All posts by shiji

1266. Minimum Time Visiting All Points

2019/11/23 shiji

On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.

You can move according to the next rules:

In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
You have to visit the points in the same order as they appear in the array.

Example 1:

Input:

 points = [[1,1],[3,4],[-1,0]]

Output:

Explanation:

One optimal path is

[1,1]

 -> [2,2] -> [3,3] ->

[3,4]

-> [2,3] -> [1,2] -> [0,1] ->

[-1,0]

   
Time from [1,1] to [3,4] = 3 seconds 
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds

Example 2:

Input:

 points = [[3,2],[-2,2]]

Output:

Constraints:

points.length == n
1 <= n <= 100
points[i].length == 2
-1000 <= points[i][0], points[i][1] <= 1000

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int result = 0;
        int prev_x = points[0][0];
        int prev_y = points[0][1];
        for (int i = 1; i < points.length; i ++){
            int x = points[i][0];
            int y = points[i][1];
            
            int diff_x = Math.abs(x - prev_x);
            int diff_y = Math.abs(y - prev_y);
            
            result += Math.min(diff_x,diff_y)+Math.abs(diff_x-diff_y);
            
            prev_x = x;
            prev_y = y;
        }
        return result;
    }
}

Leetcode

1161. Maximum Level Sum of a Binary Tree

2019/11/15 shiji

Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.

Return the smallest level X such that the sum of all the values of nodes at level X is maximal.

Example 1:

Input:

[1,7,0,7,-8,null,null]

Output:

Explanation:

Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.

Note:

The number of nodes in the given tree is between 1 and 10^4.
-10^5 <= node.val <= 10^5

Another “Follow your heart” question.

I’m using level order traversal (BFS) here.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxLevelSum(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        int max = 0;
        int lv = 1;
        
        int max_lv = 0;
        
        while (! q.isEmpty()){
            
            int size = q.size();
            int level_sum = 0;
            for (int i = 0; i < size; i ++){
                TreeNode cur = q.poll();
                if(cur.left != null)
                    q.add(cur.left);
                if (cur.right != null)
                    q.add(cur.right);
                level_sum += cur.val;
            }
            if (level_sum > max){
                max = level_sum;
                max_lv = lv;
            }
            lv ++;
            
        }
        return max_lv;
    }
    
    
}

Leetcode

904. Fruit Into Baskets

2019/11/14 shiji

In a row of trees, the i-th tree produces fruit with type tree[i].

You start at any tree of your choice, then repeatedly perform the following steps:

Add one piece of fruit from this tree to your baskets. If you cannot, stop.
Move to the next tree to the right of the current tree. If there is no tree to the right, stop.

Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.

You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.

What is the total amount of fruit you can collect with this procedure?

Example 1:

Input:

[1,2,1]

Output:

Explanation:

We can collect [1,2,1].

Example 2:

Input:

[0,1,2,2]

Output:

Explanation:

We can collect [1,2,2].
If we started at the first tree, we would only collect [0, 1].

Example 3:

Input:

[1,2,3,2,2]

Output:

Explanation:

We can collect [2,3,2,2].
If we started at the first tree, we would only collect [1, 2].

Example 4:

Input:

[3,3,3,1,2,1,1,2,3,3,4]

Output:

Explanation:

We can collect [1,2,1,1,2].
If we started at the first tree or the eighth tree, we would only collect 4 fruits.

Note:

1 <= tree.length <= 40000
0 <= tree[i] < tree.length

First Draft.

