Minimum Time to Finish K Tasks - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers tasks where each element has 3 integers. 

You are also given an integer k. 

Pick k rows from tasks, call it S, such that the following sum is minimized and return the sum:

max(S[0][0], S[1][0], ...S[k - 1][0]) +
max(S[0][1], S[1][1], ...S[k - 1][1]) +
max(S[0][2], S[1][2], ...S[k - 1][2])

In other words, each of the 3 columns contribute to a cost, and is calculated by taking the max value of 
that column in S. The max of an empty list is defined to be 0.



Constraints



k ≤ n ≤ 1,000 where n is the length of tasks

Example 1

Input

tasks = [

    [1, 2, 2],

    [3, 4, 1],

    [3, 1, 2]

]

k = 2

Output

7

Explanation

We pick the first row and the last row. And the total sum becomes



S = [[1,2,2],[3,1,2]]



max(S[0][0], S[1][0]) = 3

max(S[0][1], S[1][1]) = 2

max(S[0][2], S[1][2]) = 2



Solution :



title-img




                        Solution in C++ :

int solve(vector<vector<int>>& tasks, int k) {
    if (k == 0) return 0;
    int ret = 2e9;
    vector<pair<int, pair<int, int>>> v;
    for (auto& t : tasks) {
        v.emplace_back(t[0], make_pair(t[1], t[2]));
    }
    sort(v.begin(), v.end());
    for (int i = k - 1; i < v.size(); i++) {
        vector<pair<int, int>> twocol;
        for (int j = 0; j <= i; j++) {
            twocol.push_back(v[j].second);
        }
        sort(twocol.begin(), twocol.end());
        priority_queue<int> q;
        for (int j = 0; j < twocol.size(); j++) {
            q.push(twocol[j].second);
            if (q.size() > k) {
                q.pop();
            }
            if (q.size() == k) {
                ret = min(ret, v[i].first + twocol[j].first + q.top());
            }
        }
    }
    return ret;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, A, K):
        if not A or not K:
            return 0

        def solve_2D(B):
            B.sort()
            yheap = [-B[i][1] for i in range(K)]
            heapq.heapify(yheap)

            ans = B[K - 1][0] + (-yheap[0])
            for i in range(K, len(B)):
                x = B[i][0]
                heapq.heappushpop(yheap, -B[i][1])
                assert len(yheap) == K
                y = -yheap[0]
                ans = min(ans, x + y)

            return ans

        A.sort()
        B = [[A[i][1], A[i][2]] for i in range(K)]
        ans = A[K - 1][0] + max(y for y, z in B) + max(z for y, z in B)
        for i in range(K, len(A)):
            B.append([A[i][1], A[i][2]])
            ans = min(ans, A[i][0] + solve_2D(B))

        return ans
                    


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