**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 :

` ````
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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