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rodcutting.cpp
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78 lines (56 loc) · 1.82 KB
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#include <iostream>
using namespace std;
// RECURSION + MEMOIZATION
// int f(int index, int N, vector<int>&price, vector<vector<int>>&memo) {
// if(index == 0) {
// return price[0]*N;
// }
// if(memo[index][N] != -1) return memo[index][N];
// int notTake = 0 + f(index-1, N, price, memo);
// int take = INT_MIN;
// if(index+1 <= N)
// take = price[index] + f(index, N - index-1 ,price, memo);
// return memo[index][N] = max(notTake, take);
// }
// TABULATION METHOD
// int getProfit(vector<int>&price, int N) {
// int n = price.size();
// vector<vector<int>>memo(n, vector<int>(N+1, 0));
// for(int target = 0; target <= N; target++) {
// memo[0][target] = target*price[0];
// }
// for(int i = 1; i < n ;i++) {
// for(int target = 0; target <= N; target++) {
// int notTake = 0 + memo[i-1][target];
// int take = INT_MIN;
// if(i+1 <= target)
// take = price[i] + memo[i][target - i-1];
// memo[i][target] = max(notTake, take);
// }
// }
// return memo[n-1][N];
// }
// SPACE OPTIMIZATION
int getProfit(vector<int>&price, int N) {
int n = price.size();
vector<int>prev(N+1, 0);
for(int target = 0; target <= N; target++) {
prev[target] = target*price[0];
}
for(int i = 1; i < n; i++) {
for(int target = 0; target <= N; target++) {
int notTake = 0 + prev[target];
int take = INT_MIN;
if(i+1 <= target)
take = price[i] + prev[target - i - 1];
prev[target] = max(notTake, take);
}
}
return prev[N];
}
int main() {
vector<int>price = {2,5,7,8,10};
int N = 5;
cout << "Maximum profit by selling pieces of rod is " << getProfit(price, N) << endl;
return 0;
}