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improved time complexity
if all the price are same then we can do it in o(n). improved time complexity.
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@ -68,17 +68,71 @@ int maxProfitByCuttingRod(const std::array<int, T> &price, const int n) {
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* @brief Function to test above algorithm
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* @brief Function to test above algorithm
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* @returns void
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* @returns void
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*/
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*/
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template <size_t T>
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bool WhatIfAllPricesAreSame(const std::array<int, T> &price, const int n){
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/*
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Note that if all the prices of the different lengths of rod are same the answer will be always n*price;
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where price=price of 1 length rod
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Reason::
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cR() ---> cutRod()
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cR(4)
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/ /
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/ /
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cR(3) cR(2) cR(1) cR(0)
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/ | / |
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/ | / |
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cR(2) cR(1) cR(0) cR(1) cR(0) cR(0)
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/ | |
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/ | |
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cR(1) cR(0) cR(0) cR(0)
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/
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/
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CR(0)
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if every length has same price , you would definitely want the rod of length 1, with n quantities.
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which will give us maximum profits and maximum cuts.
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*/
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const int temp=price[0];
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for (size_t i = 1; i <n; i++) {
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if(price[i]!=temp)
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return false;
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}
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return true;
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}
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static void test() {
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static void test() {
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// Test 1
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// Test 1
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const int n1 = 8; // size of rod
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const int n1 = 8; // size of rod
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std::array<int, n1> price1 = {1, 5, 8, 9, 10, 17, 17, 20}; // price array
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std::array<int, n1> price1 = {1, 1,1,1,1,1,1,1}; // price array
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const int max_profit1 =
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const int max_profit1 =
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dynamic_programming::cut_rod::maxProfitByCuttingRod(price1, n1);
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dynamic_programming::cut_rod::maxProfitByCuttingRod(price1, n1);
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const int expected_max_profit1 = 22;
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const int expected_max_profit1 = 22;
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if( WhatIfAllPricesAreSame(price1,n1)){
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std::cout << "Maximum profit with " << n1 << " inch road is " <<(n1)*price1[0]
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<< std::endl;
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}
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else{
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assert(max_profit1 == expected_max_profit1);
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assert(max_profit1 == expected_max_profit1);
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std::cout << "Maximum profit with " << n1 << " inch road is " << max_profit1
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std::cout << "Maximum profit with " << n1 << " inch road is " << max_profit1
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<< std::endl;
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<< std::endl;
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}
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// Test 2
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// Test 2
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const int n2 = 30; // size of rod
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const int n2 = 30; // size of rod
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std::array<int, n2> price2 = {
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std::array<int, n2> price2 = {
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