Formatted question description: https://leetcode.ca/all/1638.html

# 1638. Count Substrings That Differ by One Character (Medium)

Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.

For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way.

Return the number of substrings that satisfy the condition above.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "aba", t = "baba"
Output: 6
Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character:
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
The underlined portions are the substrings that are chosen from s and t.


​​Example 2:

Input: s = "ab", t = "bb"
Output: 3
Explanation: The following are the pairs of substrings from s and t that differ by 1 character:
("ab", "bb")
("ab", "bb")
("ab", "bb")
​​​​The underlined portions are the substrings that are chosen from s and t.


Example 3:

Input: s = "a", t = "a"
Output: 0


Example 4:

Input: s = "abe", t = "bbc"
Output: 10


Constraints:

• 1 <= s.length, t.length <= 100
• s and t consist of lowercase English letters only.

Related Topics:
Hash Table, String, Trie, Rolling Hash

## Solution 1.

Intuition: We can find each pair of s[i] != t[j]. Then try to extend both sides when s[i + t] == t[j + t]. If we have left steps extended on the left side and right steps on the right side, we have (left + 1) * (right + 1) options for this { i, j } case.

Example:

s = xbabc
t = ybbbc


For i = 2 and j = 2, we have s[i] = a and t[j] = b that doesn’t match. Now look leftwards, we can extend left-side by 1 time due to b, and extend right-side by 2 times due to bc. So for this specific center { i = 2, j = 2 }, we have 2 * 3 = 6 options.

// OJ: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/

// Time: O(MN * min(M, N))
// Space: O(1)
class Solution {
public:
int countSubstrings(string s, string t) {
int M = s.size(), N = t.size(), ans = 0;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (s[i] == t[j]) continue;
int left = 1, right = 1;
while (i - left >= 0 && j - left >= 0 && s[i - left] == t[j - left]) ++left;
while (i + right < M && j + right < N && s[i + right] == t[j + right]) ++right;
ans += left * right;
}
}
return ans;
}
};


## Solution 2.

We can precompute the left and right values to save time.

// OJ: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/

// Time: O(MN)
// Space: O(MN)
class Solution {
public:
int countSubstrings(string s, string t) {
int M = s.size(), N = t.size(), ans = 0, left = {}, right = {};
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
left[i + 1][j + 1] = s[i] == t[j] ? left[i][j] + 1 : 0;
}
}
for (int i = M - 1; i >= 0; --i) {
for (int j = N - 1; j >= 0; --j) {
right[i][j] = s[i] == t[j] ? right[i + 1][j + 1] + 1 : 0;
}
}
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (s[i] != t[j]) ans += (1 + left[i][j]) * (1 + right[i + 1][j + 1]);
}
}
return ans;
}
};


## Solution 3.

Consider the following s and t and we are using x and y as the differing characters.

s=ab[x]c
t=ab[y]c


When we start from i = 0, j = 0, and reaches i = 2, j = 2, since s[i] != t[j], pre is updated as cur = 3, and cur is reset to 0. We add 3 to the answer which covers

ab[x]
ab[y]

b[x]
b[y]

[x]
[y]


When we reach i = 3, j = 3, we add pre = 3 to answer again, which covers

ab[x]c
ab[y]c

b[x]c
b[y]c

[x]c
[y]c


So the pre is the same as the left value in previous solutions. The right value is achieved through adding the pre value repetitively for repeating right-side characters.

The i and j of helper function are the starting indexes of our scanning. Note that 0, 0 should be only included once so j starts from 1 in the second loop.

// OJ: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/

// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/917985/JavaC%2B%2BPython-Time-O(nm)-Space-O(1)
class Solution {
int helper(string s, string t, int i, int j) {
int ans = 0, pre = 0, cur = 0;
for (int n = s.size(), m = t.size(); i < n && j < m; ++i, ++j) {
cur++;
if (s[i] != t[j]) pre = cur, cur = 0;
ans += pre;
}
return ans;
}
public:
int countSubstrings(string s, string t) {
int ans = 0 ;
for (int i = 0; i < s.size(); ++i) ans += helper(s, t, i, 0);
for (int j = 1; j < t.size(); ++j) ans += helper(s, t, 0, j);
return ans;
}
};


Java

class Solution {
public int countSubstrings(String s, String t) {
int count = 0;
int length = s.length();
for (int i = 0; i < length; i++) {
for (int j = i + 1; j <= length; j++) {
String substr = s.substring(i, j);
count += countStrings(substr, t);
}
}
return count;
}

public int countStrings(String s, String t) {
int count = 0;
int length = s.length();
int maxStart = t.length() - length;
for (int i = 0; i <= maxStart; i++) {
if (differByOne(s, t.substring(i, i + length)))
count++;
}
return count;
}

public boolean differByOne(String s, String t) {
if (s.length() != t.length() || s.equals(t))
return false;
int differ = 0;
int length = s.length();
for (int i = 0; i < length && differ <= 1; i++) {
if (s.charAt(i) != t.charAt(i))
differ++;
}
return differ == 1;
}
}