189. Rotate Array

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

k

 = 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

k

 = 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 ++;
            
        }
    
    }
}

 

406. Queue Reconstruction by Height

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))

911. Online Election

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. 1 <= persons.length = times.length <= 5000
  2. 0 <= persons[i] <= persons.length
  3. times is a strictly increasing array with all elements in [0, 10^9].
  4. TopVotedCandidate.q is called at most 10000 times per test case.
  5. 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);
 */

 

 

1094. Car Pooling

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 tripstrip[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

Example 2:

Input:

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

Output:

true

Example 3:

Input:

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

Output:

true

Example 4:

Input:

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

Output:

true

 

Accepted  Solution in wrong PHP syntax (don’t drink and code)

<?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;
    }
}
?>

And Here’s the solution in my mind..

<?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;
    }   
}

?>

 

1100. Find K-Length Substrings With No Repeated Characters

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;
    }
}
?>

 

917. Reverse Only Letters

Given a string S, return the “reversed” string where all characters that are not a letter stay in the same place, and all letters reverse their positions.

 

Example 1:

Input:

"ab-cd"

Output:

"dc-ba"

Example 2:

Input:

"a-bC-dEf-ghIj"

Output:

"j-Ih-gfE-dCba"

Example 3:

Input:

"Test1ng-Leet=code-Q!"

Output:

"Qedo1ct-eeLg=ntse-T!"

 

Note:

  1. S.length <= 100
  2. 33 <= S[i].ASCIIcode <= 122
  3. S doesn’t contain \ or "

 

Straightforward solution. 

  1. collect letters in reverse order.
  2. go through string from beginning, if it’s letter, replace with letter from reverse string, else keep special chars and adjust pointer accordingly
  3. of course there’s space for improvement, but not necessary
<?php
class Solution {

    /**
     * @param String $S
     * @return String
     */
    function reverseOnlyLetters($S) {
        $x = '';
        $y = '';
        for($i = strlen($S)-1; $i >=0; $i -- ){
            if(ctype_alpha($S[$i])){
                $x .= $S[$i];
            }
        }
        $xi = 0;
        for($i = 0; $i < strlen($S); $i ++){
            if(!ctype_alpha($S[$i])){
                $y .= $S[$i];
            } else {
                $y .= $x[$xi];
                $xi ++;
            }
             
        }
        
        return $y;
    }
}

?>

 

7. Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input:

 123

Output:

 321

Example 2:

Input:

 -123

Output:

 -321

Example 3:

Input:

 120

Output:

 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

class Solution {

    /**
     * @param Integer $x
     * @return Integer
     */
    function reverse($x) {
        
        $r =  (int) (strrev(trim($x,"+-")) * ($x>0?1:-1));
        
        if($r >  pow(2,31)-1 || $r < -pow(2,31)){
            return 0;
        }
        return $r;
    }
}

 

9. Palindrome Number

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input:

 121

Output:

 true

Example 2:

Input:

 -121

Output:

 false

Explanation:

 From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input:

 10

Output:

 false

Explanation:

 Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Coud you solve it without converting the integer to a string?

 

class Solution {

    /**
     * @param Integer $x
     * @return Boolean
     */
    function isPalindrome($x) {
        if($x < 0){
            return false;
        }
        return strrev($x) == $x;
    }
}

For the followup

not converting to string? Create a  stack of integers instead (like a push down automata)