Skip to content
Related Articles
Count substrings of same length differing by a single character from two given strings
• Difficulty Level : Hard
• Last Updated : 23 Apr, 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:

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 ``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

 ``
Output:
`6`

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up