给定两个字符串S1，S2的长度N和M分别与两个正整数N1和N2中，任务是找到的最大计数非重叠通过连接字符串s1其是等于S2 S1的子序列中，n1倍和字符串s2 ， n2次。
例子：
Input: S1 = “acb”, S2 = “ab”, N1 = 4, N2 = 2
Output: 2
Explanation:
Concatenating the string S1, N1 ( = 4) times modifies S1 to “acbacbacbacb”.
Concatenating the string S2, N2 ( = 2) times modifies S2 to “abab”.
Since the string S2 occurs twice as a non-overlapping subsequence in S1, the required output is 2.
Input: S1 = “abc”, S2 = “a”, N1 = 1, N2 = 1
Output: 1
方法：这个问题可以通过检查一个字符串是否是另一个字符串的子序列的概念来解决。
请按照以下步骤解决问题：
- 遍历字符串S和检查的字符，如果所有S2的字符存在的字符串S1或不在家。如果发现为假，则S1 的此类子序列不可能通过将字符串S1 , N1次和字符串S2 , N2次连接而使其等于S2 。
- 循环迭代字符串S1的字符N1次。对于每个第i个索引，检查S1 的任何子序列是否存在直到第i个索引，即等于S2或不。如果发现为真，则增加计数。
- 最后，执行上述操作后，将除以N2得到的计数打印为所需答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count maximum number of
// occurrences of s2 as subsequence in s1
// by concatenating s1, n1 times and s2, n2 times
int getMaxRepetitions(string s1, int n1,
string s2, int n2)
{
int temp1[26] = {0}, temp2[26] = {0};
for(char i:s1) temp1[i - 'a']++;
for(char i:s2) temp2[i - 'a']++;
for(int i = 0; i < 26; i++)
{
if(temp2[i] > temp1[i]) return 0;
}
// Stores number of times
// s1 is traversed
int s1_reps = 0;
// Stores number of times
// s2 is traversed
int s2_reps = 0;
// Mapping index of s2 to number
// of times s1 and s2 are traversed
map > s2_index_to_reps;
s2_index_to_reps[0] = {0, 0};
// Stores index of s1 circularly
int i = 0;
// Stores index of s2 circularly
int j = 0;
// Traverse the string s1, n1 times
while (s1_reps < n1){
// If current character of both
// the string are equal
if (s1[i] == s2[j])
// Update j
j += 1;
// Update i
i += 1;
// If j is length of s2
if (j == s2.size())
{
// Update j for
// circular traversal
j = 0;
// Update s2_reps
s2_reps += 1;
}
// If i is length of s1
if (i == s1.size())
{
// Update i for
// circular traversal
i = 0;
// Update s1_reps
s1_reps += 1;
// If already mapped j
// to (s1_reps, s2_reps)
if (s2_index_to_reps.find(j) !=
s2_index_to_reps.end())
break;
// Mapping j to (s1_reps, s2_reps)
s2_index_to_reps[j] = {s1_reps, s2_reps};
}
}
// If s1 already traversed n1 times
if (s1_reps == n1)
return s2_reps / n2;
// Otherwis, traverse string s1 by multiple of
// s1_reps and update both s1_reps and s2_reps
int initial_s1_reps = s2_index_to_reps[j].first ,
initial_s2_reps = s2_index_to_reps[j].second;
int loop_s1_reps = s1_reps - initial_s1_reps;
int loop_s2_reps = s2_reps - initial_s2_reps;
int loops = (n1 - initial_s1_reps);
// Update s2_reps
s2_reps = initial_s2_reps + loops * loop_s2_reps;
// Update s1_reps
s1_reps = initial_s1_reps + loops * loop_s1_reps;
// If s1 is traversed less than n1 times
while (s1_reps < n1)
{
// If current character in both
// the string are equal
if (s1[i] == s2[j])
// Update j
j += 1;
// Update i
i += 1;
// If i is length of s1
if (i == s1.size())
{
// Update i for circular traversal
i = 0;
// Update s1_reps
s1_reps += 1;
}
// If j is length of ss
if (j == s2.size())
{
// Update j for circular traversal
j = 0;
// Update s2_reps
s2_reps += 1;
}
}
return s2_reps / n2;
}
// Driver Code
int main()
{
string s1 = "acb";
int n1 = 4;
string s2 = "ab";
int n2 = 2;
cout << (getMaxRepetitions(s1, n1, s2, n2));
return 0;
}
// This code is contributed by mohit kumar 29. Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to count maximum number of
// occurrences of s2 as subsequence in s1
// by concatenating s1, n1 times and s2, n2 times
static int getMaxRepetitions(String s1, int n1,
String s2, int n2)
{
int temp1[] = new int[26], temp2[] = new int[26];
for(char i : s1.toCharArray())
temp1[i - 'a']++;
for(char i : s2.toCharArray())
temp2[i - 'a']++;
for(int i = 0; i < 26; i++)
{
if (temp2[i] > temp1[i])
return 0;
}
// Stores number of times
// s1 is traversed
int s1_reps = 0;
// Stores number of times
// s2 is traversed
int s2_reps = 0;
// Mapping index of s2 to number
// of times s1 and s2 are traversed
HashMap s2_index_to_reps = new HashMap<>();
s2_index_to_reps.put(0, new int[] { 0, 0 });
// Stores index of s1 circularly
int i = 0;
// Stores index of s2 circularly
int j = 0;
