Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Largest Sum of Averages in C++
Suppose we partition a row of numbers A into at most K adjacent groups, then we will set score as the sum of the average of each group. We have to find that what is the largest score that we can achieve. Suppose A = [9,1,2,3,9] and K is 3, then the result will be 20, this is because, the best choice is to partition A into [9], [1, 2, 3], [9]. So the answer is 9 + (1 + 2 + 3) / 3 + 9 = 20. We could have also partitioned A into [9, 1], [2], [3, 9],
To solve this, we will follow these steps −
- Define a matrix dp
- Define a recursive method solve(), This will take array A, index and k
- if index >= size of A, then return 0
- if k is 0, then return -100000
- if dp[index, k] is not – 1, then return dp[index, k]
- ret := -inf and sum := 0
- for i in range index to size of A – 1
- sum := sum + A[i]
- ret := max of ret and sum/(i – index + 1) + solve(A, i + 1, k – 1)
- set dp[index, k] := ret and return.
- From the main method, do the following −
- n := size of A
- dp := a matrix n x (K + 1), fill it with – 1
- return solve(A, 0, K)
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
vector < vector <double> > dp;
double solve(vector <int>& A, int idx, int k){
if(idx >= A.size()) return 0;
if(!k) return -100000;
if(dp[idx][k] != -1) return dp[idx][k];
double ret = INT_MIN;
double sum = 0;
for(int i = idx; i < A.size(); i++){
sum += A[i];
ret = max(sum / (i - idx + 1) + solve(A, i + 1, k - 1), ret);
}
return dp[idx][k] = ret;
}
double largestSumOfAverages(vector<int>& A, int K) {
int n = A.size();
dp = vector < vector <double> > (n, vector <double>(K + 1, -1));
return solve(A, 0, K);
}
};
main(){
vector<int> v = {9,1,2,3,9};
Solution ob;
cout << (ob.largestSumOfAverages(v, 3));
}Input
[9,1,2,3,9] 3
Output
20
Updated on: 04-May-2020
79 Views
Advertisements