**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][0] -> 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][0] -> 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][0] = 1 + dp[i - 1][j - 1][0];
dp[i][j][1] = dp[i - 1][j - 1][1];
} else
dp[i][j][1] = 1 +
dp[i - 1][j - 1]
[0]; // 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][1];
}
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][0] = 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
Final Answer :
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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