长度至少为 K 的给定字符串数组中等值字符串的计数

给定一个包含N个字符串和一个整数K的数组arr[] ，任务是找出长度至少为K的等值字符串的数量。

例子：

Input: arr[] = {“abcd”, “der”, “erty”}, K = 4
Output: 2
Explanation: All given strings are isograms, but only “abcd” and “erty” are of length at least K. Hence count is 2

Input: arr[] = {“ag”, “bka”, “lkmn”, “asdfg”}, K = 2
Output: 4
Explanation: All the strings are isograms and each strings is of length >=K. Hence count is 4.

方法：这个问题可以通过存储所有字符的频率或使用Set数据结构来解决。如果该字符串中没有任何字母出现多次，则该字符串是等值线。请按照以下步骤解决给定的问题。

• 遍历字符串数组arr[] ，并为每个字符串
• 创建字符的频率图。
• 只要任何字符的频率大于1 ，或者如果字符串的长度小于K ，则跳过当前字符串。
• 否则，如果没有字符的频率大于1 ，并且字符串的长度小于K ，则增加答案的计数。
• 当遍历所有字符串时，打印存储在答案中的计数。

下面是上述方法的实现。

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to check if a string
// is an isogram or not
bool isIsogram(string s)
{
    // To store the frequencies
    vector freq(26, 0);
 
    for (char c : s) {
        freq++;
 
        if (freq > 1) {
            return false;
        }
    }
 
    return true;
}
 
// Function to check if array arr contains
// all isograms or not
int allIsograms(vector& arr, int K)
{
    int ans = 0;
    for (string x : arr) {
        if (isIsogram(x) && x.length() >= K) {
            ans++;
        }
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    vector arr = { "abcd", "der", "erty" };
    int K = 4;
 
    // Function call and printing the answer
    cout << allIsograms(arr, K);
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG {
 
    // Function to check if a string
    // is an isogram or not
    static boolean isIsogram(String s)
    {
        // To store the frequencies
        int[] freq = new int[26];
 
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            freq++;
 
            if (freq > 1) {
                return false;
            }
        }
 
        return true;
    }
 
    // Function to check if array arr contains
    // all isograms or not
    static int allIsograms(String[] arr, int K)
    {
        int ans = 0;
        for (String x : arr) {
            if (isIsogram(x) && x.length() >= K) {
                ans++;
            }
        }
 
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String arr[] = { "abcd", "der", "erty" };
        int K = 4;
 
        // Function call and printing the answer
 
        System.out.println(allIsograms(arr, K));
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python code for the above approach
 
# Function to check if a string
# is an isogram or not
def isIsogram (s):
 
    # To store the frequencies
    freq = [0] * 26
 
    for c in s:
        freq[ord(c) - ord("a")] += 1
 
        if (freq[s.index(c)] > 1):
            return False
 
    return True
 
# Function to check if array arr contains
# all isograms or not
def allIsograms (arr, K):
    ans = 0
    for x in arr:
        if isIsogram(x) and len(x) >= K:
            ans += 1
 
    return ans
 
 
# Driver Code
arr = ["abcd", "der", "erty"]
K = 4
 
# Function call and printing the answer
print(allIsograms(arr, K))
 
# This code is contributed by Saurabh jaiswal

C#

// C# code for the above approach
using System;
 
class GFG{
 
// Function to check if a string
// is an isogram or not
static bool isIsogram(string s)
{
     
    // To store the frequencies
    int[] freq = new int[26];
 
    for(int i = 0; i < s.Length; i++)
    {
        char c = s[i];
        freq++;
 
        if (freq > 1)
        {
            return false;
        }
    }
    return true;
}
 
// Function to check if array arr contains
// all isograms or not
static int allIsograms(string[] arr, int K)
{
    int ans = 0;
    foreach(string x in arr)
    {
        if (isIsogram(x) && x.Length >= K)
        {
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
public static void Main(string[] args)
{
    string[] arr = { "abcd", "der", "erty" };
    int K = 4;
 
    // Function call and printing the answer
    Console.WriteLine(allIsograms(arr, K));
}
}
 
// This code is contributed by ukasp

Javascript

    // JavaScript code for the above approach
 
    // Function to check if a string
    // is an isogram or not
    const isIsogram = (s) => {
     
        // To store the frequencies
        let freq = new Array(26).fill(0);
 
        for (let c in s) {
            freq[s.charCodeAt(c) - "a".charCodeAt(0)]++;
 
            if (freq > 1) {
                return false;
            }
        }
 
        return true;
    }
 
    // Function to check if array arr contains
    // all isograms or not
    const allIsograms = (arr, K) => {
        let ans = 0;
        for (let x in arr) {
            if (isIsogram(x) && arr[x].length >= K) {
                ans++;
            }
        }
 
        return ans;
    }
 
    // Driver Code
    let arr = ["abcd", "der", "erty"];
    let K = 4;
 
    // Function call and printing the answer
    document.write(allIsograms(arr, K));
 
    // This code is contributed by rakeshsahni

输出

2

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)。