Weekly Contest 168

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Words count in article: 466 | Reading time ≈ 2

5291. Find Numbers with Even Number of Digits

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class Solution {
public:
    int findNumbers(vector<int>& nums) {
        int count = 0;
        for(auto num: nums){
            int t = num;
            int rem = 0;
            if(t == 0) continue;
            while(t){
                t /= 10;
                rem++;
            }
            if(rem % 2 == 0) count++;
        }
        return count;
    }
};

5292. Divide Array in Sets of K Consecutive Numbers

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class Solution {
public:
    bool isPossibleDivide(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        for(auto i: nums) mp[i]++;
        sort(nums.begin(), nums.end());
        for(int i=0; i<nums.size(); i++){
            int cur = nums[i];
            if(mp.find(cur) == mp.end()) continue;
            for(int j=cur; j<cur+k; j++){
                if(mp.find(j) == mp.end() || mp[j] == 0) return false;
                mp[j]--;
                if(mp[j] == 0) mp.erase(j);
            }
        }
        return mp.size() == 0;
    }
};

5293. Maximum Number of Occurrences of a Substring

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class Solution {
public:
    int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
        unordered_map<string, int> subs{};
        int result = 0;
        for(int i=0; i<=s.size()-minSize; i++){
            string sub = s.substr(i, minSize);
            set<char> chars{};
            for(auto c: sub) chars.insert(c);
            if(chars.size() <= maxLetters) result = max(result, ++subs[sub]);
        }
        return result;
    }
};

5294. Maximum Candies You Can Get from Boxes

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class Solution {
public:
    int maxCandies(vector<int>& status, vector<int>& candies, vector<vector<int>>& keys, vector<vector<int>>& containedBoxes, vector<int>& initialBoxes) {
        int result=0; //total number of candies
        vector<int> boxes(initialBoxes.begin(),initialBoxes.end()); //the boxes we have (including both open or not open)
        unordered_set<int> v; //the boxes that we have already visited
        while(true)
        {
            int flag=0; //check if the process ends
            for(int i=(int)boxes.size()-1;i>-1;i--)
            {
                 if(status[boxes[i]]==1&&v.find(boxes[i])==v.end())//is open and not visited yet
                 {
                     flag=1; //the process doesn't end
                     for(int j=0;j<keys[boxes[i]].size();j++) status[keys[boxes[i]][j]]=1; //we use the key to open the boxes
                     for(int j=0;j<containedBoxes[boxes[i]].size();j++) boxes.push_back(containedBoxes[boxes[i]][j]); //we get more boxes
                     v.insert(boxes[i]); //the box has been visited
                     result+=candies[boxes[i]]; //we get the candies
                     boxes.erase(boxes.begin()+i); //the box has been visited, we removed it from the BFS process
                 }
             }
             if(flag==0) break; //if the process ends
        }
        return result;
    }
};
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