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Posts tagged as “string”

花花酱 LeetCode 1541. Minimum Insertions to Balance a Parentheses String

Given a parentheses string s containing only the characters '(' and ')'. A parentheses string is balanced if:

  • Any left parenthesis '(' must have a corresponding two consecutive right parenthesis '))'.
  • Left parenthesis '(' must go before the corresponding two consecutive right parenthesis '))'.

For example, "())""())(())))" and "(())())))" are balanced, ")()""()))" and "(()))" are not balanced.

You can insert the characters ‘(‘ and ‘)’ at any position of the string to balance it if needed.

Return the minimum number of insertions needed to make s balanced.

Example 1:

Input: s = "(()))"
Output: 1
Explanation: The second '(' has two matching '))', but the first '(' has only ')' matching. We need to to add one more ')' at the end of the string to be "(())))" which is balanced.

Example 2:

Input: s = "())"
Output: 0
Explanation: The string is already balanced.

Example 3:

Input: s = "))())("
Output: 3
Explanation: Add '(' to match the first '))', Add '))' to match the last '('.

Example 4:

Input: s = "(((((("
Output: 12
Explanation: Add 12 ')' to balance the string.

Example 5:

Input: s = ")))))))"
Output: 5
Explanation: Add 4 '(' at the beginning of the string and one ')' at the end. The string becomes "(((())))))))".

Constraints:

  • 1 <= s.length <= 10^5
  • s consists of '(' and ')' only.

Solution: Counting

Count how many close parentheses we need.

  1. if s[i] is ‘)’, we decrease the counter.
    1. if counter becomes negative, means we need to insert ‘(‘
      1. increase ans by 1, increase the counter by 2, we need one more ‘)’
      2. ‘)’ -> ‘()’
  2. if s[i] is ‘(‘
    1. if we have an odd counter, means there is a unbalanced ‘)’ e.g. ‘(()(‘, counter is 3
      1. need to insert ‘)’, decrease counter, increase ans
      2. ‘(()(‘ -> ‘(())(‘, counter = 2
    2. increase counter by 2, each ‘(‘ needs two ‘)’s. ‘(())(‘ -> counter = 4
  3. Once done, if counter is greater than zero, we need insert that much ‘)s’
    1. counter = 5, ‘((()’ -> ‘((())))))’

Time complexity: O(n)
Space complexity: O(1)

C++

花花酱 LeetCode 1540. Can Convert String in K Moves

Given two strings s and t, your goal is to convert s into t in kmoves or less.

During the ith (1 <= i <= k) move you can:

  • Choose any index j (1-indexed) from s, such that 1 <= j <= s.length and j has not been chosen in any previous move, and shift the character at that index i times.
  • Do nothing.

Shifting a character means replacing it by the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Shifting a character by i means applying the shift operations i times.

Remember that any index j can be picked at most once.

Return true if it’s possible to convert s into t in no more than k moves, otherwise return false.

Example 1:

Input: s = "input", t = "ouput", k = 9
Output: true
Explanation: In the 6th move, we shift 'i' 6 times to get 'o'. And in the 7th move we shift 'n' to get 'u'.

Example 2:

Input: s = "abc", t = "bcd", k = 10
Output: false
Explanation: We need to shift each character in s one time to convert it into t. We can shift 'a' to 'b' during the 1st move. However, there is no way to shift the other characters in the remaining moves to obtain t from s.

Example 3:

Input: s = "aab", t = "bbb", k = 27
Output: true
Explanation: In the 1st move, we shift the first 'a' 1 time to get 'b'. In the 27th move, we shift the second 'a' 27 times to get 'b'.

Constraints:

  • 1 <= s.length, t.length <= 10^5
  • 0 <= k <= 10^9
  • st contain only lowercase English letters.

Solution: HashTable

Count how many times a d-shift has occurred.
a -> c is a 2-shift, z -> b is also 2-shift
a -> d is a 3-shift
a -> a is a 0-shift that we can skip
if a d-shift happened for the first time, we need at least d moves
However, if it happened for c times, we need at least d + 26 * c moves
e.g. we can do a 2-shift at the 2nd move, do another one at 2 + 26 = 28th move and do another at 2 + 26*2 = 54th move, and so on.
Need to find maximum move we need and make sure that one is <= k.
Since we can pick any index to shift, so the order doesn’t matter. We can start from left to right.

Time complexity: O(n)
Space complexity: O(26) = O(1)

C++

花花酱 1528. Shuffle String

Given a string s and an integer array indices of the same length.

The string s will be shuffled such that the character at the ith position moves to indices[i] in the shuffled string.

Return the shuffled string.

Example 1:

Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"
Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.

Example 2:

Input: s = "abc", indices = [0,1,2]
Output: "abc"
Explanation: After shuffling, each character remains in its position.

