给定分别为长度N和M的两个字符串S和T ，任务是计算从这两个字符串获得相同长度的子字符串的方式的数目，以使它们具有单个不同的字符。
例子：
Input: S = “ab”, T = “bb”
Output: 3
Explanation: The following are the pairs of substrings from S and T differ by a single character:
- (“a”, “b”)
- (“a”, “b”)
- (“ab”, “bb”)
Input: S = “aba”, T = “baba”
Output: 6
天真的方法：最简单的方法是从两个给定的字符串生成所有可能的子字符串，然后计算相同长度的所有可能的子字符串对，通过更改单个字符可以使它们相等。
时间复杂度： O(N 3 * M 3 )
辅助空间： O(N 2 )
高效方法：为了优化上述方法，其思想是同时遍历给定字符串的所有字符，对于每对不同的字符，从当前不同字符的下一个索引开始，对所有等长的子字符串进行计数。打印检查所有不同字符对后获得的计数。
下面是上述方法的实现：
C++
// C++ pprogram for the above approach
#include
using namespace std;
// Function to count the number of
// substrings of equal length which
// differ by a single character
int countSubstrings(string s, string t)
{
// Stores the count of
// pairs of substrings
int answ = 0;
// Traverse the string s
for(int i = 0; i < s.size(); i++)
{
// Traverse the string t
for(int j = 0; j < t.size(); j++)
{
// Different character
if (t[j] != s[i])
{
// Increment the answer
answ += 1;
int k = 1;
int z = -1;
int q = 1;
// Count equal substrings
// from next index
while (j + z >= 0 &&
0 <= i + z &&
s[i + z] ==
t[j + z])
{
z -= 1;
// Increment the count
answ += 1;
// Increment q
q += 1;
}
// Check the condtion
while (s.size() > i + k &&
j + k < t.size() &&
s[i + k] ==
t[j + k])
{
// Increment k
k += 1;
// Add q to count
answ += q;
// Decrement z
z = -1;
}
}
}
}
// Return the final count
return answ;
}
// Driver Code
int main()
{
string S = "aba";
string T = "baba";
// Function Call
cout<<(countSubstrings(S, T));
}
// This code is contributed by 29AjayKumar Java
// Java program for the above approach
class GFG{
// Function to count the number of
// subStrings of equal length which
// differ by a single character
static int countSubStrings(String s, String t)
{
// Stores the count of
// pairs of subStrings
int answ = 0;
// Traverse the String s
for(int i = 0; i < s.length(); i++)
{
// Traverse the String t
for(int j = 0; j < t.length(); j++)
{
// Different character
if (t.charAt(j) != s.charAt(i))
{
// Increment the answer
answ += 1;
int k = 1;
int z = -1;
int q = 1;
// Count equal subStrings
// from next index
while (j + z >= 0 &&
0 <= i + z &&
s.charAt(i + z) ==
t.charAt(j + z))
{
z -= 1;
// Increment the count
answ += 1;
// Increment q
q += 1;
}
// Check the condtion
while (s.length() > i + k &&
j + k < t.length() &&
s.charAt(i + k) ==
t.charAt(j + k))
{
// Increment k
k += 1;
// Add q to count
answ += q;
// Decrement z
z = -1;
}
}
}
}
// Return the final count
return answ;
}
// Driver Code
public static void main(String[] args)
{
String S = "aba";
String T = "baba";
// Function Call
System.out.println(countSubStrings(S, T));
}
}
// This code is contributed by gauravrajput1Python3
# Python3 program for the above approach
# Function to count the number of
# substrings of equal length which
# differ by a single character
def countSubstrings(s, t):
# Stores the count of
# pairs of substrings
answ = 0
# Traverse the string s
for i in range(len(s)):
# Traverse the string t
for j in range(len(t)):
# Different character
if t[j] != s[i]:
# Increment the answer
answ += 1
k = 1
z = -1
q = 1
# Count equal substrings
# from next index
while (
j + z >= 0 <= i + z and
s[i + z] == t[j + z]
):
z -= 1
# Increment the count
answ += 1
# Increment q
q += 1
# Check the condtion
while (
len(s) > i + k and
j + k < len(t) and
s[i + k] == t[j + k]
):
# Increment k
k += 1
# Add q to count
answ += q
# Decrement z
z = -1
# Return the final count
return answ
# Driver Code
S = "aba"
T = "baba"
# Function Call
print(countSubstrings(S, T))C#
// C# program for the above approach
using System;
class GFG
{
// Function to count the number of
// subStrings of equal length which
// differ by a single character
static int countSubStrings(String s, String t)
{
// Stores the count of
// pairs of subStrings
int answ = 0;
// Traverse the String s
for(int i = 0; i < s.Length; i++)
{
// Traverse the String t
for(int j = 0; j < t.Length; j++)
{
// Different character
if (t[j] != s[i])
{
// Increment the answer
answ += 1;
int k = 1;
int z = -1;
int q = 1;
// Count equal subStrings
// from next index
while (j + z >= 0 &&
0 <= i + z &&
s[i + z] ==
t[j + z])
{
z -= 1;
// Increment the count
answ += 1;
// Increment q
q += 1;
}
// Check the condtion
while (s.Length > i + k &&
j + k < t.Length &&
s[i + k] ==
t[j + k])
{
// Increment k
k += 1;
// Add q to count
answ += q;
// Decrement z
z = -1;
}
}
}
}
// Return the readonly count
return answ;
}
// Driver Code
public static void Main(String[] args)
{
String S = "aba";
String T = "baba";
// Function Call
Console.WriteLine(countSubStrings(S, T));
}
}
// This code is contributed by 29AjayKumar输出：
6时间复杂度： O(N * M)
辅助空间： O(1)