检查字符串是否为 Pangrammatic Lipogram

为了理解什么是 pangrammatic lipogram，我们将把这个术语分解为 2 个术语，即 pangram 和 lipogram

Pangram ：pangram 或全字母句子是使用给定字母表的每个字母至少一次的句子。最著名的英语 pangram 是“The quick brown fox jumps over the lazy dog”。

唇形图：唇形图是一种受限制的写作或文字游戏，由编写段落或较长的作品组成，其中避免了特定的字母或字母组——通常是常见的元音，通常是英语中最常见的字母 E。

示例：最初的“Mary Had a Little Lamb”被 A. Ross Eckler Jr. 更改为排除了字母“S”。

Original:
Mary had a little lamb
Its fleece was white as snow
And everywhere that Mary went
The lamb was sure to go

He followed her to school one day
That was against the rule
It made the children laugh and play
To see the lamb in school

Lipogram (Without "S"):
Mary had a little lamb
With fleece a pale white hue
And everywhere that Mary went
The lamb kept her in view

To academe he went with her,
Illegal, and quite rare;
It made the children laugh and play
To view the lamb in there

Pangrammatic Lipogram :
pangrammatic lipogram 是使用字母表中除一个之外的每个字母的文本。例如，“The quick brown fox jumped over the lazy dog”省略了字母 S，通常的 pangram 通过使用单词 jumps 来包含它。

给定一个字符串，我们的任务是检查这个字符串是否是一个 pangrammatic lipogram ？

这样做的想法是，我们将跟踪字符串中未找到的所有字母。

如果字母表中的所有字母都存在，那么它就是一个 pangram
如果只省略一个字母，则它是一个 pangrammatic lipogram 否则它可能只是一个 lipogram。

下面是上述想法的实现：

C++

// Javascript program to check if a string
// is Pangrammatic Lipogram
 
#include
using namespace std;
 
// collection of letters
string alphabets = "abcdefghijklmnopqrstuvwxyz";
 
 
// function to check for a Pangrammatic Lipogram
void panLipogramChecker(string s)
{  
    // convert string to lowercase
    for(int i=0; is.length())
            counter += 1;
    }
   
    if(counter == 0)
        cout<<"Pangram"<= 2)
        cout<<"Not a pangram but might a lipogram"<


Java
// Java program to check if a string
// is Pangrammatic Lipogram
import java.util.*;
 
class GFG
{
 
// collection of letters
static String alphabets = "abcdefghijklmnopqrstuvwxyz";
 
/*
    Category             No of letters unmatched
    Pangram                     0
    Lipogram                 >1
    Pangrammatic Lipogram     1
*/
 
// function to check for a Pangrammatic Lipogram
static void panLipogramChecker(char []s)
{
    // convert string to lowercase
    for(int i = 0; i < s.length; i++)
    {
        s[i] = Character.toLowerCase(s[i]);
    }
     
    // variable to keep count of all the letters
    // not found in the string
    int counter = 0 ;
     
    // traverses the string for every
    // letter of the alphabet
    for(int i = 0 ; i < 26 ; i++)
    {
        int pos = find(s, alphabets.charAt(i));
         
        // if character not found in string
        // then increment count
        if(pos<0 || pos > s.length)
            counter += 1;
    }
     
    if(counter == 0)
        System.out.println("Pangram");
    else if(counter >= 2)
        System.out.println("Not a pangram but might a lipogram");
    else
        System.out.println("Pangrammatic Lipogram");
}
 
static int find(char[]arr, char c)
{
    for(int i = 0; i < arr.length; i++)
    {
        if(c == arr[i])
            return 1;
    }
    return -1;
}
 
// Driver program to test above function
public static void main(String []args)
{
    char []str = "The quick brown fox jumped over the lazy dog".toCharArray();
    panLipogramChecker(str);
     
    str = "The quick brown fox jumps over the lazy dog".toCharArray();
    panLipogramChecker(str);
     
    str = "The quick brown fox jum over the lazy dog".toCharArray();
    panLipogramChecker(str);
}
}
 
// This code is contributed by Rajput-Ji

Python3
# Python program to check if a string
# is Pangrammatic Lipogram
 
# collection of letters
alphabets = 'abcdefghijklmnopqrstuvwxyz'
 
'''
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
'''
 
