Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), write code to calculate the number of ways

This is a recursive problem, so let’s figure out how to do makeChange(n) using prior solutions (i.e., sub-problems). Let’s say n = 100, so we want to compute the number of ways of making change of 100 cents. What’s the relationship to its sub-problems?
We know that makeChange(100):
= makeChange(100 using 0 quarters) + makeChange(100 using 1 quarter) + makeChange(100 using 2 quarter) + makeChange(100 using 3 quarter) + makeChange(100 using 4 quarter)
Can we reduce this further? Yes!
= makeChange(100 using 0 quarters) + makeChange(75 using 0 quarter) + makeChange(50 using 0 quarters) + makeChange(25 using 0 quarters) + 1
Now what? We’ve used up all our quarters, so now we can start applying our next biggest denomination: dimes.
This leads to a recursive algorithm that looks like this:



class denomination
{

public static int makeChange(int n, int denom) {
int next_denom = 0;
switch (denom) {
case 25:
next_denom = 10;
break;
case 10:
next_denom = 5;
break;
case 5:
next_denom = 1;
break;
case 1:
return 1;
}
int ways = 0;
for (int i = 0; i * denom <= n; i++) {
ways += makeChange(n - i * denom, next_denom);
}
return ways;
}

public static void main(String a[])
{

System.out.println(makeChange(100, 25));


}

}


Time Complexity O(n)