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:
- (“a”, “b”)
- (“a”, “b”)
- (“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++ program 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 Codeint 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 approachclass GFG{ // Function to count the number of// subStrings of equal length which// differ by a single characterstatic 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 Codepublic 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 characterdef 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 CodeS = "aba"T = "baba"# Function Callprint(countSubstrings(S, T)) |
C#
// C# program for the above approachusing System;class GFG{ // Function to count the number of// subStrings of equal length which// differ by a single characterstatic 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 Codepublic 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 characterfunction 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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