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

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

[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, S) = 3

max(S, S) = 2

max(S, S) = 2```

### Solution :

```                        ```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, make_pair(t, t));
}
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] for i in range(K)]
heapq.heapify(yheap)

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

return ans

A.sort()
B = [[A[i], A[i]] for i in range(K)]
ans = A[K - 1] + 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], A[i]])
ans = min(ans, A[i] + solve_2D(B))

return ans```
```

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