# For each lowercase English alphabet find the count of strings having these alphabets

Given an array of strings of lowercase English alphabets. The task is for each letter [a-z] find the count of strings having these letters.

Examples:

Input: str = { “geeks”, “for”, “code” }
Output: { 0 0 1 1 2 1 1 0 0 0 0 0 0 0 2 0 0 1 1 0 0 0 0 0 0 0 }
Explanation:
For a letter, say ‘e’, it is present in { “geeks”, “code” }, hence its count is 2.
Similarly for another letter result can be found.

Input: str = { “i”, “will”, “practice”, “everyday” }
Output: 2 0 1 1 2 0 0 0 3 0 0 0 0 0 0 1 0 2 0 1 0 1 1 0 1 0
Explanation:
For a letter, say ‘i’, it is present in { “i”, “will”, “practice” }, hence its count is 3.
Similarly for another letter result can be found.

Naive Approach:

1. Create a global counter for each letter of size 26, initialize it with 0.
2. Run a loop for each of small letter English alphabet. Increment the counter of, say letter ‘x’, if the current letter is found in the current string.
3. Repeat this for all lowercase English alphabet.

Time Complexity: O(26*N), where N is the sum of the length of all strings.

Efficient Approach:

• Instead of running a loop for each small letter English alphabet and checking whether it is present in the current string or not. We can instead run a loop on each string individually and increment the global counter for any letter present in that string.
• Also to avoid duplicate count we create a visited boolean array to mark the characters encounter so far. This approach reduces the complexity to O(N).

Below is the implementation of the above approach:

 `// C++ program to find count of strings ` `// for each letter [a-z] in english alphabet ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the countStrings ` `// for each letter [a-z] ` `void` `CountStrings(vector& str) ` `{ ` `    ``int` `size = str.size(); ` ` `  `    ``// Initialize result as zero ` `    ``vector<``int``> count(26, 0); ` ` `  `    ``// Mark all letter as not visited ` `    ``vector<``bool``> visited(26, ``false``); ` ` `  `    ``// Loop through each strings ` `    ``for` `(``int` `i = 0; i < size; ++i) ` `    ``{ ` ` `  `        ``for` `(``int` `j = 0; j < str[i].length(); ++j) ` `        ``{ ` ` `  `            ``// Increment the global counter ` `            ``// for current character of string ` `            ``if` `(visited[str[i][j]] == ``false``) ` `                ``count[str[i][j] - ``'a'``]++; ` ` `  `            ``visited[str[i][j]] = ``true``; ` `        ``} ` ` `  `        ``// Instead of re-initialising boolean ` `        ``// vector every time we just reset ` `        ``// all visited letter to false ` `        ``for` `(``int` `j = 0; j < str[i].length(); ++j) ` `        ``{ ` `            ``visited[str[i][j]] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``// Print count for each letter ` `    ``for` `(``int` `i = 0; i < 26; ++i) ` `    ``{ ` `        ``cout << count[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``// Given array of strings ` `    ``vector str = {``"i"``, ``"will"``, ` `                          ``"practice"``, ``"everyday"``}; ` ` `  `    ``// Call the countStrings function ` `    ``CountStrings(str); ` ` `  `    ``return` `0; ` `} `

Output:

```2 0 1 1 2 0 0 0 3 0 0 1 0 0 0 1 0 2 0 1 0 1 1 0 1 0
```

Time Complexity : O(N), where N is the sum of length of all strings.

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