1910. Remove All Occurrences of a Substring

Description

Given two strings s and part, perform the following operation on s until all occurrences of the substring part are removed:

• Find the leftmost occurrence of the substring part and remove it from s.

Return s after removing all occurrences of part.

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

 

Example 1:

Input: s = "daabcbaabcbc", part = "abc"
Output: "dab"
Explanation: The following operations are done:
- s = "daabcbaabcbc", remove "abc" starting at index 2, so s = "dabaabcbc".
- s = "dabaabcbc", remove "abc" starting at index 4, so s = "dababc".
- s = "dababc", remove "abc" starting at index 3, so s = "dab".
Now s has no occurrences of "abc".

Example 2:

Input: s = "axxxxyyyyb", part = "xy"
Output: "ab"
Explanation: The following operations are done:
- s = "axxxxyyyyb", remove "xy" starting at index 4 so s = "axxxyyyb".
- s = "axxxyyyb", remove "xy" starting at index 3 so s = "axxyyb".
- s = "axxyyb", remove "xy" starting at index 2 so s = "axyb".
- s = "axyb", remove "xy" starting at index 1 so s = "ab".
Now s has no occurrences of "xy".

 

Constraints:

• 1 <= s.length <= 1000
• 1 <= part.length <= 1000
• s​​​​​​ and part consists of lowercase English letters.

 

Solutions

Solution: Stack

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

 

JavaScript

/**
 * @param {string} s
 * @param {string} part
 * @return {string}
 */
const removeOccurrences = function (s, part) {
  const size = part.length;
  const result = [];
  const isMatch = end => {
    if (end < size) return false;
    let current = 0;

    for (let index = end - size; index < end; index++) {
      if (result[index] !== part[current]) return false;
      current += 1;
    }
    return true;
  };
  let current = 0;

  for (const element of s) {
    result[current++] = element;

    if (!isMatch(current)) continue;
    current -= size;
  }
  return result.slice(0, current).join('');
};