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Question:

Given a change amount, print all possible combinations using different sets of coins

Solution:

核心思想：递归

1. Sort coins from large to small based on amount

2. Determine how many large coins used

3. Calculate how much the amount left using a given number of large coins

4. Recursion to the deducted value using sub-set of the coins

5. Print out of the remaining amount dividable by the smallest coin amount

package org.fool.java.test;

public class PrintAllCoinChangeCombinationTest {
    public static void main(String[] args) {
        int[] coins = {25, 10, 5, 1};
        int[] counts = new int[coins.length];
        System.out.println("All possible coin combinations of 10 cents are: ");
        printCombination(coins, counts, 0, 10);

        System.out.println("All possible coin combinations of 25 cents are: ");
        printCombination(coins, counts, 0, 25);
    }

    // coins are the sorted coins in descending order, larger positioned more front
    // counts record the number of coins at certain location
    // start index is keep cracking of from which coin we start processing after choosing one larger coin amount
    // total amount keep track of remaining amount left processing
    private static void printCombination(int[] coins, int[] counts, int startIndex, int totalAmount) {
        if (startIndex >= coins.length) {
            // format the print out as "amount=?*25 + ?*10 + ..."
            for (int i = 0; i < coins.length; i++) {
                System.out.print("" + counts[i] + " * " + coins[i] + " + ");
            }

            System.out.println();

            return;
        }
        // notice if startIndex is the last one, we need check if it can be dividable by the smallest coin
        // if so, this is a good combination, otherwise, this is not possible combination thus discarded
        if (startIndex == coins.length - 1) {
            if (totalAmount % coins[startIndex] == 0) { // good combination
                // set the counts of coins at start index
                counts[startIndex] = totalAmount / coins[startIndex];
                // proceed to recursive call
                printCombination(coins, counts, startIndex + 1, 0); // notice startIndex + 1 and remaining amount = 0
            }
        } else {    // we still have option to choose 0-N larger coins
            for (int i = 0; i <= totalAmount / coins[startIndex]; i++) {
                // for every cycle in a loop, we choose an arbitrary number of larger coins and proceed next
                counts[startIndex] = i;
                // notice we need to update the remaining amount
                printCombination(coins, counts, startIndex + 1, totalAmount - coins[startIndex] * i);
            }
        }
    }
}

Console Output

Reference

https://www.youtube.com/watch?v=3VBjhiKUtmE&list=PLlhDxqlV_-vkak9feCSrnjlrnzzzcopSG&index=38