[Coding Made Simple] Coin Changes Number of ways to get a total

Given coins of certain denominations and a total, how many ways these coins can be combined to get the total.

 

Dynamic Programming solution 

State: T[i][j]: given the first i coins, the total number of ways these coins can be combined to get the total j.

Function: T[i][j] = T[i - 1][j] + T[i][j - coins[i - 1]],  if j >= coins[i - 1];

　　　　 T[i][j] = T[i - 1][j], if j < coins[i - 1];

Initialization:

if the total is 0, then T[i][0] = 1,i is from 0 to coins.length;  the only way is to not select any coins;

if there is no coins to select from, then T[0][j] = 0, j is from 1 to total, as there is 0 way to get a non zero total if we have no coins to select from;

Answer: T[coins.length][total]; 

 

public class CoinChange {
    public int minCoinsToGetTotal(int[] coins, int total) {
        if(total == 0) {
            return 1;
        }
        if(total != 0 && (coins == null || coins.length == 0)) {
            return 0;
        }
        int[][] T = new int[coins.length + 1][total + 1];
        for(int j = 0; j < T[0].length; j++) {
            T[0][j] = 0;
        }
        for(int i = 0; i < T.length; i++) {
            T[i][0] = 1;
        }
        for(int i = 1; i < T.length; i++) {
            for(int j = 1; j < T[0].length; j++) {
                if(j >= coins[i - 1]) {
                    T[i][j] = T[i - 1][j] + T[i][j - coins[i - 1]];
                }
                else {
                    T[i][j] = T[i - 1][j];
                }
            }
        }
        return T[coins.length][total];
    }
}

 

 

Related Problems 

Coin Changes Minimum Number of Coins