Code is accepted, but the quality is a JOKE.

class Solution {
    public int totalFruit(int[] tree) {
        
        if (tree.length <= 2){
            return tree.length;
        }
        
        int max = 2;
        int cur_max = 2;
        
        int a = tree[0];
        int b = tree[1];
        
        int a_last_index = 0;
        Integer b_last_index = null;
        
        boolean a_is_eldest = true;
        
        // get the second <Unique> number
        for(int i = 1; i < tree.length; i ++){
            if (tree[i] != a){
                b = tree[i];
                b_last_index = i;
                break;
            } else {
                // And we have to update last index of a as well
                cur_max ++;
                a_last_index = i;
                a_is_eldest = false;
            }
        }
        // update max here
        max = Math.max(max, cur_max);
        
        // if only single unique number in the array...
        if (b_last_index == null){
            return tree.length;
        }
        
        
        // Scan from the right of the second unique number
        for(int i = b_last_index+1; i < tree.length; i ++){
            int cur = tree[i];
            
            if(cur == a){

                cur_max ++;
                a_last_index = i;
                a_is_eldest = false;

            } else if (cur == b){

                cur_max ++;
                b_last_index = i;
                a_is_eldest = true;

            } else {

                // Third unique number appeared!
                // Remove eldest, and reduce current max to the last occurance of the number to be removed.
                if(a_is_eldest){
                    cur_max = i - a_last_index;
                    a = cur;
                    a_last_index = i;
                    a_is_eldest = false;
                } else {
                    cur_max = i - b_last_index;
                    b = cur;
                    b_last_index = i;
                    a_is_eldest = true;
                }
            }
            max = Math.max(max, cur_max);
        }
        return max;
    }
}

Code after refactoring:

class Solution {
    public int totalFruit(int[] tree) {
        
        if (tree.length <= 2){
            return tree.length;
        }
        
        int max = 2;
        int cur_max = 1;
        
        int a = tree[0];
        int a_last_index = 0;
        
        Integer b = null;
        Integer b_last_index = null;
        
        boolean a_is_eldest = true;
    
        
        for(int i = 1; i < tree.length; i ++){
            int cur = tree[i];
            
            if(cur == a){
                cur_max ++;
                a_last_index = i;
                a_is_eldest = false;
                
            } else if (b == null || cur == b) {
                
                if(b == null){
                    b = cur;
                }
                cur_max ++;
                b_last_index = i;
                a_is_eldest = true;
                
            } else {

                if(a_is_eldest){
                    cur_max = i - a_last_index;
                    a = cur;
                    a_last_index = i;
                    a_is_eldest = false;
                } else {
                    cur_max = i - b_last_index;
                    b = cur;
                    b_last_index = i;
                    a_is_eldest = true;
                }

            }
            max = Math.max(max, cur_max);
        }
        return max;
    }
}

There could be more things for refactoring, putting that aside, can we use any data structure?

Maybe LinkedHashMap<Type, LastSeenIndex> in access order mode? it’s an overkill for ‘collecting two types’, but could be useful for 3 or more types.

Starting by copy & paste from LRUCache. I added a variable to save the value of the most recent removed entry. The value of it is updated before the eldest entry is removed.

The modified LRUCache

class LRUCache extends LinkedHashMap<Integer, Integer>{
    private int capacity;

    // Save the latest index of the recent removed item.
    int index_removed = 0;
    
    public LRUCache(int capacity) {
        super(capacity, 0.75F, true);
        this.capacity = capacity;
    }

    public int get(int key) {
        return super.getOrDefault(key, -1);
    }

    public void put(int key, int value) {
        super.put(key, value);
    }

    @Override
    protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {

        // If the eldest has to be removed, save it's value (most recent index)
        if(size() > capacity){
            index_removed = eldest.getValue();
        }

        return size() > capacity; 
    }
}

Then, update the main function… and combine them

class Solution {

    public static final int TYPES_ALLOWED = 2;
 
    public int totalFruit(int[] tree) {
        
        if (tree.length <= TYPES_ALLOWED){
            return tree.length;
        }
        
        int cur_max = 1;
        int max = 1;
        
        // For a follow-up of more than 2 types is allowed.. just change the number here
        LRUCache map = new LRUCache(TYPES_ALLOWED);
    
        map.put(tree[0], 0);
        
        for(int i = 1; i < tree.length; i ++){
            int cur = tree[i];
            if (map.get(cur) != -1 || map.size() < 2){
                cur_max ++;
                map.put(cur, i);
            } else {
                map.put(cur, i);
                cur_max = i - map.index_removed;
            }
            
            max = Math.max(max, cur_max);
            
        }
        return max;
    }
}

class LRUCache extends LinkedHashMap<Integer, Integer>{
    private int capacity;
    int index_removed = 0;
    
    public LRUCache(int capacity) {
        super(capacity, 0.75F, true);
        this.capacity = capacity;
    }

    public int get(int key) {
        return super.getOrDefault(key, -1);
    }

    public void put(int key, int value) {
        super.put(key, value);
    }

    @Override
    protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
        if(size() > capacity){
            index_removed = eldest.getValue();
        }
        return size() > capacity; 
    }
}

Time complexity. O(n). basically going through all elements once (n), times LRUCache O(1).

