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LC1282 - Group the People Given the Group Size They Belong To

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Datastructure and Algorithm: DAY 73. Leetcode: 253/2736

Description

There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1.
You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3.
Return a list of groups such that each person i is in a group of size groupSizes[i].
Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.

Example 1:

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Input: groupSizes = [3,3,3,3,3,1,3]
Output: [[5],[0,1,2],[3,4,6]]
Explanation: 
The first group is [5]. The size is 1, and groupSizes[5] = 1.
The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].

Example 2:

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

Constraints:

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groupSizes.length == n
1 <= n <= 500
1 <= groupSizes[i] <= n

Solutions

  1. key: length of the group, value: the index of each element
  2. store them in the dictionary
  3. for loop value and use v[i:i+k] to slice the string based on the key
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from sortedcontainers import SortedList
class Solution:
    def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
        new_group = SortedList()
        for i, v in enumerate(groupSizes):
            new_group.add((v,i))
        # print(new_group)

        i = 0
        ans = []
        while i < len(groupSizes):
            size, y = new_group[i]
            group = []
            for s in range(size):
                group.append(new_group[i+s][1])
            i = i+s+1
            ans.append(group)
        return ans

#Runtime: 234 ms, faster than 5.20% of Python3 online submissions for Group the People Given the Group Size They Belong To.
#Memory Usage: 14.5 MB, less than 17.57% of Python3 online submissions for Group the People Given the Group Size They Belong To.

Optimised HashTable

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class Solution:
    def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
        ans = []
        mp = defaultdict(list)
        for i, v in enumerate(groupSizes):
            mp[v].append(i)

        for k, v in mp.items():
            for i in range(0, len(v), k):
                ans.append(v[i:i+k])
            
        return ans

#Runtime: 89 ms, faster than 82.67% of Python3 online submissions for Group the People Given the Group Size They Belong To.
#Memory Usage: 13.9 MB, less than 88.49% of Python3 online submissions for Group the People Given the Group Size They Belong To.