/**
 * @file
 * @brief Implementation of cutting a rod problem
 *
 * @details
 * Given a rod of length n inches and an array of prices that
 * contains prices of all pieces of size<=n. Determine
 * the maximum profit obtainable by cutting up the rod and selling
 * the pieces.
 *
 * ### Algorithm
 * The idea is to break the given rod into every smaller piece as possible
 * and then check profit for each piece, by calculating maximum profit for
 * smaller pieces we will build the solution for larger pieces in bottom-up
 * manner.
 *
 * @author [Anmol](https://github.com/Anmol3299)
 * @author [Pardeep](https://github.com/Pardeep009)
 */

#include <array>
#include <cassert>
#include <climits>
#include <iostream>
/**
 * @namespace dynamic_programming
 * @brief Dynamic Programming algorithms
 */
namespace dynamic_programming {
/**
 * @namespace cut_rod
 * @brief Implementation of cutting a rod problem
 */
namespace cut_rod {
/**
 * @brief Cuts the rod in different pieces and
 * stores the maximum profit for each piece of the rod.
 * @tparam T size of the price array
 * @param n size of the rod in inches
 * @param price an array of prices that contains prices of all pieces of size<=n
 * @return maximum profit obtainable for @param n inch rod.
 */
template <size_t T>
int maxProfitByCuttingRod(const std::array<int, T> &price, const int n) {
    int *profit =
        new int[n + 1];  // profit[i] will hold maximum profit for i inch rod

    profit[0] = 0;  // if length of rod is zero, then no profit

    // outer loop will select size of rod, starting from 1 inch to n inch rod.
    // inner loop will evaluate the maximum profit we can get for i inch rod by
    // making every possible cut on it and will store it in profit[i].
    for (size_t i = 1; i <= n; i++) {
        int q = INT_MIN;
        for (size_t j = 1; j <= i; j++) {
            q = std::max(q, price[j - 1] + profit[i - j]);
        }
        profit[i] = q;
    }
    int ans = profit[n];
    delete[] profit;
    return ans;  // returning maximum profit
}



template <size_t T>
bool WhatIfAllPricesAreSame(const std::array<int, T> &price, const uint64_t &n){

    /*

      Note that if all the prices of the different lengths of rod are same the answer will be always n*price;
      where price=price of 1 length rod
      Reason::
      cR() ---> cutRod()

                             cR(4)
                  /        /
                 /        /
             cR(3)       cR(2)     cR(1)   cR(0)
            /  |         /         |
           /   |        /          |
      cR(2) cR(1) cR(0) cR(1) cR(0) cR(0)
     /        |          |
    /         |          |
  cR(1) cR(0) cR(0)      cR(0)
   /
 /
CR(0)


   if every length has same price , you would definitely want the rod of length 1, with n quantities.
    which will give us maximum profits and maximum cuts.

    */

    const int temp=price[0];
     for (size_t i = 1; i <n; i++) {
         if(price[i]!=temp)
             return false;
        }

        return true;
}// checks whether all the prices are same or not


}  // namespace cut_rod
}  // namespace dynamic_programming

/**
 * @brief Function to test above algorithm
 * @returns void
 */




static void test() {
    // Test 1
    const int n1 = 8;                                           // size of rod
    std::array<int, n1> price1 = {1, 1,1,1,1,1,1,1};  // price array
    const int max_profit1 =
        dynamic_programming::cut_rod::maxProfitByCuttingRod(price1, n1);
    const int expected_max_profit1 = 22;

    if(dynamic_programming::cut_rod::WhatIfAllPricesAreSame(price1,n1)){
        std::cout << "Maximum profit with " << n1 << " inch road is " <<(n1)*price1[0]
              << std::endl;
    }
    else{

    assert(max_profit1 == expected_max_profit1);
    std::cout << "Maximum profit with " << n1 << " inch road is " << max_profit1
              << std::endl;

    }



    // Test 2
    const int n2 = 30;  // size of rod
    std::array<int, n2> price2 = {
        1,  5,  8,  9,  10, 17, 17, 20, 24, 30,  // price array
        31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
        41, 42, 43, 44, 45, 46, 47, 48, 49, 50};
    const int max_profit2 =
        dynamic_programming::cut_rod::maxProfitByCuttingRod(price2, n2);
    const int expected_max_profit2 = 90;
    assert(max_profit2 == expected_max_profit2);
    std::cout << "Maximum profit with " << n2 << " inch road is " << max_profit2
              << std::endl;
}

/**
 * @brief Main function
 * @returns 0 on exit
 */
int main() {
    // Testing
    test();
    return 0;
}