Maximum amount of capital required for selecting at most K projects

Given an integer N, representing number of projects, two arraysP[] and C[], consisting of N integers, and two integers W and K where, W is the initial capital amount, P[i] and C[i] are the profits and capital required to choose the i^th project. The task is to calculate the maximum amount of capital required to choose at most K projects, such that the profit of the chosen projects is added to W and choosing any project required at least C[i].

Examples:

Input: N = 3, W = 0, K = 2, P[] = {1, 2, 3}, C[] = {0, 1, 1}
Output: 4
Explanation:
Project 1: With the given amount of W as 0, the 0^th project can be chosen at a cost of C[0] i.e., 0, and after finishing the capital added to W i.e., W becomes W + P[0] = 1.
Project 2: With the given amount of W as 1, the 2^nd project can be chosen at a cost of C[2] i.e., 1, and after finishing the capital added to W i.e., W becomes W + P[2] = 1 + 3 = 4.

Input: N = 3, W = 1, K = 1, P[] = {10000, 2000, 3000}, C[] = {1, 1, 1}
Output: 10001

Approach: The given problem can be solved using the Greedy Algorithm and priority queue. Follow the steps below to solve the problem:

• Initialize a priority_queuePQ to store all the project's profit whose capital is at most W.
• Traverse the array C[] using the variable i as the index and push the pair {C[i], i} in a vector of pairs V.
• Sort the vector V in ascending order withrespect to the first element.
• Iterate until K is greater than 0:
  • Push profit of all the projects that have not been selected yet and whose capital is at most W in the priority queue PQ.
  • Increment the capital amount W by the maximum element of PQ i.e., W += PQ.top() and then pop the top element of PQ.
  • Decrement the K by 1.
• After completing the above steps, print the value of W as the maximum capital obtained.

Below is the implementation of the above approach:

4

Time Complexity: O((N + K) * log N)
Auxiliary Space: O(N)