[Coding Made Simple] Coin Changes Minimum Number of Coins

Posted 2020-10-03 Push your limit!

tags:

篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了[Coding Made Simple] Coin Changes Minimum Number of Coins相关的知识，希望对你有一定的参考价值。

Given coins of certain denominations and a total, how many minimum coins would you need to make this total?

Dynamic Programming solution

State: T[i][j]: given the first i coins, the min number of coins needed to make a total of j.

Function:

case 1. T[i][j] = T[i - 1][j], if j < coins[i - 1], the ith coins can\'t be used;

case 2. T[i][j] = Math.min(T[i - 1][j], 1 + T[i][j - coins[i - 1]]), if j >= coins[i - 1], the ith coins can be used;

Since T[i][j] is set to -1 when we can\'t get a total of j out of the first i coins. So in case 2, we need to check

if T[i - 1][j] or T[i][j - coins[i - 1]] is negative. If either one of them is negative, then we need to handle

them separately.

The following implementation also includes how to reconstruct one of the optimal solution.

 1 import java.util.ArrayList;
 2 
 3 public class CoinChange {
 4     private ArrayList<Integer> pickedCoins;
 5     public int minCoinsToGetTotal(int[] coins, int total) {
 6         if(total == 0) {
 7             return 0;
 8         }
 9         if(total != 0 && (coins == null || coins.length == 0)) {
10             return -1;
11         }
12         int[][] T = new int[coins.length + 1][total + 1];
13         for(int j = 0; j < T[0].length; j++) {
14             T[0][j] = -1;
15         }
16         for(int i = 0; i < T.length; i++) {
17             T[i][0] = 0;
18         }
19         for(int i = 1; i < T.length; i++) {
20             for(int j = 1; j < T[0].length; j++) {
21                 if(j >= coins[i - 1]) {
22                     if(T[i - 1][j] >= 0 && T[i][j - coins[i - 1]] >= 0) {
23                         T[i][j] = Math.min(T[i - 1][j], 1 + T[i][j - coins[i - 1]]);    
24                     }
25                     else if(T[i - 1][j] < 0 && T[i][j - coins[i - 1]] >= 0) {
26                         T[i][j] = 1 + T[i][j - coins[i - 1]];                        
27                     }
28                     else {
29                         T[i][j] = T[i - 1][j];
30                     }
31                 }
32                 else {
33                     T[i][j] = T[i - 1][j];
34                 }
35             }
36         }
37         int i = coins.length, j = total;
38         pickedCoins = new ArrayList<Integer>();
39         if(T[i][j] > 0) {
40             while(T[i][j] != 0) {
41                 if(j >= coins[i - 1] && T[i][j - coins[i - 1]] >= 0 && 1 + T[i][j - coins[i - 1]] == T[i][j]) {
42                     pickedCoins.add(coins[i - 1]);
43                     j -= coins[i - 1];                                        
44                 }
45                 else{
46                     i--;
47                 }
48             }
49         }
50         return T[coins.length][total];
51     }
52 }