1638. Count Substrings That Differ by One Character

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

Solution: Continuous Matching

• Time complexity: O(mn)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {string} s
 * @param {string} t
 * @return {number}
 */
const countSubstrings = function (s, t) {
  const m = s.length;
  const n = t.length;
  let result = 0;

  const checkSubstring = (a, b) => {
    let current = (previous = 0);

    while (a < m && b < n) {
      current += 1;

      if (s[a] !== t[b]) {
        previous = current;
        current = 0;
      }
      result += previous;
      a += 1;
      b += 1;
    }
  };

  for (let index = 0; index < m; index++) checkSubstring(index, 0);
  for (let index = 1; index < n; index++) checkSubstring(0, index);
  return result;
};