README_EN.md

1638. Count Substrings That Differ by One Character

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Description

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.

 

Constraints:

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

Solutions

Python3

class Solution:
    def countSubstrings(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        ans = 0
        for i in range(m):
            for j in range(n):
                if s[i] != t[j]:
                    l = r = 1
                    while i - l >= 0 and j - l >= 0 and s[i - l] == t[j - l]:
                        l += 1
                    while i + r < m and j + r < n and s[i + r] == t[j + r]:
                        r += 1
                    ans += l * r
        return ans

Java

class Solution {
    public int countSubstrings(String s, String t) {
        int m = s.length(), n = t.length();
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (s.charAt(i) != t.charAt(j)) {
                    int l = 1, r = 1;
                    while (i - l >= 0 && j - l >= 0 && s.charAt(i - l) == t.charAt(j - l)) {
                        ++l;
                    }
                    while (i + r < m && j + r < n && s.charAt(i + r) == t.charAt(j + r)) {
                        ++r;
                    }
                    ans += l * r;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countSubstrings(string s, string t) {
        int m = s.size(), n = t.size();
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (s[i] == t[j]) continue;
                int l = 1, r = 1;
                while (i - l >= 0 && j - l >= 0 && s[i - l] == t[j - l]) ++l;
                while (i + r < m && j + r < n && s[i + r] == t[j + r]) ++r;
                ans += l * r;
            }
        }
        return ans;
    }
};

Go

func countSubstrings(s string, t string) int {
	m, n := len(s), len(t)
	ans := 0
	for i := range s {
		for j := range t {
			if s[i] == t[j] {
				continue
			}
			l, r := 1, 1
			for i-l >= 0 && j-l >= 0 && s[i-l] == t[j-l] {
				l++
			}
			for i+r < m && j+r < n && s[i+r] == t[j+r] {
				r++
			}
			ans += l * r
		}
	}
	return ans
}

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