// Traverse the string s1, n1 times
while (s1_reps < n1)
{
// If current character of both
// the string are equal
if (s1.charAt(i) == s2.charAt(j))
// Update j
j += 1;
// Update i
i += 1;
// If j is length of s2
if (j == s2.length())
{
// Update j for
// circular traversal
j = 0;
// Update s2_reps
s2_reps += 1;
}
// If i is length of s1
if (i == s1.length())
{
// Update i for
// circular traversal
i = 0;
// Update s1_reps
s1_reps += 1;
// If already mapped j
// to (s1_reps, s2_reps)
if (!s2_index_to_reps.containsKey(j))
break;
// Mapping j to (s1_reps, s2_reps)
s2_index_to_reps.put(
j, new int[] { s1_reps, s2_reps });
}
}
// If s1 already traversed n1 times
if (s1_reps == n1)
return s2_reps / n2;
// Otherwis, traverse string s1 by multiple of
// s1_reps and update both s1_reps and s2_reps
int initial_s1_reps = s2_index_to_reps.get(j)[0],
initial_s2_reps = s2_index_to_reps.get(j)[1];
int loop_s1_reps = s1_reps - initial_s1_reps;
int loop_s2_reps = s2_reps - initial_s2_reps;
int loops = (n1 - initial_s1_reps);
// Update s2_reps
s2_reps = initial_s2_reps + loops * loop_s2_reps;
// Update s1_reps
s1_reps = initial_s1_reps + loops * loop_s1_reps;
// If s1 is traversed less than n1 times
while (s1_reps < n1)
{
// If current character in both
// the string are equal
if (s1.charAt(i) == s2.charAt(j))
// Update j
j += 1;
// Update i
i += 1;
// If i is length of s1
if (i == s1.length())
{
// Update i for circular traversal
i = 0;
// Update s1_reps
s1_reps += 1;
}
// If j is length of ss
if (j == s2.length())
{
// Update j for circular traversal
j = 0;
// Update s2_reps
s2_reps += 1;
}
}
return s2_reps / n2;
}
// Driver Code
public static void main(String[] args)
{
String s1 = "acb";
int n1 = 4;
String s2 = "ab";
int n2 = 2;
System.out.println(
getMaxRepetitions(s1, n1, s2, n2));
}
}
// This code is contributed by Kingash Python3
# Python3 program for the above approach
# Function to count maximum number of
# occurrences of s2 as subsequence in s1
# by concatenating s1, n1 times and s2, n2 times
def getMaxRepetitions(s1, n1, s2, n2):
# If all the characters of s2 are not present in s1
if any(c for c in set(s2) if c not in set(s1)):
return 0
# Stores number of times
# s1 is traversed
s1_reps = 0
# Stores number of times
# s2 is traversed
s2_reps = 0
# Mapping index of s2 to number
# of times s1 and s2 are traversed
s2_index_to_reps = { 0 : (0, 0)}
# Stores index of s1 circularly
i = 0
# Stores index of s2 circularly
j = 0
# Traverse the string s1, n1 times
while s1_reps < n1:
# If current character of both
# the string are equal
if s1[i] == s2[j]:
# Update j
j += 1
# Update i
i += 1
# If j is length of s2
if j == len(s2):
# Update j for
# circular traversal
j = 0
# Update s2_reps
s2_reps += 1
# If i is length of s1
if i == len(s1):
# Update i for
# circular traversal
i = 0
# Update s1_reps
s1_reps += 1
# If already mapped j
# to (s1_reps, s2_reps)
if j in s2_index_to_reps:
break
# Mapping j to (s1_reps, s2_reps)
s2_index_to_reps[j] = (s1_reps, s2_reps)
# If s1 already traversed n1 times
if s1_reps == n1:
return s2_reps // n2
# Otherwis, traverse string s1 by multiple of
# s1_reps and update both s1_reps and s2_reps
initial_s1_reps, initial_s2_reps = s2_index_to_reps[j]
loop_s1_reps = s1_reps - initial_s1_reps
loop_s2_reps = s2_reps - initial_s2_reps
loops = (n1 - initial_s1_reps)
# Update s2_reps
s2_reps = initial_s2_reps + loops * loop_s2_reps
# Update s1_reps
s1_reps = initial_s1_reps + loops * loop_s1_reps
# If s1 is traversed less than n1 times
while s1_reps < n1:
# If current character in both
# the string are equal
if s1[i] == s2[j]:
# Update j
j += 1
# Update i
i += 1
# If i is length of s1
if i == len(s1):
# Update i for circular traversal
i = 0
# Update s1_reps
s1_reps += 1
# If j is length of ss
if j == len(s2):
# Update j for circular traversal
j = 0
# Update s2_reps
s2_reps += 1
return s2_reps // n2
# Driver Code
if __name__ == '__main__':
s1 = "acb"
n1 = 4
s2 = "ab"
n2 = 2
print(getMaxRepetitions(s1, n1, s2, n2))输出：
2时间复杂度： O((N + M) * n1)
辅助空间： O(1)
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