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 }
Related Problems
Coin Changes Number of ways to get a total