给定字符串数组中的等值线字符串计数

给定一个包含N个字符串的数组arr[] ，任务是找出等值图的字符串数。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。

例子：

Input: arr[] = {“abcd”, “derg”, “erty”}
Output: 3
Explanation: All given strings are isograms. In all the strings no character 
is present more than once. Hence count is 3

Input: arr[] = {“agka”, “lkmn”}
Output: 1
Explanation: Only string “lkmn” is isogram. In the string “agka” 
the character ‘a’ is present twice. Hence count is 1.

方法：贪心方法可以用来解决这个问题。遍历给定字符串数组中的每个字符串并检查它是否是等图线。为此，请按照以下步骤操作：

• 遍历字符串数组并对每个字符串执行以下步骤：
• 创建字符的频率图。
• 无论任何字符的频率大于 1 ，跳过当前字符串并移至下一个字符串。
• 如果没有字符的频率超过 1，则将答案计数加1。
• 当遍历所有字符串时，返回存储在 answer 中的计数。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
// Function to check
// if a string is an isogram
bool isIsogram(string s)
{
    // Loop to check
    // if string is isogram or not
    vector freq(26, 0);
    for (char c : s) {
        freq++;
        if (freq > 1) {
            return false;
        }
    }
    return true;
}
 
// Function to count the number of isograms
int countIsograms(vector& arr)
{
    int ans = 0;
 
    // Loop to iterate the string array
    for (string x : arr) {
        if (isIsogram(x)) {
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    vector arr = { "abcd", "derg", "erty" };
 
    // Count of isograms in string array arr[]
    cout << countIsograms(arr) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
 
class GFG {
 
    // Function to check
    // if a String is an isogram
    static boolean isIsogram(String s) {
 
        // Loop to check
        // if String is isogram or not
        int[] freq = new int[26];
        for (int i = 0; i < 26; i++) {
            freq[i] = 0;
        }
 
        for (char c : s.toCharArray()) {
            freq++;
            if (freq > 1) {
                return false;
            }
        }
        return true;
    }
 
    // Function to count the number of isograms
    static int countIsograms(ArrayList arr) {
        int ans = 0;
 
        // Loop to iterate the String array
        for (String x : arr) {
            if (isIsogram(x)) {
                ans++;
            }
        }
        return ans;
    }
 
    // Driver Code
    public static void main(String args[]) {
        ArrayList arr = new ArrayList();
 
        arr.add("abcd");
        arr.add("derg");
        arr.add("erty");
 
        // Count of isograms in String array arr[]
        System.out.println(countIsograms(arr));
    }
}
 
// This code is contributed by gfgking

Python3

# Function to check
# if a string is an isogram
def isIsogram(s):
 
    # Loop to check
    # if string is isogram or not
    freq = [0]*(26)
    for c in s:
        freq[ord(c) - ord('a')] += 1
        if (freq[ord(c) - ord('a')] > 1):
            return False
 
    return True
 
# Function to count the number of isograms
def countIsograms(arr):
    ans = 0
 
    # Loop to iterate the string array
    for x in arr:
        if (isIsogram(x)):
            ans += 1
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    arr = ["abcd", "derg", "erty"]
 
    # Count of isograms in string array arr[]
    print(countIsograms(arr))
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections;
 
class GFG
{
     
// Function to check
// if a string is an isogram
static bool isIsogram(string s)
{
   
    // Loop to check
    // if string is isogram or not
    int []freq = new int[26];
    for(int i = 0; i < 26; i++) {
        freq[i] = 0;
    }
     
    foreach (char c in s) {
        freq++;
        if (freq > 1) {
            return false;
        }
    }
    return true;
}
 
// Function to count the number of isograms
static int countIsograms(ArrayList arr)
{
    int ans = 0;
 
    // Loop to iterate the string array
    foreach (string x in arr) {
        if (isIsogram(x)) {
            ans++;
        }
    }
    return ans;
}
 
// Driver Code
public static void Main()
{
    ArrayList arr = new ArrayList();
     
    arr.Add("abcd");
    arr.Add("derg");
    arr.Add("erty");
 
    // Count of isograms in string array arr[]
    Console.WriteLine(countIsograms(arr));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
    // Function to check
    // if a string is an isogram
    const isIsogram = (s) => {
     
        // Loop to check
        // if string is isogram or not
        let freq = new Array(26).fill(0);
        for (let c in s) {
            freq[s.charCodeAt(c) - "0".charCodeAt(0)]++;
            if (freq > 1) {
                return false;
            }
        }
        return true;
    }
 
    // Function to count the number of isograms
    const countIsograms = (arr) => {
        let ans = 0;
 
        // Loop to iterate the string array
        for (let x in arr) {
            if (isIsogram(x)) {
                ans++;
            }
        }
        return ans;
    }
 
    // Driver Code
    let arr = ["abcd", "derg", "erty"];
 
    // Count of isograms in string array arr[]
    document.write(countIsograms(arr));
 
    // This code is contributed by rakeshsahni
 

输出

3

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)