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.
Example:
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]]
Solution:
The idea is to first sort by height which is the first index and then sort by the number of people before them.
For the above input [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]].
We will first sort by decreasing height which will result in
Arrays.sort(people, (a,b)->a[0]==b[0]?a[1]-b[1]:b[0]-a[0])[[7,0],[7,1],[6,1],[5,0],[4,4]]Now we need to place them in the order of the weight, so we loop through the array and depending on the weight we will add to the list at the position of the weight.
Below code adds the people to the location with respect to their weight.
for(int i=0;i<people.length;i++){
list.add(people[i][1],people[i]);
}Complete code:
class Solution {
public int[][] reconstructQueue(int[][] people) {
Arrays.sort(people, (a,b) -> a[0]==b[0]?a[1]-b[1]:b[0]-a[0]);
List<int[]> returnVal = new ArrayList<>();
for(int i=0;i<people.length;i++){
returnVal.add(people[i][1],people[i]);
}
return returnVal.toArray(new int[people.length][2]);
}
}The code run in O(N*N) as the sort takes N*N comparing both the height and weight.