This is basically a sliding window solution, while the Map is used for maintaining the valid window status.

Well, we can also use other Collections, giving the accepted types is two. The search and insertion time have to be counted after the number of accepted types is increased. So a Hash* will have a better performance.

Can we use a LinkedHashSet? Yes, then giving the type to be removed, find the closest index before the current index.

Leetcode

973. K Closest Points to Origin

2019/11/13 shiji

We have a list of points on the plane. Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:

Input:

points = [[1,3],[-2,2]], K = 1

Output:

[[-2,2]]

Explanation:

The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input:

points = [[3,3],[5,-1],[-2,4]], K = 2

Output:

[[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:

1 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000

A acceptable but definitely no the best solution….

In PHP:

a ** b means power (a,b)
<=>, the spaceship operator, used by ‘comparator’
1. if left < right return -1
2. if left > right return 1
3. if left == right return 0

class Solution {

    /**
     * @param Integer[][] $points
     * @param Integer $K
     * @return Integer[][]
     */
    function kClosest($points, $K) {
        usort($points, function ($a, $b){
            return ($a[0]**2 + $a[1]**2) <=> ($b[0]**2 + $b[1]**2);
        });
        
        $result = [];
        $i = 0;
        foreach($points as $p){
            if($i >= $K){
                break;   
            }
            $result[] = $p;
            $i ++;
    
        }
        return $result;
    }
}

Leetcode

102. Binary Tree Level Order Traversal

2019/11/13 shiji

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Straight forward solution:
root node in => left and right children out . -- Lv1
lv 1 nodes in => lv 2 nodes out......
until no more next lv.

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($value) { $this->val = $value; }
 * }
 */
class Solution {

    /**
     * @param TreeNode $root
     * @return Integer[][]
     */
    function levelOrder($root) {
        $q = [];
        $result = [];
        
        if($root === null){
            return $result;
        }
        $q[] = $root;
        while (count($q) > 0){
            $current_level = [];
            $current_result = [];
            foreach ($q as $node){
                if ($node->left !== null) 
                    $current_level[] = $node->left;
                
                $current_result[] = $node->val;
                
                if ($node->right !== null) 
                    $current_level[] = $node->right;
            }
            $q = $current_level;
            $result [] = $current_result;
        }

        return $result;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> r = new ArrayList<>();
        
        if(root == null){
            return r;
        }
        
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        
        while(!q.isEmpty()){
            
            int sizeSnapshot = q.size();
            List<Integer> current_level = new ArrayList<>();
            
            
            for(int i = 0; i < sizeSnapshot; i ++){
                TreeNode t = q.poll();
                if(t.left != null){
                    q.add(t.left);
                }
                
                current_level.add(t.val);
                
                if(t.right != null){
                    q.add(t.right);
                }
            }
            r.add(current_level);
        }
        
        return r;
    }
}

Leetcode

189. Rotate Array

2019/10/11 shiji

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input:

[1,2,3,4,5,6,7] and

= 3

Output:

[5,6,7,1,2,3,4]

Explanation:

rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input:

[-1,-100,3,99] and

= 2

Output:

 [3,99,-1,-100]

Explanation:

 rotate 1 steps to the right: [99,-1,-100,3] rotate 2 steps to the right: [3,99,-1,-100]

Note:

Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Could you do it in-place with O(1) extra space?

