill(0);

    // For every string
    for (var i = 0; i < n; i++) {
        var str = arr[i];

        // Hash array to set which character is
        // present in the current string
        var hash = Array(MAX).fill(0);
        for (var j = 0; j < str.length; j++) {
            hash[str[j].charCodeAt(0) - 'a'.charCodeAt(0)] = true;
        }

        for (var j = 0; j < MAX; j++) {

            // If current character appears in the
            // string then update its count
            if (hash[j])
                count[j]++;
        }
    }

    return count.reduce((a,b)=>Math.max(a,b));
}

// Driver code
var arr = ["ab", "bc", "de" ];
var n = arr.length;
document.write( largestSubSeq(arr, n));

</script>

Output:

2