# Number of Substrings with Single Character Difference - Microsoft Top Interview Questions

### Problem Statement :

```You are given two lowercase alphabet strings s and t. Return the number of pairs of substrings across s and t such that if we replace a single character to a different character, it becomes a substring of t.

Constraints

0 ≤ n ≤ 100 where n is the length of s

0 ≤ m ≤ 100 where m is the length of t

Example 1

Input

s = "ab"

t = "db"

Output

4

Explanation

We can have the following substrings:

"a" changed to "d"

"a" changed to "b"

"b" changed to "d"

"ab" changed to "db"```

### Solution :

```                        ```Solution in C++ :

int solve(string s, string t) {
int m = s.length(), n = t.length();

/*
* dp[i][j] -> number of same substrings ending at the ith character in s and j th chaaracter
* in t such that no change has been made to the string s 'yet'. dp[i][j] -> number of same
* substrings ending at the ith character in s and j th chaaracter in t such that exactly one
* change has been made to the string.
*/
vector<vector<vector<int>>> dp(m + 1, vector<vector<int>>(n + 1, vector<int>(2)));

for (int i = 1; i <= m; ++i) {       // ending at i - 1th character in s
for (int j = 1; j <= n; ++j) {   // ending at the j - 1 th character in t
if (s[i - 1] == t[j - 1]) {  // if the characters are same
/*
* number of equal substrings would be:
* 1. With no change -> 1 + number of equal substrings ending at s[i - 2] and t[j -
2] with no change 1 because s[i - 1] == t[j - 1], ie substring of length 1 is also
valid
* 2. With change -> number of equal substrings ending at s[i - 2] and t[j - 2] with
exatly one change.
*/
dp[i][j] = 1 + dp[i - 1][j - 1];
dp[i][j] = dp[i - 1][j - 1];
} else
dp[i][j] = 1 +
dp[i - 1][j - 1]
;  // if charactes are not same then we have only one option, ie
// to change it. So, number of equal substrings ending here ->
// 1 + number of equal substrings ending at s[i - 1] and t[j -
// 1] with no change. 1 because when s[i - 1] is changed to
// t[j - 1] we have equal substring of length 1.
}
}

// count all substrings ending at some s[i - 1] and t[j - 1] and with exactly one change
int res = 0;
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= n; ++j) res += dp[i][j];
}

return res;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(String s, String t) {
/*
definitions :
let string s = s1 s2 s3 ..... si-1 si si+1 ..... sm-2 sm-1 sm
let string t = t1 t2 t3 ..... tj-1 tj tj+1 ..... tn-2 tn-1 tn
LEFT[i][j] = max size equal substring ending at position i in string s and position
j in string t when traversing s and t from left side RIGHT[i][j] = max size equal
substring ending at position i in string s and position j in string t when traversing s
and t from right side

Base Cases :
LEFT = 0
RIGH[M + 1][N + 1] = 0

Transitive Equations :
LEFT[i][j] = (1 + LEFT[i - 1][j - 1])   ; si == tj
LEFT[i][j] = 0                          ; si != tj
RIGHT[i][j] = (1 + RIGHT[i + 1][j + 1]) ; si == tj
RIGHT[i][j] = 0                         ; si != tj

count += (1 + LEFT[i - 1][j - 1] + RIGHT[i + 1][j + 1] + (LEFT[i - 1][j - 1] *
RIGHT[i + 1][j + 1])) ; si != tj
*/
int M = s.length(), N = t.length(), count = 0;
int[][] LEFT = new int[M + 1][N + 1], RIGHT = new int[M + 2][N + 2];
for (int i = 1; i <= M; i++) {
for (int j = 1; j <= N; j++) {
if (s.charAt(i - 1) == t.charAt(j - 1)) {
LEFT[i][j] = (1 + LEFT[i - 1][j - 1]);
}
}
}
for (int i = M; i >= 1; i--) {
for (int j = N; j >= 1; j--) {
if (s.charAt(i - 1) == t.charAt(j - 1)) {
RIGHT[i][j] = (1 + RIGHT[i + 1][j + 1]);
} else {
count += (1 + LEFT[i - 1][j - 1] + RIGHT[i + 1][j + 1]
+ (LEFT[i - 1][j - 1] * RIGHT[i + 1][j + 1]));
}
}
}
return count;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, s, t):
ans = 0

for i in range(len(s)):
for j in range(len(t)):
mismatches = 0
for k in range(min(len(s) - i, len(t) - j)):
mismatches += s[i + k] != t[j + k]
if mismatches == 1:
ans += 1

return ans```
```

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