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As a result, dynamic programming algorithms are highly optimized. Your code has many minor problems, and two major design flaws. You have two options for each coin: include it or exclude it. So, Time Complexity = O (A^m), where m is the number of coins given (Think!) Now that you have grasped the concept of dynamic programming, look at the coin change problem. See the following recursion tree for coins[] = {1, 2, 3} and n = 5. Reference:https://algorithmsndme.com/coin-change-problem-greedy-algorithm/, https://algorithmsndme.com/coin-change-problem-greedy-algorithm/. Hence, 2 coins. To make 6, the greedy algorithm would choose three coins (4,1,1), whereas the optimal solution is two coins (3,3) Hence, we need to check all possible combinations. Is it possible to create a concave light? Unlike Greedy algorithm [9], most of the time it gives the optimal solution as dynamic . Small values for the y-axis are either due to the computation time being too short to be measured, or if the number of elements is substantially smaller than the number of sets ($N \ll M$). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Coin change problem : Algorithm1. Is it suspicious or odd to stand by the gate of a GA airport watching the planes? Why does the greedy coin change algorithm not work for some coin sets? Now, take a look at what the coin change problem is all about. Similarly, the third column value is 2, so a change of 2 is required, and so on. The main caveat behind dynamic programming is that it can be applied to a certain problem if that problem can be divided into sub-problems. Subtract value of found denomination from amount. The algorithm only follows a specific direction, which is the local best direction. O(numberOfCoins*TotalAmount) is the space complexity. Therefore, to solve the coin change problem efficiently, you can employ Dynamic Programming. Minimum coins required is 2 Time complexity: O (m*V). int findMinimumCoinsForAmount(int amount, int change[]){ int numOfCoins = sizeof(coins)/sizeof(coins[0]); int count = 0; while(amount){ int k = findMaxCoin(amount, numOfCoins); if(k == -1) printf("No viable solution"); else{ amount-= coins[k]; change[count++] = coins[k]; } } return count;} int main(void) { int change[10]; // This needs to be dynamic int amount = 34; int count = findMinimumCoinsForAmount(amount, change); printf("\n Number of coins for change of %d : %d", amount, count); printf("\n Coins : "); for(int i=0; i