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improved time complexity

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if all the price are same then we can do it in o(n).
improved time complexity.
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ABHISHEK-821005 2020-10-25 20:55:58 +05:30 committed by GitHub
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dynamic_programming/cut_rod.cpp
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@ -68,17 +68,71 @@ int maxProfitByCuttingRod(const std::array<int, T> &price, const int n) {
* @brief Function to test above algorithm
* @returns void
*/
template <size_t T>
bool WhatIfAllPricesAreSame(const std::array<int, T> &price, const int 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;
}
static void test() {
// Test 1
const int n1 = 8; // size of rod
std::array<int, n1> price1 = {1, 5, 8, 9, 10, 17, 17, 20}; // price array
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( 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 = {