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Add method for minimum number of coins required for given amount by bansalKeshav · Pull Request #468 · TheAlgorithms/Java · GitHub

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39 changes: 34 additions & 5 deletions Dynamic Programming/CoinChange.java
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Original file line number Diff line number Diff line change
Expand Up @@ -10,9 +10,11 @@ public class CoinChange {
public static void main(String[] args) {

int amount = 12;
int[] coins = {1, 2, 5};
int[] coins = {2, 4, 5};

System.out.println("Number of combinations of getting change for " + amount + " is: " + change(coins, amount));
System.out.println("Minimum number of coins required for amount :" + amount + " is: " + minimumCoins(coins, amount));

}

/**
Expand All @@ -29,22 +31,49 @@ public static int change(int[] coins, int amount) {

for (int coin : coins) {
for (int i=coin; i<amount+1; i++) {
combinations[i] += combinations[i-coin];
combinations[i] += combinations[i-coin];
}
// Uncomment the below line to see the state of combinations for each coin
// printAmount(combinations);
}

return combinations[amount];
}
/**
* This method finds the minimum number of coins needed for a given amount.
*
* @param coins The list of coins
* @param amount The amount for which we need to find the minimum number of coins.
* Finds the the minimum number of coins that make a given value.
**/
public static int minimumCoins(int[] coins, int amount) {
//minimumCoins[i] will store the minimum coins needed for amount i
int[] minimumCoins = new int[amount+1];

minimumCoins[0] = 0;

for(int i=1;i<=amount;i++){
minimumCoins[i]=Integer.MAX_VALUE;
}
for(int i=1;i<=amount;i++){
for (int coin :coins){
if(coin <=i){
int sub_res = minimumCoins[i-coin];
if (sub_res != Integer.MAX_VALUE && sub_res + 1 < minimumCoins[i])
minimumCoins[i] = sub_res + 1;
}
}
}
// Uncomment the below line to see the state of combinations for each coin
//printAmount(minimumCoins);
return minimumCoins[amount];
}

// A basic print method which prints all the contents of the array
public static void printAmount(int[] arr) {

for (int i=0; i<arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}

}
}

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