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Count substrings of same length differing by a single character from two given strings

  • Difficulty Level : Hard
  • Last Updated : 30 Sep, 2021

Given two strings S and T of length N and M respectively, the task is to count the number of ways of obtaining same-length substring from both the strings such that they have a single different character.

Examples:

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Input: S = “ab”, T = “bb”
Output: 3
Explanation: The following are the pairs of substrings from S and T differ by a single character:



  1. (“a”, “b”)
  2. (“a”, “b”)
  3. (“ab”, “bb”)

Input: S = “aba”, T = “baba”
Output: 6

Naive Approach: The simplest approach is to generate all possible substrings from both the given strings and then count all possible pairs of substrings of the same lengths which can be made equal by changing a single character. 

Time Complexity: O(N3*M3)
Auxiliary Space: O(N2)

Efficient Approach: To optimize the above approach, the idea is to iterate over all characters of both the given strings simultaneously and for each pair of different characters, count all those substrings of equal length starting from the next index of the current different character. Print the count obtained after checking for all pairs of different characters.

Below is the implementation of the above approach:

C++




// C++ pprogram for the above approach
#include <bits/stdc++.h>
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 condition
                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 condition
                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 condition
                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 condition
                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




<script>
 
// Javascript program to implement
// the above approach
 
// Function to count the number of
// subStrings of equal length which
// differ by a single character
function countSubStrings(s, t)
{
      
    // Stores the count of
    // pairs of subStrings
    let answ = 0;
  
    // Traverse the String s
    for(let i = 0; i < s.length; i++)
    {
          
        // Traverse the String t
        for(let j = 0; j < t.length; j++)
        {
              
            // Different character
            if (t[j] != s[i])
            {
                  
                // Increment the answer
                answ += 1;
                  
                let k = 1;
                let z = -1;
                let 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 condition
                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
    let S = "aba";
    let T = "baba";
  
    // Function Call
    document.write(countSubStrings(S, T));
 
// This code is contributed by souravghosh0416.
</script>
Output: 
6

 

Time Complexity: O(N*M*max(N,M))
Auxiliary Space: O(1)




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