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))

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.

["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]]

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

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.

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);
 */

You are driving a vehicle that has capacity empty seats initially available for passengers.  The vehicle only drives east (ie. it cannot turn around and drive west.)

Given a list of trips, trip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off.  The locations are given as the number of kilometers due east from your vehicle’s initial location.

Return true if and only if it is possible to pick up and drop off all passengers for all the given trips.

Example 1:

Input:

trips = [[2,1,5],[3,3,7]], capacity = 4

Output:

false

trips = [[2,1,5],[3,3,7]], capacity = 5

true

trips = [[2,1,5],[3,5,7]], capacity = 3

true

trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11

<?php
class Solution {

     public $dic=[];
    /**
     * @param Integer[][] $trips
     * @param Integer $capacity
     * @return Boolean
     */
    function carPooling($trips, $capacity) {
        $max = 0;
        $sum = 0;
        $this->$dic = [];
        foreach ($trips as $c){
            if(!isset($this->$dic[$c[1]])){
                $this->$dic[$c[1]] = 0;
            }
            $this->$dic[$c[1]] += $c[0];
            $this->$dic[$c[2]] -= $c[0];
        }
        // 
        ksort($this->$dic);
    
        // Calculate least expected capacity
        foreach($this->$dic as $d){
            $sum += $d;
            if($sum > $max){
                $max  = $sum;
            }
        }
   
        return $max <= $capacity;
    }
}
?>

<?php
class Solution {

     public $dic=[];
    /**
     * @param Integer[][] $trips
     * @param Integer $capacity
     * @return Boolean
     */
    function carPooling($trips, $capacity) {
        $max = 0;
        $sum = 0;
        $this->dic = [];
        foreach ($trips as $c){
            if(!isset($this->dic[$c[1]])){
                $this->dic[$c[1]] = 0;
            }
            $this->dic[$c[1]] += $c[0];
            $this->dic[$c[2]] -= $c[0];
        }

        ksort($this->dic);
    
        // I'm finding the min capacity here, but it's not necessary 
        foreach($this->dic as $d){
            $sum += $d;
            if($sum > $max){
                $max  = $sum;
            }
        }
   
        return $max <= $capacity;
    }   
}

?>

Given a string S, return the number of substrings of length K with no repeated characters.

Example 1:

Input:

S = "havefunonleetcode", K = 5

Output:

6

Explanation:

There are 6 substrings they are : 'havef','avefu','vefun','efuno','etcod','tcode'.

Example 2:

Input:

S = "home", K = 5

Output:

0

Explanation:

Notice K can be larger than the length of S. In this case is not possible to find any substring.

Note:

1. 1 <= S.length <= 10^4
2. All characters of S are lowercase English letters.
3. 1 <= K <= 10^4

Quite straight forward solution.

the brute force way is foreach K chars, check dupes.

to optimize that, push next char and remove last instead, so same work won’t be done multiple times

<?php
class Solution {

    /**
     * @param String $S
     * @param Integer $K
     * @return Integer
     */
    function numKLenSubstrNoRepeats($S, $K) {
        if($K > strlen($S)){
            return 0;
        }
        // This is not necessary in PHP
        $arr = str_split($S);
        $dic = [];
        $result = 0;
        
        // Helper, any repeated chars so far?
        $unique = static function() use (&$dic) {
            foreach($dic as $v){
                if ($v > 1){
                    return false;
                }
            }
            return true;
        };

        // Load first K chars
        for($i = 0; $i < $K; $i ++){
            if(isset($dic[$arr[$i]])){
                $dic[$arr[$i]]++;
            } else {
                $dic[$arr[$i]] = 1;
            }
            
        }
        // First K has repeat? update counter
        if($unique()){
             $result++;
        }

        // Starting from K, take next char and remove the last one, update dictionary, and check after each round
        for($in = $K; $in < strlen($S); $in ++){
            if(isset($dic[$arr[$in]])){
                $dic[$arr[$in]]++;
            } else {
                $dic[$arr[$in]] = 1;
            }
            
            $out = $in-$K;
            $dic[$arr[$out]]--;
            
            if($unique()){
                $result++;
            }
        }
            
        return $result;
    }
}
?>