We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

R = 1, C = 2, r0 = 0, c0 = 0

[[0,0],[0,1]]
Explanation: The distances from (r0, c0) to other cells are: [0,1]

R = 2, C = 2, r0 = 0, c0 = 1

[[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.

R = 2, C = 3, r0 = 1, c0 = 2

[[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].

class Solution {

    /**
     * @param Integer $R
     * @param Integer $C
     * @param Integer $r0
     * @param Integer $c0
     * @return Integer[][]
     */
    function allCellsDistOrder($R, $C, $r0, $c0) {

        $a =[];

        for($x = 0; $x < $R; $x ++){
           for($y = 0; $y < $C; $y ++){
             $a[] = [$x,$y];
           } 
        }
        usort($a, function($arr1, $arr2) use ($r0,$c0) {
            return (abs($arr1[0]-$r0)+ abs($arr1[1]-$c0) 
                    > 
                    abs($arr2[0]-$r0)+ abs($arr2[1]-$c0)
                   ) ? 1 : -1;
        });
        
        return $a;
        
    }
    

}

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly Npeople arrive in each city.

Example 1:

Input:

[[10,20],[30,200],[400,50],[30,20]]

Output:

110

Explanation:

The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Note:

1. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

class Solution {

    /**
     * @param Integer[][] $costs
     * @return Integer
     */
    function twoCitySchedCost($costs) {
        usort($costs, function($arr1, $arr2) {
            return (abs($arr1[1]-$arr1[0]) < abs($arr2[1]-$arr2[0]) ) ? 1 : -1;
        });

        $a = 0;
        $b = 0;
        $total = 0;
        foreach($costs as $c){
            if($c[0] < $c[1]){

                if($a < count($costs)/2){
                    $total += $c[0];
                    $a ++;
                } else {
                    $total += $c[1];
                    $b ++;
                }
               
            } else{
                if($b < count($costs)/2){
                    $total += $c[1];
                    $b ++;
                } else {
                    $total += $c[0];
                    $a ++;
                }
                
            } 
        } 

        return $total;
    }
}

16ms.

Thoughts:

• This code was used for the contest, so it’s not well structured (in terms of redundancy)
• The basic idea:
  • Sort by switching cost (price diff), DESC.
  • then for each person, assign to lower cost city until full.
    • in case of full, assign to city with higher cost
  • Since sorted by price diff, at the time one city is full,  the cost to switch remaining people to the second choice is the lowest

If I’m a interviewer:

• Let’s try three cities….
• What about the capacity of each city is specified?
• And what about the capacity is flexible and within a range provided?

A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.  For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

Example 1:

Input:

"(()())(())"

Output:

"()()()"

Explanation:

The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

"(()())(())(()(()))"

"()()()()(())"

The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

"()()"

""

The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

Note:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

class Solution {
    public String removeOuterParentheses(String S) {
        int lv = 0;
        String result = "";
        
        for(char c : S.toCharArray()){
            if(c == '('){
                if(lv > 0){
                    result += c;
                }
                lv ++;
                
            } else if(c == ')'){
                lv--;
                if(lv > 0){
                    result += c;
                }
            }
        }
        return result;
    }
}

* Using a StringBuilder will somehow improve the performance.

Thoughts:

Since we can assume the input is valid, just use a counter to mark the depth. Send char to result only when depth >= 1

If I’m a Interviewer

1. What if the input string is not always valid?
   • ref: LC20
2. What if we upgrade parentheses to brackets ()[]{} ?
   • Use Stack and Stack.size() instead of level counter
3. What if we remove inner most (max depth) parentheses instead?
   • Go through twice, 1) find max depth, 2) copy to result and skip max depth.

Well we can always construct an array of ints (levels) as a form of representation.

For example, “(()())(())(()(()))” -> [1,2,2,1,2,1,2,2,3]

Then we can easily remove any levels (outer/inner/somewhere in the middle).

• Remove Outer: remove 1, everything else —
• Remove Inner: remove max
• Remove Level2: remove 2, everything > 2 decrease by 1.