Today I have done Reduce Array Size to The Half on leetcode and leetcode's December LeetCoding Challenge with cpp.

Reduce Array Size to The Half

Description

Given an array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

Example 3:

Input: arr = [1,9]
Output: 1

Example 4:

Input: arr = [1000,1000,3,7]
Output: 1

Example 5:

Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5

Constraints:

• 1 <= arr.length <= 10^5
• arr.length is even.
• 1 <= arr[i] <= 10^5

Solution

nothing to say

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        unordered_map<int, int> m;
        for(auto i : arr) {
            m[i] += 1;
        }
        if(m.size() < 3) return 1;
        int len = arr.size();
        vector<int> cnt;
        for(auto [i, c] : m) {
            cnt.push_back(c);
        }
        sort(cnt.begin(), cnt.end(), greater<int>());
        if(cnt[0] >= len/2) return 1;
        for(int i = 1; i < cnt.size(); ++i) {
            cnt[i] += cnt[i-1];
            if(cnt[i] >= len/2) return i+1;
        }
        return len;
    }
};

December LeetCoding Challenge 7

Description

Spiral Matrix II

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

Example 1:

Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]

Example 2:

Input: n = 1
Output: [[1]]

Constraints:

• 1 <= n <= 20

Solution

in fact I used pick one and have already done it on this morning, so I just do another random problem.

class Solution {
    int move[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int sat(vector<vector<int>> &result, int newH, int newW, int n) {
        return newH >= 0 && newW >= 0 && newH < n && newW < n && !result[newH][newW];
    }
public:
    vector<vector<int>> generateMatrix(int n) {
        auto result = vector<vector<int>>(n, vector<int>(n, 0));
        result[0][0] = 1;
        if(n == 1) return result;
        int h = 0;
        int w = 1;
        int direction = 0;
        while(true) {
            result[h][w] = result[h-move[direction][0]][w-move[direction][1]] + 1;
            int turn = 0;
            while(turn < 4 && !sat(result, h + move[direction][0], w + move[direction][1], n)) {
                direction = (direction + 1) % 4;
                turn += 1;
            }
            if(turn == 4) break;
            h += move[direction][0];
            w += move[direction][1];
        }
        return result;
    }
};