Всем привет! Надеемся, что ваши выходные прошли максимально приятно.
В первый рабочий день этой недели публикуем по одному варианту решения для каждой задачи.
Комментарии, как и раньше, не переводим.

1. Простое O(N) Java решение с использованием Insert Index

// Shift non-zero values as far forward as possible
// Fill remaining space with zeros
 
public void moveZeroes(int[] nums) {
	if (nums == null || nums.length == 0) return;    	
 
	int insertPos = 0;
	for (int num: nums) {
    	if (num != 0) nums[insertPos++] = num;
	}    	
 
	while (insertPos < nums.length) {
    	nums[insertPos++] = 0;
	}
}

2. O(n) C++ решение с использованием unordered_map

class Solution {
public:
	int maxSubArrayLen(vector<int>& nums, int k) {
    	unordered_map<int, int> sums;
    	int cur_sum = 0;
    	int max_len = 0;
    	for (int i = 0; i < nums.size(); i++) {
        	cur_sum += nums[i];
        	if (cur_sum == k) {
            	max_len = i + 1;
        	} else if (sums.find(cur_sum - k) != sums.end()) {
            	max_len = max(max_len, i - sums[cur_sum - k]);
        	}
        	if (sums.find(cur_sum) == sums.end()) {
            	sums[cur_sum] = i;
        	}        	
    	}
    	return max_len;
	}
};

3. Java BFS решение

The idea is straightforward, with the input string s, we generate all possible states by removing one ( or ), check if they are valid, if found valid ones on the current level, put them to the final result list and we are done, otherwise, add them to a queue and carry on to the next level.
The good thing of using BFS is that we can guarantee the number of parentheses that need to be removed is minimal, also no recursion call is needed in BFS.
Time complexity:
In BFS we handle the states level by level, in the worst case, we need to handle all the levels, we can analyze the time complexity level by level and add them up to get the final complexity.
On the first level, there's only one string which is the input string s, let's say the length of it is n, to check whether it's valid, we need O(n) time. On the second level, we remove one ( or ) from the first level, so there are C(n, n-1) new strings, each of them has n-1 characters, and for each string, we need to check whether it's valid or not, thus the total time complexity on this level is (n-1) x C(n, n-1). Come to the third level, total time complexity is (n-2) x C(n, n-2), so on and so forth…
Finally we have this formula:
T(n) = n x C(n, n) + (n-1) x C(n, n-1) +… + 1 x C(n, 1) = n x 2^(n-1).
Following is the Java solution:

public class Solution {
	public List<String> removeInvalidParentheses(String s) {
  	List<String> res = new ArrayList<>();
  	
  	// sanity check
  	if (s == null) return res;
  	
  	Set<String> visited = new HashSet<>();
  	Queue<String> queue = new LinkedList<>();
  	
  	// initialize
  	queue.add(s);
  	visited.add(s);
  	
  	boolean found = false;
  	
  	while (!queue.isEmpty()) {
    	s = queue.poll();
    	
    	if (isValid(s)) {
      	// found an answer, add to the result
     	 res.add(s);
      	found = true;
    	}
  	
    	if (found) continue;
  	
    	// generate all possible states
    	for (int i = 0; i < s.length(); i++) {
      	// we only try to remove left or right paren
      	if (s.charAt(i) != '(' && s.charAt(i) != ')') continue;
    	
      	String t = s.substring(0, i) + s.substring(i + 1);
    	
      	if (!visited.contains(t)) {
        	// for each state, if it's not visited, add it to the queue
        	queue.add(t);
        	visited.add(t);
      	}
    	}
  	}
  	
  	return res;
	}
	
	// helper function checks if string s contains valid parantheses
	boolean isValid(String s) {
  	int count = 0;
	
  	for (int i = 0; i < s.length(); i++) {
    	char c = s.charAt(i);
    	if (c == '(') count++;
    	if (c == ')' && count-- == 0) return false;
  	}
	
  	return count == 0;
	}
}