Example 3:

Input: s = "aiohn", indices = [3,1,4,2,0]
Output: "nihao"

Example 4:

Input: s = "aaiougrt", indices = [4,0,2,6,7,3,1,5]
Output: "arigatou"

Example 5:

Input: s = "art", indices = [1,0,2]
Output: "rat"

Constraints:

  • s.length == indices.length == n
  • 1 <= n <= 100
  • s contains only lower-case English letters.
  • 0 <= indices[i] < n
  • All values of indices are unique (i.e. indices is a permutation of the integers from 0 to n - 1).

Solution: Simulation

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

C++

Java

Pyhton3

花花酱 LeetCode 1520. Maximum Number of Non-Overlapping Substrings

Given a string s of lowercase letters, you need to find the maximum number of non-empty substrings of s that meet the following conditions:

  1. The substrings do not overlap, that is for any two substrings s[i..j] and s[k..l], either j < k or i > l is true.
  2. A substring that contains a certain character c must also contain all occurrences of c.

Find the maximum number of substrings that meet the above conditions. If there are multiple solutions with the same number of substrings, return the one with minimum total length. It can be shown that there exists a unique solution of minimum total length.

Notice that you can return the substrings in any order.

Example 1:

Input: s = "adefaddaccc"
Output: ["e","f","ccc"]
Explanation: The following are all the possible substrings that meet the conditions:
[
  "adefaddaccc"
  "adefadda",
  "ef",
  "e",
  "f",
  "ccc",
]
If we choose the first string, we cannot choose anything else and we'd get only 1. If we choose "adefadda", we are left with "ccc" which is the only one that doesn't overlap, thus obtaining 2 substrings. Notice also, that it's not optimal to choose "ef" since it can be split into two. Therefore, the optimal way is to choose ["e","f","ccc"] which gives us 3 substrings. No other solution of the same number of substrings exist.

Example 2:

Input: s = "abbaccd"
Output: ["d","bb","cc"]
Explanation: Notice that while the set of substrings ["d","abba","cc"] also has length 3, it's considered incorrect since it has larger total length.

Constraints:

  • 1 <= s.length <= 10^5
  • s contains only lowercase English letters.

Solution: Greedy

Observation: If a valid substring contains shorter valid strings, ignore the longer one and use the shorter one.
e.g. “abbeefba” is a valid substring, however, it includes “bbeefb”, “ee”, “f” three valid substrings, thus it won’t be part of the optimal solution, since we can always choose a shorter one, with potential to have one or more non-overlapping substrings. For “bbeefb”, again it includes “ee” and “f”, so it won’t be optimal either. Thus, the optimal ones are “ee” and “f”.

  1. We just need to record the first and last occurrence of each character
  2. When we meet a character for the first time we must include everything from current pos to it’s last position. e.g. “abbeefba” | ccc, from first ‘a’ to last ‘a’, we need to cover “abbeefba”
  3. If any character in that range has larger end position, we must extend the string. e.g. “abcabbcc” | efg, from first ‘a’ to last ‘a’, we have characters ‘b’ and ‘c’, so we have to extend the string to cover all ‘b’s and ‘c’s. Our first valid substring extended from “abca” to “abcabbcc”.
  4. If any character in the covered range has a smallest first occurrence, then it’s an invalid substring. e.g. ab | “cbc”, from first ‘c’ to last ‘c’, we have ‘b’, but ‘b’ is not fully covered, thus “cbc” is an invalid substring.
  5. For the first valid substring, we append it to the ans array. “abbeefba” => ans = [“abbeefba”]
  6. If we find a shorter substring that is full covered by the previous valid substring, we replace that substring with the shorter one. e.g.
    “abbeefba” | ccc => ans = [“abbeefba”]
    abbeefba” | ccc => ans = [“bbeefb”]
    “abbeefba” | ccc => ans = [“ee”]
  7. If the current substring does not overlap with previous one, append it to ans array.
    “abbeefba” | ccc => ans = [“ee”]
    “abbeefba” | ccc => ans = [“ee”, “f”]
    “abbeefbaccc” => ans = [“ee”, “f”, “ccc”]

Time complexity: O(n)
Space complexity: O(1)

C++

花花酱 LeetCode 1507. Reformat Date

Given a date string in the form Day Month Year, where:

  • Day is in the set {"1st", "2nd", "3rd", "4th", ..., "30th", "31st"}.
  • Month is in the set {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"}.
  • Year is in the range [1900, 2100].

Convert the date string to the format YYYY-MM-DD, where:

  • YYYY denotes the 4 digit year.
  • MM denotes the 2 digit month.
  • DD denotes the 2 digit day.

Example 1:

Input: date = "20th Oct 2052"
Output: "2052-10-20"

Example 2:

Input: date = "6th Jun 1933"
Output: "1933-06-06"

Example 3:

Input: date = "26th May 1960"
Output: "1960-05-26"

Constraints:

  • The given dates are guaranteed to be valid, so no error handling is necessary.

Solution: String + HashTable

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

C++

Java

Python