N 个字符串中最长的公共字谜子序列

给定 N字符串。从这 N 个字符串中的每一个中找到最长的可能子序列，使得它们彼此是字谜。任务是打印所有子序列中字典顺序最大的子序列。

例子：

Input: s[] = { geeks, esrka, efrsk }
Output: ske
First string has “eks”, Second string has “esk”, third string has “esk”. These three are anagrams. “ske” is lexoigrapically large.

Input: string s[] = { loop, lol, olive }
Output: ol

编程需要懂一点英语

方法：

制作一个n*26的二维数组来存储字符串中每个字符的频率。
制作频率数组后，对每个数字进行反向遍历，找到该类型字符数最少的字符串。
完成反向遍历后，打印出现次数最少的字符，因为它给出了字典上最大的字符串。

下面是上述方法的实现。

C++

// C++ program to find longest possible
// subsequence anagram of N strings.
#include 
using namespace std;
const int MAX_CHAR = 26;
 
// function to store frequency of
// each character in each string
void frequency(int fre[][MAX_CHAR], string s[], int n)
{
    for (int i = 0; i < n; i++) {
        string str = s[i];
        for (int j = 0; j < str.size(); j++)
            fre[i][str[j] - 'a']++;       
    }
}
 
// function to Find longest possible sequence of N
// strings which is anagram to each other
void LongestSequence(int fre[][MAX_CHAR], int n)
{
    // to get lexicographical largest sequence.
    for (int i = MAX_CHAR-1; i >= 0; i--) {
 
        // find minimum of that character
        int mi = fre[0][i];
        for (int j = 1; j < n; j++)
            mi = min(fre[j][i], mi);       
 
        // print that character
        // minimum number of times
        while (mi--)
            cout << (char)('a' + i);       
    }
}
 
// Driver code
int main()
{
 
    string s[] = { "loo", "lol", "olive" };
    int n = sizeof(s)/sizeof(s[0]);
 
    // to store frequency of each character in each string
    int fre[n][26] = { 0 };
 
    // to get frequency of each character
    frequency(fre, s, n);
 
    // function call
    LongestSequence(fre, n);
 
    return 0;
}

Java

// Java program to find longest
// possible subsequence anagram
// of N strings.
class GFG
{
final int MAX_CHAR = 26;
 
// function to store frequency
// of each character in each
// string
static void frequency(int fre[][],
                      String s[], int n)
{
    for (int i = 0; i < n; i++)
    {
        String str = s[i];
        for (int j = 0;
                 j < str.length(); j++)
            fre[i][str.charAt(j) - 'a']++;    
    }
}
 
// function to Find longest
// possible sequence of N
// strings which is anagram
// to each other
static void LongestSequence(int fre[][],
                            int n)
{
    // to get lexicographical
    // largest sequence.
    for (int i = 24; i >= 0; i--)
    {
 
        // find minimum of
        // that character
        int mi = fre[0][i];
        for (int j = 1; j < n; j++)
            mi = Math.min(fre[j][i], mi);    
 
        // print that character
        // minimum number of times
        while (mi--!=0)
            System.out.print((char)('a' + i));    
    }
}
 
// Driver code
public static void main(String args[])
{
 
    String s[] = { "loo", "lol", "olive" };
    int n = s.length;
 
    // to store frequency of each
    // character in each string
    int fre[][] = new int[n][26] ;
 
    // to get frequency
    // of each character
    frequency(fre, s, n);
 
    // function call
    LongestSequence(fre, n);
}
}
 
// This code is contributed
// by Arnab Kundu

Python3

# Python3 program to find longest possible
# subsequence anagram of N strings.
 
# Function to store frequency of
# each character in each string
def frequency(fre, s, n):
 
    for i in range(0, n):
        string = s[i]
        for j in range(0, len(string)):
            fre[i][ord(string[j]) - ord('a')] += 1       
 
# Function to Find longest possible sequence 
# of N strings which is anagram to each other
def LongestSequence(fre, n):
 
    # to get lexicographical largest sequence.
    for i in range(MAX_CHAR-1, -1, -1):
 
        # find minimum of that character
        mi = fre[0][i]
        for j in range(1, n):
            mi = min(fre[j][i], mi)        
 
        # print that character
        # minimum number of times
        while mi:
            print(chr(ord('a') + i), end = "")
            mi -= 1
     
# Driver code
if __name__ == "__main__":
 
    s = ["loo", "lol", "olive"]
    n = len(s)
    MAX_CHAR = 26
 
    # to store frequency of each
    # character in each string
    fre = [[0 for i in range(26)]
              for j in range(n)]
 
    # To get frequency of each character
    frequency(fre, s, n)
 
    # Function call
    LongestSequence(fre, n)
 
# This code is contributed by
# Rituraj Jain

C#

// c# program to find longest
// possible subsequence anagram
// of N strings.
using System;
 
class GFG
{
public readonly int MAX_CHAR = 26;
 
// function to store frequency
// of each character in each
// string
public static void frequency(int[,] fre,
                             string[] s, int n)
{
    for (int i = 0; i < n; i++)
    {
        string str = s[i];
        for (int j = 0;
                 j < str.Length; j++)
        {
            fre[i, str[j] - 'a']++;
        }
    }
}
 
// function to Find longest
// possible sequence of N
// strings which is anagram
// to each other
public static void LongestSequence(int[, ] fre,
                                   int n)
{
    // to get lexicographical
    // largest sequence.
    for (int i = 24; i >= 0; i--)
    {
 
        // find minimum of
        // that character
        int mi = fre[0, i];
        for (int j = 1; j < n; j++)
        {
            mi = Math.Min(fre[j, i], mi);
        }
 
        // print that character
        // minimum number of times
        while (mi--!=0)
        {
            Console.Write((char)('a' + i));
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
 
    string[] s = new string[] {"loo", "lol", "olive"};
    int n = s.Length;
 
    // to store frequency of each
    // character in each string
    int[, ] fre = new int[n, 26];
 
    // to get frequency
    // of each character
    frequency(fre, s, n);
 
    // function call
    LongestSequence(fre, n);
}
}
 
// This code is contributed by Shrikanth13

Javascript

输出：

ol