1392. Longest Happy Prefix

A string is called a happy prefix if is a non-empty prefix which is also a suffix (excluding itself).

Given a string s, return the longest happy prefix of s. Return an empty string “” if no such prefix exists.

1910. Remove All Occurrences of a Substring

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.

214. Shortest Palindrome

You are given a string s. You can convert s to a palindrome by adding characters in front of it.

Return the shortest palindrome you can find by performing this transformation.

459. Repeated Substring Pattern

Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.

846. Hand of Straights

Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize, and consists of groupSize consecutive cards.

Given an integer array hand where hand[i] is the value written on the ith card and an integer groupSize, return true if she can rearrange the cards, or false otherwise.

1296. Divide Array in Sets of K Consecutive Numbers

Given an array of integers nums and a positive integer k, check whether it is possible to divide this array into sets of k consecutive numbers.

Return true if it is possible. Otherwise, return false.

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class Solution {
public:
    bool isPossibleDivide(vector<int>& nums, int k) {
        if(nums.size() % k) return false;
        map<int, int> count;
        for(auto n : nums) count[n]++;
        while(!count.empty()) {
          auto be = count.begin();
          if(be->second) {
              auto nxt = next(be);
                for(int i = 1; i < k; i++,nxt++) {
                    if(nxt == count.end()) return false;
                    if(nxt->first != be->first + i) return false;
                    if(nxt->second < be->second)return false;
                    nxt->second -= be->second;

                }
                
            }
            count.erase(be);
            
        } 
        return true;
    }
};

159. Longest Substring with At Most Two Distinct Characters

Given a string s, return the length of the longest substring that contains at most two distinct characters.

1055. Shortest Way to Form String

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., “ace” is a subsequence of “abcde” while “aec” is not).

Given two strings source and target, return the minimum number of subsequences of source such that their concatenation equals target. If the task is impossible, return -1

284. Peeking Iterator

Design an iterator that supports the peek operation on an existing iterator in addition to the hasNext and the next operations.

Implement the PeekingIterator class:

• PeekingIterator(Iterator nums) Initializes the object with the given integer iterator iterator.
• int next() Returns the next element in the array and moves the pointer to the next element.
• boolean hasNext() Returns true if there are still elements in the array.
• int peek() Returns the next element in the array without moving the pointer.

Note: Each language may have a different implementation of the constructor and Iterator, but they all support the int next() and boolean hasNext() functions.

544. Output Contest Matches

During the NBA playoffs, we always set the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting.

Given n teams, return their final contest matches in the form of a string.

The n teams are labeled from 1 to n, which represents their initial rank (i.e., Rank 1 is the strongest team and Rank n is the weakest team).

We will use parentheses ‘(‘, and ‘)’ and commas ‘,’ to represent the contest team pairing. We use the parentheses for pairing and the commas for partition. During the pairing process in each round, you always need to follow the strategy of making the rather strong one pair with the rather weak one.