LEETCODE JUNE 2021 DAILY CODING CHALLENGE

 LEETCODE JUNE 2021 DAILY CODING CHALLENGE

JUNE 1

JUNE 2

JUNE 3

JUNE 4

JUNE 5

Maximum Performance of a Team

LINK

You are given two integers n and k and two integer arrays speed and efficiency both of length n. There are n engineers numbered from 1 to n. speed[i] and efficiency[i] represent the speed and efficiency of the ith engineer respectively.

Choose at most k different engineers out of the n engineers to form a team with the maximum performance.

The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.

Return the maximum performance of this team. Since the answer can be a huge number, return it modulo 109 + 7.

 

const int mod = 1e9 + 7;

    int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {

        vector<pair<int, int>>ord;

        

        for(int i = 0; i < n; i++){

            ord.push_back({efficiency[i], speed[i]});

        }

        sort(ord.rbegin(), ord.rend());

        priority_queue<int>speedPq;

        

        long totalSpeed = 0, res = 0;

        

        for(auto& p : ord){

            int spd = p.second;

            speedPq.push(-spd);

            if(speedPq.size() <= k)

                totalSpeed += spd;

            else{

                totalSpeed += spd + speedPq.top();

                speedPq.pop();

            }

            res = max(res, totalSpeed*p.first);

        }

        return res% mod;

    }

LEETCODE DAILY CODING CHALLENGES SOLUTIONS

 MAY DAILY CODING CHALLENGE

1 MAY

2 MAY

3 MAY

4 MAY

5 MAY

6 MAY

7 MAY

8 MAY

9 MAY

CONSTRUCT TARGET ARRAY WITH MULTIPLE SUMS

REFER

Given an array of integers target. From a starting array, A consisting of all 1's, you may perform the following procedure :

• let x be the sum of all elements currently in your array.
• choose index i, such that 0 <= i < target.size and set the value of A at index i to x.
• You may repeat this procedure as many times as needed.

Return True if it is possible to construct the target array from A otherwise return False.

bool isPossible(vector<int>& A) {

        long total = 0;

        int n = A.size(), x;

        priority_queue<int>pq(A.begin(), A.end());

        for(int a : A)

            total += a;

        while(true){

            x = pq.top();

            pq.pop();

            total -= x;

            if(x == 1 || total == 1)

                return true;

            if(x < total || total == 0 || x % total == 0)

                return false;

            x = x % total;

            total += x;

            pq.push(x);

        }

    }

10 MAY