给定 2 个字符串X和Y数组，任务是找到X字符串的数量。
A string s is said to be a Superstring, if each string present in array Y is a subsequence of string s .
例子：
Input: X = {“ceo”, “alco”, “caaeio”, “ceai”}, Y = {“ec”, “oc”, “ceo”}
Output: 2
Explanation: Strings “ceo” and “caaeio” are superstrings as each string of array Y is a subset of these 2 strings. Other strings are not included in answer all strings of array Y are not
sub-sets of them.
Input: X = {“iopo”, “oaai”, “iipo”}, Y = {“oo”}
Output: 1
方法：思路是利用Hashing的概念来存储字符的频率来解决问题。请按照以下步骤解决问题：
- 初始化大小为 26 的数组，以存储数组Y中每个字符串中每个字符[az] 的最大出现次数。
- 现在考虑 X 中的每个字符串s ，
- 检查s中每个字符的频率是否大于等于上一步得到的频率
下面是上述方法的实现：
C++
// C++ implementation for the above approach
#include
using namespace std;
// Function to find total number of superstrings
int superstring(string X[], string Y[], int N, int M)
{
// Array to store max frequency
// Of each letter
int maxFreq[26];
for (int i = 0; i < 26; i++)
maxFreq[i] = 0;
for (int j = 0; j < M; j++) {
int temp[26];
for (int i = 0; i < 26; i++)
temp[i] = 0;
for (int k = 0; k < Y[j].size(); k++) {
temp[Y[j][k] - 'a']++;
}
for (int i = 0; i < 26; i++) {
maxFreq[i] = max(maxFreq[i], temp[i]);
}
}
int ans = 0;
for (int j = 0; j < N; j++) {
// Array to find frequency of each letter in string
// x
int temp[26];
for (int i = 0; i < 26; i++)
temp[i] = 0;
for (int k = 0; k < X[j].size(); k++) {
temp[X[j][k] - 'a']++;
}
int i = 0;
for (i = 0; i < 26; i++) {
// If any frequency is less in string x than
// maxFreq, then it can't be a superstring
if (temp[i] < maxFreq[i]) {
break;
}
}
if (i == 26) {
// Increment counter of x is a superstring
ans++;
}
}
return ans;
}
// Driver code
int main()
{
// Size of array X
int N = 4;
// Size of array Y
int M = 3;
string X[N] = { "ceo", "alco", "caaeio", "ceai" };
string Y[M] = { "ec", "oc", "ceo" };
cout << superstring(X, Y, N, M); // Function call
return 0;
} Java
// Java implementation for the above approach
import java.io.*;
class GFG {
// Function to find total number of superstrings
public static int superString(String X[], String Y[],
int N, int M)
{
// Array to store max frequency
// Of each letter
int[] maxFreq = new int[26];
for (int i = 0; i < 26; i++)
maxFreq[i] = 0;
for (int j = 0; j < M; j++) {
int[] temp = new int[26];
for (int k = 0; k < Y[j].length(); k++) {
temp[Y[j].charAt(k) - 'a']++;
}
for (int i = 0; i < 26; i++) {
maxFreq[i] = Math.max(maxFreq[i], temp[i]);
}
}
int ans = 0;
for (int j = 0; j < N; j++) {
// Array to find frequency of each letter in
// string x
int[] temp = new int[26];
for (int i = 0; i < 26; i++)
temp[i] = 0;
for (int k = 0; k < X[j].length(); k++) {
temp[X[j].charAt(k) - 'a']++;
}
int i = 0;
for (i = 0; i < 26; i++) {
// If any frequency is less in string x than
// maxFreq, then it can't be a superstring
if (temp[i] < maxFreq[i]) {
break;
}
}
if (i == 26) {
// Increment counter of x is a superstring
ans++;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
String[] X = new String[] { "ceo", "alco", "caaeio",
"ceai" };
String[] Y = new String[] { "ec", "oc", "ceo" };
System.out.println(
superString(X, Y, X.length, Y.length));
}
}Python3
# Python3 implementation for the above approach
# Function to find total number of superstrings
def superstring(X, Y, N, M):
# Array to store max frequency
# Of each letter
maxFreq = [0] * 26
for j in range(M):
temp = [0] * 26
for k in range(len(Y[j])):
temp[ord(Y[j][k]) - ord('a')] += 1
for i in range(26):
maxFreq[i] = max(maxFreq[i], temp[i])
ans = 0
for j in range(N):
# Array to find frequency of each letter
# in string x
temp = [0] * 26
for k in range(len(X[j])):
temp[ord(X[j][k]) - ord('a')] += 1
i = 0
while i < 26:
# If any frequency is less in x than
# maxFreq, then it can't be a superstring
if (temp[i] < maxFreq[i]):
break
i += 1
if (i == 26):
# Increment counter of x is a superstring
ans += 1
return ans
# Driver code
if __name__ == '__main__':
# Size of array X
N = 4
# Size of array Y
M = 3
X = ["ceo", "alco", "caaeio", "ceai"]
Y = [ "ec", "oc", "ceo" ]
print(superstring(X, Y, N, M)) #Function call
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
class GFG
{
// Function to find total number of superstrings
public static int superString(string[] X, string[] Y,
int N, int M)
{
// Array to store max frequency
// Of each letter
int[] maxFreq = new int[26];
for (int i = 0; i < 26; i++)
maxFreq[i] = 0;
for (int j = 0; j < M; j++) {
int[] temp = new int[26];
for (int k = 0; k < Y[j].Length; k++) {
temp[Y[j][k] - 'a']++;
}
for (int i = 0; i < 26; i++) {
maxFreq[i] = Math.Max(maxFreq[i], temp[i]);
}
}
int ans = 0;
for (int j = 0; j < N; j++) {
// Array to find frequency of each letter in
// string x
int[] temp = new int[26];
int i =0;
for ( i = 0; i < 26; i++)
temp[i] = 0;
for (int k = 0; k < X[j].Length; k++) {
temp[X[j][k] - 'a']++;
}
for ( i = 0; i < 26; i++) {
// If any frequency is less in string x than
// maxFreq, then it can't be a superstring
if (temp[i] < maxFreq[i]) {
break;
}
}
if ( i == 26) {
// Increment counter of x is a superstring
ans++;
}
}
return ans;
}
// Driver code
static void Main()
{
string[] X = new String[] { "ceo", "alco", "caaeio",
"ceai" };
string[] Y = new String[] { "ec", "oc", "ceo" };
Console.Write(
superString(X, Y, X.Length, Y.Length));
}
}
// This code is contributed b sanjoy_62.Javascript
输出：
2时间复杂度： O(N*N1 + M*M1)，其中 N = 数组 X 的大小，N1 = Maxlength(x)，M = 数组 Y 的大小，M1 = Maxlength(y)，
空间复杂度： O(1)