2023-09-11 Daily Challenge
Today I have done leetcode's September LeetCoding Challenge with cpp
.
September LeetCoding Challenge 11
Description
Group the People Given the Group Size They Belong To
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:
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:
Input: groupSizes = [2,1,3,3,3,2] Output: [[1],[0,5],[2,3,4]]
Constraints:
groupSizes.length == n
1 <= n <= 500
1 <= groupSizes[i] <= n
Solution
class Solution {
public:
vector<vector<int>> groupThePeople(vector<int>& groupSizes) {
unordered_map<int, vector<int>> mp;
for(int i = 0; i < groupSizes.size(); ++i) {
mp[groupSizes[i]].push_back(i);
}
vector<vector<int>> answer;
for(auto &[sz, people] : mp) {
for(int i = 0; i * sz < people.size(); ++i) {
vector<int> result;
for(int j = 0; j < sz; ++j) {
result.push_back(people[i * sz + j]);
}
answer.emplace_back(move(result));
}
}
return answer;
}
};
// Accepted
// 103/103 cases passed (11 ms)
// Your runtime beats 57.7 % of cpp submissions
// Your memory usage beats 46.45 % of cpp submissions (13.2 MB)