给定三个字符串str ， str1和str2 ，任务是将str1和str2出现的对数作为字符串str中的子字符串进行计数，以使每对字符串中str1的起始索引均小于或等于str2 。
例子：
Input: str = “geeksforgeeksfor”, str1 = “geeks”, str2 = “for”
Output: 3
Explanation:
Starting indices of str1 are { 0, 8 }
Starting indices of str2 are { 5, 13 }
Possible pairs such that start index of str1 less than or equal to str2 are { (0, 5), (0, 13), (8, 13) }
Therefore, the required output is 3.
Input: str = “geeksforgeeks”, str1 = “geek”, str2 = “for”
Output: 1
方法：这个想法是在字符串str中找到子字符串str2的起始索引，并且在字符串str2的每个起始索引处，将对数增加str1的起始索引数，直到str的第i个索引。最后，打印获得的对的总数。请按照以下步骤解决问题：
- 初始化变量，说cntPairs，以存储对str1和STR2的的计数，使得STR1的起始索引小于或等于STR2。
- 初始化一个变量cntStr1 ，以将子字符串str1的计数存储在起始索引小于或等于i的str中。
- 在[0，N – 1]范围内迭代。对于字符串str的每个第i个索引，检查子字符串str1是否存在于字符串str中，其起始索引等于i 。如果确定为true，则将cntStr1的值增加1 。
- 对于每个第i个索引，检查子字符串str2是否存在于起始索引等于i的字符串。如果发现为true，则将cntPairs的值增加cntStr1 。
- 最后，打印cntPairs的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the count of pairs of str1
// and str2 in str such that starting index of
// str1 less than or equal to str2
int CountSubstring(string str, string str1, string str2)
{
// Stores count of pairs of str1 and str2
// such that the start index of str1 is
// less than or equal to str2
int cntPairs = 0;
// Stores count of substring str1 in
// str whose start index is up to i
int cntStr1 = 0;
// Iterate over the range [0, N -1]
for (int i = 0; i < str.length(); i++) {
// If substring str1 whose
// start index is i
if (str.substr(i, str1.length())
== str1) {
// Update cntStr1
cntStr1++;
}
// If substring str2 whose
// start index is i
else if (str.substr(i, str2.length())
== str2) {
// Update cntPairs
cntPairs += cntStr1;
}
}
return cntPairs;
}
// Driver Code
int main()
{
string str = "geeksforgeeksfor";
string str1 = "geeks", str2 = "for";
cout << CountSubstring(str, str1, str2);
return 0;
} Java
// Java program to implement
// the above approach
public class GFG {
// Driver Code
public static void main(String[] args)
{
String str = new String("geeksforgeeksfor");
String str1 = new String("geeks");
String str2 = new String("for");
System.out.println(CountSubstring(
str, str1, str2));
}
// Function to find the count of pairs of str1
// and str2 in str such that starting index of
// str1 less than or equal to str2
static int CountSubstring(String str,
String str1,
String str2)
{
// Stores count of pairs of str1 and str2
// such that the start index of str1 is
// less than or equal to str2
int cntPairs = 0;
// Stores count of substring str1 in
// str whose start index is up to i
int cntStr1 = 0;
// Stores length of str
int N = str.length();
// Stores length of str1
int s1 = str1.length();
// Stores length of str2
int s2 = str2.length();
// Iterate over the range [0, N -1]
for (int i = 0; i < N; i++) {
// If substring str1 whose
// starting index is i
if (((i + s1) <= N)
&& (str.substring(
i, (i + s1)))
.compareTo(str1)
== 0) {
// Update cntStr1
cntStr1++;
}
// If substring str2 whose
// starting index is i
else if (((i + s2) <= N)
&& (str.substring(
i, (i + s2)))
.compareTo(str2)
== 0) {
// Update cntPairs
cntPairs += cntStr1;
}
}
return cntPairs;
}
}Python3
# Python3 program to implement
# the above approach
# Function to find the count of pairs of str1
# and str2 in str such that starting index of
# str1 less than or equal to str2
def CountSubstring(str, str1, str2):
# Stores count of pairs of str1 and str2
# such that the start index of str1 is
# less than or equal to str2
cntStr1 = 0
# Stores count of substring str1 in
# str whose start index is up to i
cntPairs = 0
# Iterate over the range [0, N -1]
for i in range(len(str)):
# If substring str1 whose
# start index is i
if str[i : i + len(str1)] == str1:
# Update cntStr1
cntStr1 += 1
# If substring str2 whose
# start index is i
elif str[i : i + len(str2)] == str2:
# Update cntPairs
cntPairs += cntStr1
return cntPairs
# Driver Code
str = 'geeksforgeeksfor'
str1 = 'geeks'
str2 = 'for'
print(CountSubstring(str, str1, str2))
# This code is contributed by dharanendralv23C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find the count of pairs of str1
// and str2 in str such that starting index of
// str1 less than or equal to str2
static int CountSubstring(String str,
String str1,
String str2)
{
// Stores count of pairs of str1 and str2
// such that the start index of str1 is
// less than or equal to str2
int cntPairs = 0;
// Stores count of substring str1 in
// str whose start index is up to i
int cntStr1 = 0;
// Stores length of str
int N = str.Length;
// Stores length of str1
int s1 = str1.Length;
// Stores length of str2
int s2 = str2.Length;
// Iterate over the range [0, N -1]
for(int i = 0; i < N; i++)
{
// If substring str1 whose
// starting index is i
if (((i + s1) <= N) &&
(str.Substring(i, s1)).Equals(str1))
{
// Update cntStr1
cntStr1++;
}
// If substring str2 whose
// starting index is i
else if (((i + s2) <= N) &&
(str.Substring(i, s2)).Equals(str2))
{
// Update cntPairs
cntPairs += cntStr1;
}
}
return cntPairs;
}
// Driver code
static public void Main()
{
String str = "geeksforgeeksfor";
String str1 = "geeks";
String str2 = "for";
Console.Write(CountSubstring(str, str1, str2));
}
}
// This code is contributed by dharanendralv23输出：
3时间复杂度： O(| str | *(| str1 | + | str2 |))
辅助空间： O(1)