Count substrings of same length differing by a single character from two given strings
Last Updated :
01 Feb, 2022
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 Code
int main()
{
string S = "aba";
string T = "baba";
// Function Call
cout<<(countSubstrings(S, T));
}
// This code is contributed by 29AjayKumar
// 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 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# 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
<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>
Time Complexity: O(N*M*max(N,M))
Auxiliary Space: O(1)
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