# function to check for a Pangrammatic Lipogram
def panLipogramChecker(s):
    s.lower()
     
    # variable to keep count of all the letters
    # not found in the string
    counter = 0
     
    # traverses the string for every
    # letter of the alphabet
    for ch in alphabets:
        # character not found in string then increment count
        if(s.find(ch) < 0):
            counter += 1
 
    if(counter == 0):
        result = "Pangram"
    else if(counter == 1):
        result = "Pangrammatic Lipogram"
    else:
        result = "Not a pangram but might a lipogram"
 
    return result
 
# Driver program to test above function
def main():
    print(panLipogramChecker("The quick brown fox \
                            jumped over the lazy dog"))
     
    print(panLipogramChecker("The quick brown fox \
                              jumps over the lazy dog"))
 
    print(panLipogramChecker("The quick brown fox jum\
                                     over the lazy dog"))
     
 
if __name__ == '__main__':
    main()

C#
// C# program to check if a string
// is Pangrammatic Lipogram
using System;
     
class GFG
{
 
// collection of letters
static String alphabets = "abcdefghijklmnopqrstuvwxyz";
 
/*
    Category             No of letters unmatched
    Pangram                     0
    Lipogram                 >1
    Pangrammatic Lipogram     1
*/
 
// function to check for a Pangrammatic Lipogram
static void panLipogramChecker(char []s)
{
    // convert string to lowercase
    for(int i = 0; i < s.Length; i++)
    {
        s[i] = char.ToLower(s[i]);
    }
     
    // variable to keep count of all the letters
    // not found in the string
    int counter = 0 ;
     
    // traverses the string for every
    // letter of the alphabet
    for(int i = 0 ; i < 26 ; i++)
    {
        int pos = find(s, alphabets[i]);
         
        // if character not found in string
        // then increment count
        if(pos<0 || pos > s.Length)
            counter += 1;
    }
     
    if(counter == 0)
        Console.WriteLine("Pangram");
    else if(counter >= 2)
        Console.WriteLine("Not a pangram but might a lipogram");
    else
        Console.WriteLine("Pangrammatic Lipogram");
}
 
static int find(char[]arr, char c)
{
    for(int i = 0; i < arr.Length; i++)
    {
        if(c == arr[i])
            return 1;
    }
    return -1;
}
 
// Driver program to test above function
public static void Main(String []args)
{
    char []str = "The quick brown fox jumped over the lazy dog".
                                                  ToCharArray();
    panLipogramChecker(str);
     
    str = "The quick brown fox jumps over the lazy dog".   
                                          ToCharArray();
    panLipogramChecker(str);
     
    str = "The quick brown fox jum over the lazy dog".
                                        ToCharArray();
    panLipogramChecker(str);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

C++
// C++ program to check for a Pangrammatic
// Lipogram O(n) approach
 
/*
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
*/
 
#include 
using namespace std;
 
// function to check for Pangrammatic Lipogram
void panLipogramChecker(string s)
{
 
    // using map to keep count of the
    // occurence of each letter
    unordered_map mp;
 
    for (char c = 'a'; c <= 'z'; c++) {
        mp = 0;
    }
 
    transform(s.begin(), s.end(), s.begin(), ::tolower);
 
    int i, n = s.length();
    for (i = 0; i <= n - 1; i++) {
        if (isalpha(s[i])) {
 
            // increment count of characters in dictionary
            mp[s[i]]++;
        }
    }
    int count_zero = 0;
 
    for (auto it : mp) {
        if (it.second == 0) {
            count_zero++;
        }
    }
    if (count_zero > 1) {
        cout << "Not a pangram, but might be a lipogram.\n";
    }
    else if (count_zero == 1) {
        cout << "Pangrammatic Lipogram.\n";
    }
    else if (count_zero < 1) {
        cout << "Pangram.\n";
    }
}
 
// Driver program to test above function
int main()
{
 
    panLipogramChecker("The quick brown fox \
                        jumped over the lazy dog");
    panLipogramChecker("The quick brown fox \
                        jumps over the lazy dog");
    panLipogramChecker("The quick brown fox \
                        jum over the lazy dog");
    return 0;
}
 
// This code is contributed by rajsanghavi9.

