Given an array arr[] of N integers and an integer K. Also given are Q queries which have two numbers L and R. For every query, you can increase all the elements of the array in the index range [L, R] by 1. The task is to choose exactly K queries out of Q queries such that the sum of the array at the end is maximized. Print the sum after performing K such queries.
Examples:
Input: arr[] = {1, 1, 2, 2, 2, 3},
que[] = {{0, 4}, {1, 2}, {2, 5}, {2, 3}, {2, 4}},
K = 3
Output: 23
We choose the first, third and the fifth query.
After performing first query -> arr[] = {2, 2, 3, 3, 3, 3}
After performing third query -> arr[] = {2, 2, 4, 4, 4, 4}
After performing fifth query -> arr[] = {2, 2, 5, 5, 5, 4}
And the array sum is 2 + 2 + 5 + 5 + 5 + 4 = 23.Input: arr[] = {4, 5, 4, 21, 22},
que[] = {{1, 2}, {2, 2}, {2, 4}, {2, 2}},
K = 2
Output: 61
Naive approach: A naive approach is to use Dynamic Programming and Combinatorics, in which we choose any K queries out of Q. The combination which gives the maximum sum of the array will be the answer.
Time Complexity: O(N*N*K)
Efficient Approach: Since we need to maximize the sum of the array at the end. We just need to choose those queries that affect the maximum number of elements from the array i.e. with bigger ranges. Every query contributes (R – L + 1) to the increase in the sum if it is chosen. The sum of the array elements after performing such queries will be (initial sum of the array + (Contribution of K queries)).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to perform K queries out // of Q to maximize the final sum int getFinalSum(int a[], int n, pair<int, int> queries[], int q, int k) { int answer = 0; // Get the initial sum // of the array for (int i = 0; i < n; i++) answer += a[i]; vector<int> contribution; // Stores the contriution of every query for (int i = 0; i < q; i++) { contribution.push_back(queries[i].second - queries[i].first + 1); } // Sort the contribution of queries // in descending order sort(contribution.begin(), contribution.end(), greater<int>()); int i = 0; // Get the K most contributions while (i < k) { answer += contribution[i]; i++; } return answer; } // Driver code int main() { int a[] = { 1, 1, 2, 2, 2, 3 }; int n = sizeof(a) / sizeof(a[0]); pair<int, int> queries[] = { { 0, 4 }, { 1, 2 }, { 2, 5 }, { 2, 3 }, { 2, 4 } }; int q = sizeof(queries) / sizeof(queries[0]); int k = 3; cout << getFinalSum(a, n, queries, q, k); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { //pair class static class pair { int first,second; pair(int f,int s) { first = f; second = s; } } // Function to perform K queries out // of Q to maximize the final sum static int getFinalSum(int a[], int n, pair queries[], int q, int k) { int answer = 0; // Get the initial sum // of the array for (int i = 0; i < n; i++) answer += a[i]; Vector<Integer> contribution = new Vector<Integer>(); // Stores the contriution of every query for (int i = 0; i < q; i++) { contribution.add(queries[i].second - queries[i].first + 1); } //compartor Comparator<Integer> Comp = new Comparator<Integer>() { public int compare(Integer e1,Integer e2) { if(e1 > e2) return -1; else return 1; } }; // Sort the contribution of queries // in descending order Collections.sort(contribution,Comp); int i = 0; // Get the K most contributions while (i < k) { answer += (int) contribution.get(i); i++; } return answer; } // Driver code public static void main(String args[]) { int a[] = { 1, 1, 2, 2, 2, 3 }; int n = a.length; pair queries[] = new pair[5]; queries[0] = new pair( 0, 4 ); queries[1] = new pair( 1, 2 ); queries[2] = new pair( 2, 5 ); queries[3] = new pair( 2, 3 ); queries[4] = new pair( 2, 4 ); int q = queries.length; int k = 3; System.out.println( getFinalSum(a, n, queries, q, k)); } } // This code is contributed by Arnab Kundu |
Python3
# Python 3 implementation of the approach # Function to perform K queries out # of Q to maximize the final sum def getFinalSum(a, n, queries, q, k): answer = 0 # Get the initial sum # of the array for i in range(n): answer += a[i] contribution = [] # Stores the contriution of every query for i in range(q): contribution.append(queries[i][1]- queries[i][0] + 1) # Sort the contribution of queries # in descending order contribution.sort(reverse = True) i = 0 # Get the K most contributions while (i < k): answer += contribution[i] i += 1 return answer # Driver code if __name__ == '__main__': a = [1, 1, 2, 2, 2, 3] n = len(a) queries = [[0, 4], [1, 2], [2, 5], [2, 3], [2, 4]] q = len(queries); k = 3 print(getFinalSum(a, n, queries, q, k)) # This code is contributed by # Surendra_Gangwar |
23
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