从字符串的定Array字符串的总长度使用给定字符组成的

给定一个字符列表和一个字符串数组，找出可以使用给定字符组成的字符串数组中所有字符串的总长度。
例子：

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. 
 

方法：
要解决上述问题，我们必须按照以下步骤操作：

• 我们可以从给定的字符串，是“字符”，而形成一个字符串使用的字符。我们还可以重用使用过的字符来形成下一个字符串
• 通过跟踪字符字符串中每个字符的频率，维护一个以字符为键和值的无序映射。
• 每次我们从字符串列表中扫描字符，我们都会减少无序映射中字符的频率，但我们必须保留原始映射的副本，以便检查第二个字符串。
• 如果映射中不存在该键，它会创建一个默认值为 0 的键，而不是抛出错误。

下面是上述方法的实现：

C++

// C++ implementation to find total length
// of string composed of given characters
// formed from given Array of strings
 
#include 
using namespace std;
 
// Function to count the total length
int countCharacters(
    vector& strings,
    string chars)
{
    int res = 0;
 
    // Unordered_map for
    // keeping frequency of characters
    unordered_map 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 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 strings,
                                String chars)
{
    int res = 0;
 
    // Map for
    // keeping frequency of characters
    Map 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 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 strings
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG{
 
// Function to count the total length
static int countCharacters(List strings,
                                string chars)
{
    int res = 0;
     
    // Dictionary for keeping frequency
    // of characters
    Dictionary freq = new Dictionary();
     
    // 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 code
public static void Main(string[] args)
{
    string []tmp = { "hi", "data",
                     "geeksforgeeks" };
                      
    List strings = tmp.ToList();
 
    string chars = "tiadhae";
 
    Console.Write(countCharacters(strings, chars));
}
}
 
// This code is contributed by rutvik_56

Javascript

 
 
// 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.

输出：

6

时间复杂度： O(n * m) ，其中n是字符的长度， m是字符串的长度。
辅助空间复杂度： O(1) ，因为无序映射的大小仅为 26。

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