Java
// Java program to check for a Pangrammatic
// Lipogram O(n) approach
 
/*
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
*/
 
import java.util.*;
 
class GFG {
 
    // function to check for Pangrammatic Lipogram
    static void panLipogramChecker(String s)
    {
 
        // using map to keep count of the
        // occurence of each letter
        HashMap mp = new HashMap<>();
 
        for (char c = 'a'; c <= 'z'; c++) {
            mp.put(c, 0);
        }
 
        s = s.toLowerCase();
 
        int i, n = s.length();
        for (i = 0; i <= n - 1; i++) {
            if (Character.isAlphabetic(s.charAt(i))) {
 
                // increment count of characters in
                // dictionary
                mp.put(s.charAt(i),
                       mp.get(s.charAt(i)) + 1);
            }
        }
        int count_zero = 0;
 
        // Getting an iterator
        Iterator hmIterator = mp.entrySet().iterator();
 
        while (hmIterator.hasNext()) {
            Map.Entry mapElement
                = (Map.Entry)hmIterator.next();
            int marks = ((int)mapElement.getValue());
 
            if (marks == 0)
                count_zero++;
        }
        if (count_zero > 1) {
            System.out.println(
                "Not a pangram, but might be a lipogram.");
        }
        else if (count_zero == 1) {
            System.out.println("Pangrammatic Lipogram.");
        }
        else if (count_zero < 1) {
            System.out.println("Pangram.");
        }
    }
 
    public static void main(String[] args)
    {
        panLipogramChecker(
            "The quick brown fox jumped over the lazy dog");
        panLipogramChecker(
            "The quick brown fox jumps over the lazy dog");
        panLipogramChecker(
            "The quick brown fox jum over the lazy dog");
    }
}
 
// This code is contributed by rajsanghavi9.

Python3
# Python program to check for a Pangrammatic
# Lipogram O(n) approach
 
'''
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
'''
 
# function to check for Pangrammatic Lipogram
 
 
def panLipogramChecker(s):
 
    # dictionary to keep count of the
    # occurence of each letter
    counter = {'a': 0, 'b': 0, 'c': 0, 'd': 0, 'e': 0,
               'f': 0, 'g': 0, 'h': 0, 'i': 0, 'j': 0,
               'k': 0, 'l': 0, 'm': 0, 'n': 0, 'o': 0,
               'p': 0, 'q': 0, 'r': 0, 's': 0, 't': 0,
               'u': 0, 'v': 0, 'w': 0, 'x': 0, 'y': 0,
               'z': 0}
 
    s = s.lower()
 
    # increment count of characters in dictionary
    for i in s:
        if (i.isalpha()):
            counter[i] += 1
 
    # returns a list containing the values of all
    # the keys in h=the dictionary
    b = list(counter.values())
 
    if (b.count(0) > 1):
        print("Not a pangram, but might be a lipogram.")
    else if (b.count(0) == 1):
        print("Pangrammatic Lipogram.")
    else if (b.count(0) < 1):
        print("Pangram.")
 
 
# Driver program to test above function
def main():
    panLipogramChecker("The quick brown fox \
                        jumped over the lazy dog")
    panLipogramChecker("The quick brown fox \
                        jumps over the lazy dog")
    panLipogramChecker("The quick brown fox \
                        jum over the lazy dog")
 
 
if __name__ == '__main__':
    main()

输出：

Pangrammatic Lipogram
Pangram
Not a Pangram but might a Lipogram

时间复杂度： O(26 * N) ，这里 N 是要检查的字符串中的字符数，26 表示字母的总数。

有效的方法：一种有效的方法不是遍历字母表中的所有字母，而是我们可以维护一个散列数组或映射来存储输入字符串中每个字母表的出现次数。最初计数的所有字母将被初始化为零。我们将开始遍历字符串并增加字符数。一旦我们完成了对字符串的遍历，我们将遍历映射或散列数组以查找有多少字符计数为零。感谢 Ravi Teja Gannavarapu 提出这种方法。

下面是上述想法的实现。

`C++`

// C++ program to check for a Pangrammatic
// Lipogram O(n) approach
 
/*
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
*/
 
#include 
using namespace std;
 
// function to check for Pangrammatic Lipogram
void panLipogramChecker(string s)
{
 
    // using map to keep count of the
    // occurence of each letter
    unordered_map mp;
 
    for (char c = 'a'; c <= 'z'; c++) {
        mp = 0;
    }
 
    transform(s.begin(), s.end(), s.begin(), ::tolower);
 
    int i, n = s.length();
    for (i = 0; i <= n - 1; i++) {
        if (isalpha(s[i])) {
 
