Given an array of strings of lowercase English alphabets. The task is for each letter [a-z] to find the count of strings having these letters.
Examples:
Input: str = { “neveropen”, “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 { “neveropen”, “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.
Method : Brute Force
- Create a global counter array for each letter of size 26, initialize it with 0.
- For every lowercase alphabet check whether it exists in a current string
- If the alphabet exists in the string then increment the count.
- Repeat this for all strings of input.
C++
#include <iostream>#include <vector>#include <string>using namespace std;// Function to find the countStrings// for each letter [a-z]void CountStrings(vector<string>& str){ int size = str.size(); // Initialize result as zero vector<int> count(26, 0); // Loop through each Strings for (int i = 0; i < size; ++i) { // looping all the characters from a to z for (int j = 0; j < 26; j++) { char c = static_cast<char>(j + 'a'); // checking whether the current string // contains the present Character if (str[i].find(c) != string::npos) { count[j]++; } } } // Print count for each letter for (int i = 0; i < 26; ++i) { cout << count[i] << " "; }}// Driver programint main(){ // Given array of Strings vector<string> str = { "i", "will", "practice", "everyday" }; // Call the CountStrings function CountStrings(str); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;// Java program to find count of Strings// for each letter [a-z] in english alphabetclass GFG { // Function to find the countStrings // for each letter [a-z] static void CountStrings(String[] str) { int size = str.length; // Initialize result as zero int[] count = new int[26]; // Loop through each Strings for (int i = 0; i < size; ++i) { // looping all the characters from a to z for (int j = 0; j < 26; j++) { char c = (char)(j + 'a'); // checking whether the current string // contains the present Character if (str[i].contains("" + c)) { count[j]++; } } } // Print count for each letter for (int i = 0; i < 26; ++i) { System.out.print(count[i] + " "); } } // Driver program public static void main(String[] args) { // Given array of Strings String[] str = { "i", "will", "practice", "everyday" }; // Call the countStrings function CountStrings(str); }}// This code is contributed by aeroabrar_31 |
Python3
# Python program to find count of Strings# for each letter [a-z] in the English alphabet# Function to find the countStrings# for each letter [a-z]def count_strings(strings): size = len(strings) # Initialize result as zero count = [0] * 26 # Loop through each string for i in range(size): # Loop through all the characters from 'a' to 'z' for j in range(26): c = chr(j + ord('a')) # Checking whether the current string # contains the present character if c in strings[i]: count[j] += 1 # Print count for each letter for i in range(26): print(count[i], end=' ')# Driver programif __name__ == '__main__': # Given list of strings strings = ["i", "will", "practice", "everyday"] # Call the count_strings function count_strings(strings) |
C#
using System;using System.Collections.Generic;class Program{ // Function to find the countStrings // for each letter [a-z] static void CountStrings(List<string> str) { int size = str.Count; // Initialize result as zero int[] count = new int[26]; // Loop through each Strings for (int i = 0; i < size; ++i) { // looping all the characters from a to z for (int j = 0; j < 26; j++) { char c = (char)(j + 'a'); // checking whether the current string // contains the present Character if (str[i].Contains(c)) { count[j]++; } } } // Print count for each letter for (int i = 0; i < 26; ++i) { Console.Write(count[i] + " "); } } // Driver program static void Main() { // Given array of Strings List<string> str = new List<string> { "i", "will", "practice", "everyday" }; // Call the CountStrings function CountStrings(str); }} |
Javascript
// Function to find the countStrings// for each letter [a-z]function CountStrings(str) { const size = str.length; // Initialize result as zero const count = new Array(26).fill(0); // Loop through each Strings for (let i = 0; i < size; ++i) { // looping all the characters from a to z for (let j = 0; j < 26; j++) { const c = String.fromCharCode(j + 'a'.charCodeAt(0)); // checking whether the current string // contains the present Character if (str[i].includes(c)) { count[j]++; } } } // Print count for each letter for (let i = 0; i < 26; ++i) { process.stdout.write(count[i] + " "); }}// Driver programfunction main() { // Given array of Strings const str = ["i", "will", "practice", "everyday"]; // Call the CountStrings function CountStrings(str);}main(); |
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*26*(len))
