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📜  计算与两个给定字符串相差一个字符的相同长度的子字符串

📅  最后修改于: 2021-09-07 04:52:25             🧑  作者: Mango

给定两个长度分别为NM 的字符串ST ,任务是计算从两个字符串中获得相同长度子字符串的方法的数量,以便它们具有单个不同的字符。

例子:

朴素的方法:最简单的方法是从给定的字符串生成所有可能的子字符串,然后计算所有可能的相同长度的子字符串对,这些子字符串可以通过更改单个字符来变得相等。

时间复杂度: O(N 3 *M 3 )
辅助空间: O(N 2 )

有效的方法:为了优化上述方法,其思想是同时迭代给定字符串的所有字符,并且对于每对不同的字符,从当前不同字符的下一个索引开始计算所有这些长度相等的子字符串。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 gauravrajput1


Python3
# 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


Javascript


输出:
6

时间复杂度: O(N*M)
辅助空间: O(1)

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