In place rotate

class Solution {

    /**
     * @param Integer[] $nums
     * @param Integer $k
     * @return NULL
     */
    function rotate(&$nums, $k) {
        $c = count($nums);
        $steps = 0;
        $start = 0;
        
        while($steps < $c){  
            $i = $start;
            $prev = $nums[$i];
            do{         
                $next = ($i+$k)%$c;
                $temp = $nums[$next];
                $nums[$next] = $prev;
                $prev = $temp;
                $i += $k;
                $i = $i % $c;
                $steps++;
            }while($start != $i);
            $start ++;
            
        }
    
    }
}

Leetcode

406. Queue Reconstruction by Height

2019/10/01 shiji

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

Note:
The number of people is less than 1,100.

Input: [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]] Output: [[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

Example

Solution:

<?php

class Solution {

    /**
     * @param Integer[][] $people
     * @return Integer[][]
     */
    function reconstructQueue($people) {
        usort($people, function($p1, $p2){
            return ($p1[0] !== $p2[0])?
                 $p2[0] <=> $p1[0] : $p1[1] <=> $p2[1];
        });
        $result = [];
        foreach($people as $p){
            array_splice($result, $p[1], 0, [$p]);
        }

        return $result;
    }
}

?>

[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

The higher the person is, the more importance it has for other persons

So starting sort by height, DESC
Then for each height level, sort by # of higher heads in the front, ASC
(Just like ‘ORDER BY h DESC, k ASC’ in SQL)

Now we have:

More Important to Others <===> Less Important to Others

[7,0] [7,1] [6,1] [5,0] [5,2] [4,4]

Then set group height = 7 to the result

[7,0] [7,1] ; TBD: [6,1] [5,0] [5,2] [4,4]

Now set group height = 6. It could be places any where, since h=7 can’t see them, so the position is determined by their value of k. for example: [6,1]. there’s one higher/equal height person ahead, we can simply insert it right after k is matched, and push everyone else backwards safely since they don’t care about lower person moving around.

   0     1     2     3     4  ..... (index)

[7,0] [6, 1] [7,1]  ;                TBD:    [5,0]    [5,2]   [4,4]

[5,0] [7,0] [6, 1] [7,1]  ;          TBD:    [5,2]   [4,4]

[5,0] [7,0] [5,2] [6, 1] [7,1]  ;    TBD:    [4,4]

[5,0] [7,0] [5,2] [6, 1] [4,4]  [7,1]

In PHP, array_splice($result, $p[1], 0, [$p]); means insert p to position p[i] and move people behind backwards

In Java, LinkedList seems easier.

In Python, there’s even easier method (list.insert(index, value))

Leetcode

911. Online Election

2019/09/29 shiji

In an election, the i-th vote was cast for persons[i] at time times[i].

Now, we would like to implement the following query function: TopVotedCandidate.q(int t) will return the number of the person that was leading the election at time t.

Votes cast at time t will count towards our query. In the case of a tie, the most recent vote (among tied candidates) wins.

Example 1:

Input:

["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]

Output:

[null,0,1,1,0,0,1]

Explanation:

At time 3, the votes are [0], and 0 is leading.
At time 12, the votes are [0,1,1], and 1 is leading.
At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
This continues for 3 more queries at time 15, 24, and 8.

Note:

1 <= persons.length = times.length <= 5000
0 <= persons[i] <= persons.length
times is a strictly increasing array with all elements in [0, 10^9].
TopVotedCandidate.q is called at most 10000 times per test case.
TopVotedCandidate.q(int t) is always called with t >= times[0].

Here’s my rejected solution

<?php
class TopVotedCandidate {
    
    public $dic=[];
    public $r =[];

    public $keys = [];
    /**
     * @param Integer[] $persons
     * @param Integer[] $times
     */
    function __construct($persons, $times) {
        $max = 0;
        $prev_person = null;
        for($i = 0; $i < count($times); $i ++){
            $this->inc($persons[$i]);
            arsort($this->$dic);
  
            foreach($this->$dic as $k=>$v){
                $max_this_round = $k;
                break;
            }
            $person_this_round = $persons[$i];

            if($i > 0){
                if($max_this_round === $max){
                    $person_this_round = $prev_person;
                } 
            }
            echo $person_this_round;
            $this->$r[$times[$i]] = $person_this_round;
            $max = $max_this_round;
            $prev_person = $persons[$i];
        }
        var_dump($this->$r);
        ksort($this->$r);
        var_dump($this->$r);
        $this->$keys = array_keys($this->$r);
        