where N is length of input array and len is maximum length of 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++
// C++ program to find count of strings// for each letter [a-z] in english alphabet#include <bits/stdc++.h>using namespace std;// Function to find the countStrings// for each letter [a-z]void CountStrings(vector<string>& 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 programint main(){ // Given array of strings vector<string> str = {"i", "will", "practice", "everyday"}; // Call the countStrings function CountStrings(str); return 0;} |
Java
// Java program to find count of Strings// for each letter [a-z] in english alphabetclass GFG{// Function to find the countStrings// for each letter [a-z]static void CountStrings(String []str){ int size = str.length; // Initialize result as zero int []count = new int[26]; // Mark all letter as not visited boolean []visited = new boolean[26]; // 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].charAt(j) - 'a'] == false) count[str[i].charAt(j) - 'a']++; visited[str[i].charAt(j) - 'a'] = 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].charAt(j) - 'a'] = false; } } // Print count for each letter for (int i = 0; i < 26; ++i) { System.out.print(count[i] + " "); }}// Driver programpublic static void main(String[] args){ // Given array of Strings String []str = {"i", "will", "practice", "everyday"}; // Call the countStrings function CountStrings(str);}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program to find count of # strings for each letter [a-z] in # english alphabet# Function to find the countStrings # for each letter [a-z] def CountStrings(s): size = len(s) # Initialize result as zero count = [0] * 26 # Mark all letter as not visited visited = [False] * 26 # Loop through each strings for i in range(size): for j in range(len(s[i])): # Increment the global counter # for current character of string if visited[ord(s[i][j]) - ord('a')] == False: count[ord(s[i][j]) - ord('a')] += 1 visited[ord(s[i][j]) - ord('a')] = True # Instead of re-initialising boolean # vector every time we just reset # all visited letter to false for j in range(len(s[i])): visited[ord(s[i][j]) - ord('a')] = False # Print count for each letter for i in range(26): print(count[i], end = ' ')# Driver codeif __name__=='__main__': # Given array of strings s = [ "i", "will", "practice", "everyday" ] # Call the countStrings function CountStrings(s)# This code is contributed by rutvik_56 |
C#
// C# program to find count of Strings// for each letter [a-z] in english alphabetusing System;class GFG{// Function to find the countStrings// for each letter [a-z]static void CountStrings(String []str){ int size = str.Length; // Initialize result as zero int []count = new int[26]; // Mark all letter as not visited bool []visited = new bool[26]; // 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] - 'a'] == false) count[str[i][j] - 'a']++; visited[str[i][j] - 'a'] = true; } // Instead of re-initialising bool // vector every time we just reset // all visited letter to false for(int j = 0; j < str[i].Length; ++j) { visited[str[i][j] - 'a'] = false; } } // Print count for each letter for(int i = 0; i < 26; ++i) { Console.Write(count[i] + " "); }}// Driver codepublic static void Main(String[] args){ // Given array of Strings String []str = { "i", "will", "practice", "everyday"}; // Call the countStrings function CountStrings(str);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program to find count of strings// for each letter [a-z] in english alphabet// Function to find the countStrings// for each letter [a-z]function CountStrings(str){ let size = str.length; // Initialize result as zero let count = new Array(26).fill(0); // Mark all letter as not visited let visited = new Array(26).fill(false); // Loop through each strings for (let i = 0; i < size; ++i) { for (let j = 0; j < str[i].length; ++j) { // Increment the global counter // for current character of string if(visited[str[i].charCodeAt(j) - 97] == false) count[str[i].charCodeAt(j) - 97]++; visited[str[i].charCodeAt(j) - 97] = true; } // Instead of re-initialising boolean // vector every time we just reset // all visited letter to false for (let j = 0; j < str[i].length; ++j) { visited[str[i].charCodeAt(j) - 97] = false; } } // Print count for each letter for (let i = 0; i < 26; ++i) { document.write(count[i]," "); }}// Driver program// Given array of stringslet str = ["i", "will","practice", "everyday"];// Call the countStrings functionCountStrings(str);// This code is contributed by shinjanpatra</script> |
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 the length of all strings.
Auxiliary Space: O(1)
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