            // increment count of characters in dictionary
            mp[s[i]]++;
        }
    }
    int count_zero = 0;
 
    for (auto it : mp) {
        if (it.second == 0) {
            count_zero++;
        }
    }
    if (count_zero > 1) {
        cout << "Not a pangram, but might be a lipogram.\n";
    }
    else if (count_zero == 1) {
        cout << "Pangrammatic Lipogram.\n";
    }
    else if (count_zero < 1) {
        cout << "Pangram.\n";
    }
}
 
// Driver program to test above function
int main()
{
 
    panLipogramChecker("The quick brown fox \
                        jumped over the lazy dog");
    panLipogramChecker("The quick brown fox \
                        jumps over the lazy dog");
    panLipogramChecker("The quick brown fox \
                        jum over the lazy dog");
    return 0;
}
 
// This code is contributed by rajsanghavi9.

`Java`

// Java program to check for a Pangrammatic
// Lipogram O(n) approach
 
/*
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
*/
 
import java.util.*;
 
class GFG {
 
    // function to check for Pangrammatic Lipogram
    static void panLipogramChecker(String s)
    {
 
        // using map to keep count of the
        // occurence of each letter
        HashMap mp = new HashMap<>();
 
        for (char c = 'a'; c <= 'z'; c++) {
            mp.put(c, 0);
        }
 
        s = s.toLowerCase();
 
        int i, n = s.length();
        for (i = 0; i <= n - 1; i++) {
            if (Character.isAlphabetic(s.charAt(i))) {
 
                // increment count of characters in
                // dictionary
                mp.put(s.charAt(i),
                       mp.get(s.charAt(i)) + 1);
            }
        }
        int count_zero = 0;
 
        // Getting an iterator
        Iterator hmIterator = mp.entrySet().iterator();
 
        while (hmIterator.hasNext()) {
            Map.Entry mapElement
                = (Map.Entry)hmIterator.next();
            int marks = ((int)mapElement.getValue());
 
            if (marks == 0)
                count_zero++;
        }
        if (count_zero > 1) {
            System.out.println(
                "Not a pangram, but might be a lipogram.");
        }
        else if (count_zero == 1) {
            System.out.println("Pangrammatic Lipogram.");
        }
        else if (count_zero < 1) {
            System.out.println("Pangram.");
        }
    }
 
    public static void main(String[] args)
    {
        panLipogramChecker(
            "The quick brown fox jumped over the lazy dog");
        panLipogramChecker(
            "The quick brown fox jumps over the lazy dog");
        panLipogramChecker(
            "The quick brown fox jum over the lazy dog");
    }
}
 
// This code is contributed by rajsanghavi9.

`Python3`

# Python program to check for a Pangrammatic
# Lipogram O(n) approach
 
'''
Category                No of letters unmatched
Pangram                     0
Lipogram                    >1
Pangrammatic Lipogram       1
'''
 
# function to check for Pangrammatic Lipogram
 
 
def panLipogramChecker(s):
 
    # dictionary to keep count of the
    # occurence of each letter
    counter = {'a': 0, 'b': 0, 'c': 0, 'd': 0, 'e': 0,
               'f': 0, 'g': 0, 'h': 0, 'i': 0, 'j': 0,
               'k': 0, 'l': 0, 'm': 0, 'n': 0, 'o': 0,
               'p': 0, 'q': 0, 'r': 0, 's': 0, 't': 0,
               'u': 0, 'v': 0, 'w': 0, 'x': 0, 'y': 0,
               'z': 0}
 
    s = s.lower()
 
    # increment count of characters in dictionary
    for i in s:
        if (i.isalpha()):
            counter[i] += 1
 
    # returns a list containing the values of all
    # the keys in h=the dictionary
    b = list(counter.values())
 
    if (b.count(0) > 1):
        print("Not a pangram, but might be a lipogram.")
    else if (b.count(0) == 1):
        print("Pangrammatic Lipogram.")
    else if (b.count(0) < 1):
        print("Pangram.")
 
 
# Driver program to test above function
def main():
    panLipogramChecker("The quick brown fox \
                        jumped over the lazy dog")
    panLipogramChecker("The quick brown fox \
                        jumps over the lazy dog")
    panLipogramChecker("The quick brown fox \
                        jum over the lazy dog")
 
 
if __name__ == '__main__':
    main()

输出：

Pangrammatic Lipogram
Pangram
Not a Pangram but might a Lipogram

时间复杂度： O(N)，其中 N 是输入字符串中的字符数。

参考： https://www.wikiwand.com/en/Lipogram