    }
  
    /**
     * @param Integer $t
     * @return Integer
     */
    function q($t) {
        $k =  $this->binarySearch($t);
        return $this->$r[$k];
    }
    
    function inc($p){
        if(isset($this->$dic[$p])){
            $this->$dic[$p] ++;
        } else {
            $this->$dic[$p] = 1;
        }
    }
    
    
    function binarySearch($x) { 
        $arr =  $this->$keys;
        $low = 0; 
        $high = count($arr) - 1; 
        

        while ($low <= $high) { 

            // compute middle index 
            $mid = floor(($low + $high) / 2); 

            // element found at mid 
            if($arr[$low] <= $x && $arr[$mid] >= $x &&  $mid - $low <= 1) { 
                return $low; 
            } 

            if ($x < $arr[$mid]) { 
                // search the left side of the array 
                $high = $mid -1; 
            } 
            else { 
                // search the right side of the array 
                $low = $mid + 1; 
            } 
            echo "$low-$mid-$high\n";
        } 

        return false; 
    }
}

/**
 * Your TopVotedCandidate object will be instantiated and called as such:
 * $obj = TopVotedCandidate($persons, $times);
 * $ret_1 = $obj->q($t);
 */

?>

Here are the problems:

Just like the previous problem, my PHP syntax is wrong, and the prev problem is accepted. I didn’t even ever question about the syntax
And I wasted too much time on debugging, caused by syntax.
And my modified binary search isn’t working
And yes, that’s totally a mess

Here’s the my accepted solution

<?php
class TopVotedCandidate {
    
    public $dic=[];
    public $r =[];

    public $keys = [];
    /**
     * @param Integer[] $persons
     * @param Integer[] $times
     */
    function __construct($persons, $times) {
        $prev_max = 0;
        $prev_winner = null;

        for($i = 0; $i < count($times); $i ++){
            $this->inc($persons[$i]);

            arsort($this->dic);

            reset($this->dic);
            $winner = key($this->dic);
            
            $max = $this->dic[$winner];
            if($prev_winner !== null){
                if($max == $prev_max){
                    $winner = $prev_winner;
                }
            }
            if($this->dic[$persons[$i]] == $this->dic[$winner]){
                 $winner = $persons[$i];
            }

            $prev_max = max($prev_max, $max);
            $prev_winner = $winner;
            $this->r[$times[$i]] = $winner;
            
        }
        ksort($this->r);
        $this->keys = array_keys($this->r);
        
    }
  
    /**
     * @param Integer $t
     * @return Integer
     */
    function q($t) {
        $k =  $this->binarySearch($t);
        echo "$t => $k";
        return $this->r[$k];
    }
    
    function inc($p){
        if(isset($this->dic[$p])){
            $this->dic[$p] ++;
        } else {
            $this->dic[$p] = 1;
        }
    }
    
    // This is a modified binary search, finding the nearest smaller index 
    // just like floorKey in Java's TreeMap
    function binarySearch($x) { 
        $arr =  $this->keys;
        $low = 0; 
        $high = count($arr) - 1; 
        
        
        if($x < $arr[$low]){
            return false;
        }
        
        while ($high >= $low) { 
            $mid = $low + floor(($high - $low) / 2); 

            if($x >= $arr[$high]){
                return (int) $arr[$high];
            }
            
            if($arr[$mid] == $x) { 
                return (int) $arr[$mid]; 
            } 

            if ($x < $arr[$mid]) { 
                $high = $mid-1; 
            } else { 
                $low = $mid+1; 
            } 
            
        } 
        return (int) $arr[$low-1];

    }
}
?>
/**
 * Your TopVotedCandidate object will be instantiated and called as such:
 * $obj = TopVotedCandidate($persons, $times);
 * $ret_1 = $obj->q($t);
 */