每个字符的频率最多为 X 且长度至少为 Y 的字符串计数

给定一个由字符串和整数X和Y组成的数组arr[] ，任务是找到每个字符的频率最多为 X且字符串长度至少为 Y的字符串计数。

例子：

Input: arr[] = { “ab”, “derdee”, “erre” }, X = 2, Y = 4
Output: 1
Explanation: Strings with character frequency at most 2 and 
length at least 4 is “erre”. Hence count is 1

Input: arr[] = {“ag”, “ka”, “nanana”}, X = 3, Y = 2
Output: 3

方法：按照下面提到的方法解决问题：

• 遍历字符串数组，对每个字符串执行以下步骤。
• 创建字符的频率图。
• 每当任何字符的频率大于 X或长度小于 Y时，跳过当前字符串。
• 如果没有字符的频率超过 X，长度至少为 Y，则增加答案计数。
• 当遍历所有字符串时，返回存储在 answer 中的计数。

C++

#include 
using namespace std;
 
// Function to check if
// the string has
// frequency of each character
// less than X
bool isValid(string s, int X)
{
    vector freq(26, 0);
 
    // Loop to check the frequency
    // of each character in the string
    for (char c : s) {
        freq++;
    }
 
    // Loop to check
    // if the frequency of all characters
    // are at most X
    for (int i = 0; i < 26; i++)
        if (freq[i] > X)
            return false;
    return true;
}
 
// Function to calculate the count of strings
int getCount(vector& arr, int X, int Y)
{
    int ans = 0;
 
    // Loop to iterate the string array
    for (string st : arr) {
        if (isValid(st, X) && st.length() >= Y) {
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    vector arr = { "ab", "derdee", "erre" };
    int X = 2, Y = 4;
 
    // Function call to get count for arr[]
    cout << getCount(arr, X, Y);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG
{
 
  // Function to check if
  // the string has
  // frequency of each character
  // less than X
  static boolean isValid(String s, int X)
  {
    int freq[] = new int[26];
 
    // Loop to check the frequency
    // of each character in the string
    for (int i=0;i X)
        return false;
    return true;
  }
 
  // Function to calculate the count of strings
  static int getCount(String[] arr, int X, int Y)
  {
    int ans = 0;
 
    // Loop to iterate the string array
    for (String st : arr) {
      if (isValid(st, X) && st.length() >= Y) {
        ans++;
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    String arr[] = { "ab", "derdee", "erre" };
    int X = 2, Y = 4;
 
    // Function call to get count for arr[]
 
    System.out.println(getCount(arr, X, Y));
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Function to check if
# the string has
# frequency of each character
# less than X
def isValid (s, X) :
    freq = [0] * 26
 
    # Loop to check the frequency
    # of each character in the string
    for c in s:
        freq[ord(c)  - ord("a")] += 1
     
 
    # Loop to check
    # if the frequency of all characters
    # are at most X
    for i in range(26):
        if (freq[i] > X):
            return False
    return True
 
 
# Function to calculate the count of strings
def getCount (arr, X, Y):
    ans = 0
 
    # Loop to iterate the string array
    for st in arr:
        if (isValid(st, X) and len(st) >= Y):
            ans += 1
    return ans
 
 
# Driver Code
 
arr = ["ab", "derdee", "erre"]
X = 2
Y = 4
 
# Function call to get count for arr[]
print(getCount(arr, X, Y))
 
# This code is contributed by gfgking.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if the string
// has frequency of each character
// less than X
static bool isValid(String s, int X)
{
    int []freq = new int[26];
     
    // Loop to check the frequency
    // of each character in the string
    for(int i = 0; i < s.Length; i++)
    {
        char c = s[i];
        freq++;
    }
     
    // Loop to check if the frequency
    // of all characters are at most X
    for(int i = 0; i < 26; i++)
        if (freq[i] > X)
            return false;
             
        return true;
}
 
// Function to calculate the count of strings
static int getCount(String[] arr, int X, int Y)
{
    int ans = 0;
     
    // Loop to iterate the string array
    foreach (String st in arr)
    {
        if (isValid(st, X) && st.Length >= Y)
        {
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    String []arr = { "ab", "derdee", "erre" };
    int X = 2, Y = 4;
     
    // Function call to get count for []arr
    Console.WriteLine(getCount(arr, X, Y));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
    // Function to check if
    // the string has
    // frequency of each character
    // less than X
    const isValid = (s, X) => {
        let freq = new Array(26).fill(0);
 
        // Loop to check the frequency
        // of each character in the string
        for (let c in s) {
            freq[s.charCodeAt(c) - "a".charCodeAt(0)]++;
        }
 
        // Loop to check
        // if the frequency of all characters
        // are at most X
        for (let i = 0; i < 26; i++)
            if (freq[i] > X)
                return false;
        return true;
    }
 
    // Function to calculate the count of strings
    const getCount = (arr, X, Y) => {
        let ans = 0;
 
        // Loop to iterate the string array
        for (let st in arr) {
            if (isValid(arr[st], X) && arr[st].length >= Y) {
                ans++;
            }
        }
        return ans;
    }
 
    // Driver Code
 
    let arr = ["ab", "derdee", "erre"];
    let X = 2, Y = 4;
 
    // Function call to get count for arr[]
    document.write(getCount(arr, X, Y));
 
    // This code is contributed by rakeshsahni
 

输出

1

时间复杂度： O(N*M)，其中 N 是数组的大小，M 是最长字符串的大小
辅助空间： O(1)