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coin_change_topdown.cpp
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coin_change_topdown.cpp
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/**
* @file
* @brief [Minimum coins](https://leetcode.com/problems/coin-change/) change
* problem is a problem used to find the minimum number of coins required to
* completely reach a target amount.
*
* @details
* This problem can be solved using 2 methods:
* 1. Top down approach
* 2. Bottom up appraoch
* Top down approach involves a vector with all elements initialised to 0.
* It is based on optimal substructure and overlapping subproblems.
* Overall time complexity of coin change problem is O(n*t)
* For example: example 1:-
* Coins: {1,7,10}
* Target:15
* Therfore minimum number of coins required = 3 of denomination 1,7 and 7.
* @author [Divyansh Kushwaha](https://github.com/webdesignbydivyansh)
*/
#include <cassert> // for assert
#include <climits> // for INT_MAX
#include <iostream> // for io operations
#include <vector> // for std::vector
/**
* @namespace dynamic_programming
* @brief Dynamic Programming algorithm
*/
namespace dynamic_programming {
/**
* @namespace mincoins_topdown
* @brief Functions for [minimum coin
* exchange](https://leetcode.com/problems/coin-change/) problem
*/
namespace mincoins_topdown {
/**
* @brief This implementation is for finding minimum number of coins .
* @param T template-type to use any kind of value
* @param n amount to be reached
* @param coins vector of coins
* @param t deontes the number of coins
* @param dp initilised to 0
* @returns minimum number of coins
*/
template <typename T>
int64_t mincoins(const T &n, const std::vector<T> &coins, const int16_t &t,
std::vector<T> dp) {
if (n == 0) {
return 0;
}
if (dp[n] != 0) {
return dp[n];
}
int ans = INT_MAX; // variable to store min coins
for (int i = 0; i < t; i++) {
if (n - coins[i] >= 0) { // if after subtracting the current
// denomination is it greater than 0 or not
int sub = mincoins(n - coins[i], coins, t, dp);
ans = std::min(ans, sub + 1);
}
}
dp[n] = ans;
return dp[n]; // returns minimum number of coins
}
} // namespace mincoins_topdown
} // namespace dynamic_programming
/**
* @brief Test implementations
* @returns void
*/
static void test() {
// example 1: number of coins=3 and minimum coins required=3(7,7,1)
const int64_t n1 = 15;
const int8_t t1 = 3, a1 = 0;
std::cout << "\nTest 1...";
std::vector<int64_t> arr1{1, 7, 10};
std::vector<int64_t> dp1(n1 + 1);
fill(dp1.begin(), dp1.end(), a1);
assert(dynamic_programming::mincoins_topdown::mincoins(n1, arr1, t1, dp1) ==
3);
std::cout << "Passed\n";
}
/**
* @brief Main function
* @returns 0 on exit
*/
int main() {
test(); // execute the test
return 0;
}