Sliding Window Second Maximum

Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. 

For example:
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3. 
Then, return the second max sliding window [1, -1, -1, 3, 5, 6] .

/* time: amortized O(n), space:O(n) */
import java.util.*;

public class Solution {
public static int[] test(int[] nums, int k) {
if (nums == null || nums.length <= 1 || k <= 1) {
return nums;
}
if (k > nums.length) {
k = nums.length;
}
int[] res = new int[nums.length - k + 1];
Deque<Integer> deque = new ArrayDeque<>();
int maxNumIndex = 0;
int index = 0;
for (int i = 1; i < nums.length; i++) {
if (!deque.isEmpty() && deque.peek() < i - k + 1) {
deque.poll();
}
if (maxNumIndex < i - k + 1) {
maxNumIndex = deque.poll();
}
if (nums[i] > nums[maxNumIndex]) {
int temp = maxNumIndex;
maxNumIndex = i;
while (!deque.isEmpty() && temp > deque.peek()) {
deque.poll();
}
deque.offerFirst(temp);
} else {
while (!deque.isEmpty() && nums[i] > nums[deque.peekLast()]) {
deque.pollLast();
}
deque.offer(i);
}
if (i >= k - 1) {
res[index++] = nums[deque.peek()];
}
}
return res;
}
public static void main(String[] args) {
int[] nums = {1, 3, -1, -3, 5, 3, 6, 7};
int[] t0 = test(nums, 1);
int[] t1 = test(nums, 2);
int[] t2 = test(nums, 3);
int[] t3 = test(nums, 4);
int[] t4 = test(nums, 5);
int[] t5 = test(nums, 6);
int[] t6 = test(nums, 7);
int[] t7 = test(nums, 8);
int[] t8 = test(nums, 9);
int[] t9 = test(nums, 10);
System.out.println(Arrays.toString(t0));
System.out.println(Arrays.toString(t1));
System.out.println(Arrays.toString(t2));
System.out.println(Arrays.toString(t3));
System.out.println(Arrays.toString(t4));
System.out.println(Arrays.toString(t5));
System.out.println(Arrays.toString(t6));
System.out.println(Arrays.toString(t7));
System.out.println(Arrays.toString(t8));
System.out.println(Arrays.toString(t9));
}
}

Word Search

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.


/* method 1: 
 * time:O(m*n*4^L), space:O(m*n), m: board's row, n: board's column, L: word size
 * Imagine this is a tree,  this quadtree's height is L(L is length of word) and its total node number is
 * 4^0 + 4^1 + ... + 4^L = 1/3 * ( 4^(L+1) - 1 ). So the time complexity of this dfs solution is  * mnO(4^L).
 */
public class Solution {
    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        int width = board.length;
        int length = board[0].length;
        boolean[][] visited = new boolean[width][length];
        for (int y = 0; y < width; y++) {
        for (int x = 0; x < length; x++) {
        if (exist(board, y, x, words, 0, visited)) {
                                return true;
                        }
        }
        }
        return false;
    }
    
    private boolean exist(char[][] board, int y, int x, char[] words, int i, boolean[][] visited) {
    if (i == words.length) {
            return true;
        }
    if (y < 0 || x < 0 || y == board.length || x == board[y].length || visited[y][x] == true) {
            return false;
        }
    if (board[y][x] != words[i]) {
            return false;
        }
    visited[y][x] = true;
    boolean exist = exist(board, y, x + 1, words, i + 1, visited)
               || exist(board, y, x - 1, words, i + 1, visited)
               || exist(board, y + 1, x, words, i + 1, visited)
               || exist(board, y - 1, x, words, i + 1, visited);
    visited[y][x] = false;
    return exist;
    }

}



/* method2:
 * time:O(m*n*4^L), space:O(1)
 *
 * board[y][x] ^= 256 it's a marker that the letter at position x,y is a part of word we search.
 * After board[y][x] ^= 256 the char became not a valid letter. After second board[y][x] ^= 256
 * it became a valid letter again.
 */

public class Solution {
    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        int row = board.length;
        int column = board[0].length;
        for (int y = 0; y < row; y++) {
         for (int x = 0; x < column; x++) {
         if (exist(board, y, x, words, 0)) {
                              return true;
                        }
         }
        }
        return false;
    }
    
    private boolean exist(char[][] board, int y, int x, char[] words, int i) {
     if (i == words.length) {
            return true;
        }
     if (y < 0 || x < 0 || y == board.length || x == board[y].length) {
            return false;
        }
     if (board[y][x] != words[i]) {
            return false;
        }
     board[y][x] ^= 256;
     boolean exist = exist(board, y, x + 1, words, i + 1)
                 || exist(board, y, x - 1, words, i + 1)
                 || exist(board, y + 1, x, words, i + 1)
                 || exist(board, y - 1, x, words, i + 1);
     board[y][x] ^= 256;
     return exist;
    }

}

K closest points

/*
 * Find the K closest points to the origin in 2D plane, given an array containing N points.
 * You can assume K is much smaller than N and N is very large.
 * You need only use standard math operators (addition, subtraction, multiplication, and division).
 */
import java.util.*;

class Point {
    double x;
    double y;
    public Point(double x, double y) {
        this.x = x;
        this.y = y;
    }
}
public class Solution {
private static double distance(Point a, Point b) {
        return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
    }
public static Point[] closestPoint(Point[] array, final Point origin, int k) {
        if (k >= array.length) {
        // return array;
        k = array.length;
        }
        if (k <= 0) {
        return new Point[0];
        }
        Point[] res = new Point[k];
        PriorityQueue<Point> maxHeap = new PriorityQueue<>(k + 1, new Comparator<Point>() {
            @Override
            public int compare(Point a, Point b) {
                return Double.compare(distance(b, origin), distance(a, origin));
            }
        });
        for(Point p : array) {
        maxHeap.offer(p);
        if (maxHeap.size() > k) {
        maxHeap.poll();
        }
        }
        for(int i = k - 1; i >= 0; i--) {
        res[i] = maxHeap.poll();
        }
        return res;
    }

    public static void main(String[] args) {
        Point origin = new Point(0, 0);
        Point[] input = new Point[]{new Point(0, 2), new Point(1, 1), new Point(-1, 0), new Point(2, 0), new Point(3, 0)};
        Point[] output = closestPoint(input, origin, 3);
        System.out.println("input");
        for(Point i : input) System.out.print("("+i.x+", "+i.y+") ");
        System.out.println("");
        System.out.println("output");
        for(Point i : output) System.out.print("("+i.x+", "+i.y+") ");
    }
}