# 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

2

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