Queue Reconstruction by Height – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

You are given an array of people, people, which are the attributes of some people in a queue (not necessarily in order). Each people[i] = [hi, ki] represents the ith person of height hi with exactly ki other people in front who have a height greater than or equal to hi.

Reconstruct and return the queue that is represented by the input array people. The returned queue should be formatted as an array queue, where queue[j] = [hj, kj] is the attributes of the jth person in the queue (queue[0] is the person at the front of the queue).

Example 1:

Input: people = [[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]]
Explanation:
Person 0 has height 5 with no other people taller or the same height in front.
Person 1 has height 7 with no other people taller or the same height in front.
Person 2 has height 5 with two persons taller or the same height in front, which is person 0 and 1.
Person 3 has height 6 with one person taller or the same height in front, which is person 1.
Person 4 has height 4 with four people taller or the same height in front, which are people 0, 1, 2, and 3.
Person 5 has height 7 with one person taller or the same height in front, which is person 1.
Hence [[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]] is the reconstructed queue.

Example 2:

Input: people = [[6,0],[5,0],[4,0],[3,2],[2,2],[1,4]]
Output: [[4,0],[5,0],[2,2],[3,2],[1,4],[6,0]]

Constraints:

• 1 <= people.length <= 2000
• 0 <= hi <= 106
• 0 <= ki < people.length
• It is guaranteed that the queue can be reconstructed.

C++ Queue Reconstruction by Height LeetCode Solution

public class Solution {
    public int[][] reconstructQueue(int[][] people) {
        //pick up the tallest guy first
        //when insert the next tall guy, just need to insert him into kth position
        //repeat until all people are inserted into list
        Arrays.sort(people,new Comparator<int[]>(){
           @Override
           public int compare(int[] o1, int[] o2){
               return o1[0]!=o2[0]?-o1[0]+o2[0]:o1[1]-o2[1];
           }
        });
        List<int[]> res = new LinkedList<>();
        for(int[] cur : people){
            res.add(cur[1],cur);       
        }
        return res.toArray(new int[people.length][]);
    }
}`

Java Queue Reconstruction by Height LeetCode Solution

 
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        sort(people.begin(), people.end(),[](pair<int,int> p1, pair<int,int> p2){
            return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second);
        });
        vector<pair<int,int>> sol;
        for (auto person : people){
            sol.insert(sol.begin() + person.second, person);
        }
        return sol;
    }

Python 3 Queue Reconstruction by Height LeetCode Solution

class Solution:
    def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
        # sort the people from tall to short
        # insert from tall to short (insert at index = p[1])
        people.sort(key=lambda p: (-p[0], p[1]))
        res = []
        for p in people:
            res.insert(p[1], p)
        return res

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264

Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195

Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140

Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96

Backtracking-92
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Sliding Window-73
Linked List-69
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Ordered Set-48
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Enumeration-39
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Queue-33
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Topological Sort-31

Geometry-30
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Game Theory-24
Hash Function-24
Binary Indexed Tree-21

Interactive-18
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Number Theory-16
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Concurrency-9
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Shell-4

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