Given a list of characters and an array of strings, find the total length of all strings in the array of strings that can be composed using the given characters.
Examples:
Input: string = [“mouse”, “me”, “bat”, “lion”], chars = “eusamotb”
Output: 10
Explanation:
The strings that can be formed using the characters “eusamotb” are “mouse” and “me” and “bat”.
Length of “mouse” is 5, length of “me” is 2, and length of “bat” is 3
Sum of all lengths = 5 + 2 + 3 = 10.
Input: string = [“hi”, “data”, “geeksforgeeks”], chars = “tiadha”
Output: 6
Explanation:
The strings that can be formed using the characters “tiadha” are “hi” and “data”. Where length of “hi” is 2, length of “data” is 4, the sum of all is 2 + 4 = 6.
Approach:
To solve the problem mentioned above we have to follow the steps given below:
- We can use characters from the given character string that is ‘chars’ while forming a string. We can also reuse the used characters for forming the next string
- Maintain an unordered map with character as a key and the value by keeping track of the frequency of each character from the string of chars.
- Every time we scan characters from the list of string we reduce the frequency of character from the unordered map but we have to maintain the copy of the original map so as to check the second string .
- If the key is not present in the map it creates one with default value as zero rather than throwing an error.
Below is the implementation of the above approach:
C++
// C++ implementation to find total length // of string composed of given characters // formed from given Array of strings #include <bits/stdc++.h> using namespace std;
// Function to count the total length int countCharacters(
vector<string>& strings,
string chars)
{ int res = 0;
// Unordered_map for
// keeping frequency of characters
unordered_map<char, int> freq;
// Calculate the frequency
for (int i = 0; i < chars.length(); i++)
freq[chars[i]] += 1;
// Iterate in the N strings
for (auto st : strings) {
bool flag = true;
// Iterates in the string
for (auto c : st) {
// Checks if given character of string
// string appears in it or not
if (!freq) {
flag = false;
break;
}
}
// Adds the length of string
// if all characters are present
if (flag)
res += st.length();
}
// Return the final result
return res;
} // Driver code int main()
{ vector<string> strings
= { "hi", "data",
"geeksforgeeks" };
string chars = "tiadhae";
cout << countCharacters(strings, chars);
return 0;
} |
Java
// Java implementation to find total length // of string composed of given characters // formed from given Array of strings import java.util.*;
class GFG {
// Function to count the total length static int countCharacters(List<String> strings,
String chars)
{ int res = 0;
// Map for
// keeping frequency of characters
Map<Character, Integer> freq = new HashMap<>();
// Calculate the frequency
for (int i = 0; i < chars.length(); i++)
{
freq.put(chars.charAt(i),
freq.getOrDefault(chars.charAt(i), 0) + 1);
}
// Iterate in the N strings
for (String st : strings)
{
boolean flag = true;
// Iterates in the string
for (char c : st.toCharArray())
{
// Checks if given character of string
// string appears in it or not
if (!freq.containsKey(c))
{
flag = false;
break;
}
}
// Adds the length of string
// if all characters are present
if (flag)
res += st.length();
}
// Return the final result
return res;
} // Driver code public static void main(String[] args)
{ List<String> strings = Arrays.asList("hi", "data",
"geeksforgeeks");
String chars = "tiadhae";
System.out.println(countCharacters(strings, chars));
} } // This code is contributed by offbeat |
Python3
# Python3 implementation to find total length # of string composed of given characters # formed from given Array of strings # Function to count the total length def countCharacters(arr, chars):
res = 0
# Unordered_map for
# keeping frequency of characters
freq = dict()
# Calculate the frequency
for i in range(len(chars)):
freq[chars[i]] = freq.get(chars[i], 0)+1
# Iterate in the N strings
for st in arr:
flag = True
# Iterates in the string
for c in st:
# Checks if given character of string
# string appears in it or not
if (c not in freq):
flag = False
break
# Adds the length of string
# if all characters are present
if (flag):
res += len(st)
# Return the final result
return res
# Driver code if __name__ == '__main__':
arr =["hi", "data", "geeksforgeeks"]
chars = "tiadhae"
print(countCharacters(arr, chars))
# This code is contributed by mohit kumar 29 |
C#
// C# implementation to find total length// of string composed of given characters// formed from given Array of stringsusing System;
using System.Collections.Generic;
using System.Linq;
class GFG{
// Function to count the total lengthstatic int countCharacters(List<string> strings,
string chars)
{ int res = 0;
// Dictionary for keeping frequency
// of characters
Dictionary<char,
int> freq = new Dictionary<char,
int>();
// Calculate the frequency
for(int i = 0; i < chars.Length; i++)
{
if(freq.ContainsKey(chars[i]))
{
freq[chars[i]]++;
}
else
{
freq.Add(chars[i],
freq.GetValueOrDefault(
chars[i], 0) + 1);
}
}
// Iterate in the N strings
foreach(string st in strings)
{
bool flag = true;
// Iterates in the string
foreach (char c in st.ToCharArray())
{
// Checks if given character of string
// string appears in it or not
if (!freq.ContainsKey(c))
{
flag = false;
break;
}
}
// Adds the length of string
// if all characters are present
if (flag)
res += st.Length;
}
// Return the final result
return res;
}// Driver codepublic static void Main(string[] args)
{ string []tmp = { "hi", "data",
"geeksforgeeks" };
List<string> strings = tmp.ToList();
string chars = "tiadhae";
Console.Write(countCharacters(strings, chars));
}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript implementation to find total length // of string composed of given characters // formed from given Array of strings // Function to count the total length function countCharacters( strings, chars)
{ var res = 0;
// Unordered_map for
// keeping frequency of characters
var freq = new Map();
// Calculate the frequency
for (var i = 0; i < chars.length; i++)
{
if(freq.has(chars[i]))
freq.set(chars[i], freq.get(chars[i])+1)
else
freq.set(chars[i], 1)
}
// Iterate in the N strings
strings.forEach(st => {
var flag = true;
// Iterates in the string
st.split('').forEach(c => {
// Checks if given character of string
// string appears in it or not
if (!freq.has(c)) {
flag = false;
}
});
// Adds the length of string
// if all characters are present
if (flag)
res += st.length;
});
// Return the final result
return res;
} // Driver code var strings
= ["hi", "data",
"geeksforgeeks"];
var chars = "tiadhae";
document.write( countCharacters(strings, chars)); // This code is contributed by noob2000.</script> |
Output:
6
Time Complexity: O(n * m), where n is the length of the char and m is the length of the string.
Auxiliary Space Complexity: O(1), as the unordered map will be of size 26 only.