I am solving a problem to find out all the 4 digit Vampire numbers.
A Vampire Number v=x*y is defined as a number with 'n' even number of digits formed by multiplying a pair of 'n/2'-digit numbers (where the digits are taken from the original number in any order)x and y together. If v is a vampire number, then x&y and are called its "fangs."
Examples of vampire numbers are:
1. 1260=21*60
2. 1395=15*93
3. 1530=30*51
I have tried the brute force algorithm to combine different digits of a given number and multiply them together . But this method is highly inefficient and takes up a lot of time.
Is there a more efficient algorithmic solution to this problem?
Or you can use a property of vampire numbers described on this page (linked from Wikipedia) :
An important theoretical result found by Pete Hartley:
If x·y is a vampire number then x·y == x+y (mod 9)
Proof: Let mod be the binary modulo operator and d(x) the sum of the decimal
digits of x. It is well-known that d(x) mod 9 = x mod 9, for all x.
Assume x·y is a vampire. Then it contains the same digits as x and y,
and in particular d(x·y) = d(x)+d(y). This leads to:
(x·y) mod 9 = d(x·y) mod 9 = (d(x)+d(y)) mod 9 = (d(x) mod 9 + d(y) mod 9) mod 9
= (x mod 9 + y mod 9) mod 9 = (x+y) mod 9
The solutions to the congruence are (x mod 9, y mod 9) in {(0,0),
(2,2), (3,6), (5,8), (6,3), (8,5)}
So your code could look like this :
for(int i=18; i<100; i=i+9){ // 18 is the first multiple of 9 greater than 10
for(int j=i; j<100; j=j+9){ // Start at i because as #sh1 said it's useless to check both x*y and y*x
checkVampire(i,j);
}
}
for(int i=11; i<100; i=i+9){ // 11 is the first number greater than 10 which is = 2 mod 9
for(int j=i; j<100; j=j+9){
checkVampire(i,j);
}
}
for(int i=12; i<100; i=i+9){
for(int j=i+3; j<100; j=j+9){
checkVampire(i,j);
}
}
for(int i=14; i<100; i=i+9){
for(int j=i+3; j<100; j=j+9){
checkVampire(i,j);
}
}
// We don't do the last 2 loops, again for symmetry reasons
Since they are 40 elements in each of the sets like {(x mod 9, y mod 9) = (0,0); 10 <= x <= y <= 100}, you only do 4*40 = 160 iterations, when a brute-force gives you 10ˆ4 iterations. You can do even less operations if you take into account the >= 1000 constraint, for instance you can avoid checking if j < 1000/i.
Now you can easily scale up to find vampires with more than 4 digits =)
Iterate over all possible fangs (100 x 100 = 10000 possibilities), and find if their product has the same digits as the fangs.
Yet another brute force (C) version, with a free bubble sort to boot...
#include <stdio.h>
static inline void bubsort(int *p)
{ while (1)
{ int s = 0;
for (int i = 0; i < 3; ++i)
if (p[i] > p[i + 1])
{ s = 1;
int t = p[i]; p[i] = p[i + 1]; p[i + 1] = t;
}
if (!s) break;
}
}
int main()
{ for (int i = 10; i < 100; ++i)
for (int j = i; j < 100; ++j)
{ int p = i * j;
if (p < 1000) continue;
int xd[4];
xd[0] = i % 10;
xd[1] = i / 10;
xd[2] = j % 10;
xd[3] = j / 10;
bubsort(xd);
int x = xd[0] + xd[1] * 10 + xd[2] * 100 + xd[3] * 1000;
int yd[4];
yd[0] = p % 10;
yd[1] = (p / 10) % 10;
yd[2] = (p / 100) % 10;
yd[3] = (p / 1000);
bubsort(yd);
int y = yd[0] + yd[1] * 10 + yd[2] * 100 + yd[3] * 1000;
if (x == y)
printf("%2d * %2d = %4d\n", i, j, p);
}
return 0;
}
Runs pretty much instantaneously. Variable names aren't too descriptive, but should be pretty obvious...
The basic idea is to start with two potential fangs, break them down into digits, and sort the digits for easy comparison. Then we do the same with the product - break it down to digits and sort. Then we re-constitute two integers from the sorted digits, and if they're equal, we have a match.
Possible improvements: 1) start j at 1000 / i instead of i to avoid having to do if (p < 1000) ..., 2) maybe use insertion sort instead of bubble sort (but who's gonna notice those 2 extra swaps?), 3) use a real swap() implementation, 4) compare the arrays directly rather than building a synthetic integer out of them. Not sure any of those would make any measurable difference, though, unless you run it on a Commodore 64 or something...
Edit: Just out of curiosity, I took this version and generalized it a bit more to work for the 4, 6 and 8 digit cases - without any major optimization, it can find all the 8-digit vampire numbers in < 10 seconds...
This is an ugly hack (brute force, manual checking for permutations, unsafe buffer operations, produces dupes, etc.) but it does the job. Your new exercise is to improve it :P
Wikipedia claims that there are 7 vampire numbers which are 4 digits long. The full code has found them all, even some duplicates.
Edit: Here's a slightly better comparator function.
Edit 2: Here's a C++ version that uniques results (therefore it avoids duplicates) using an std::map (and stores the last occurrence of the particular vampire number along with its factors in it). It also meets the criterion that at least one of the factors should not end with 0, i. e. a number is not a vampire number if both of the multiplicands are divisible by then. This test looks for 6-digit vampire numbers and it does indeed find exactly 148 of them, in accordance with what Wikipedia sates.
The original code:
#include <stdio.h>
void getdigits(char buf[], int n)
{
while (n) {
*buf++ = n % 10;
n /= 10;
}
}
int is_vampire(const char n[4], const char i[2], const char j[2])
{
/* maybe a bit faster if unrolled manually */
if (i[0] == n[0]
&& i[1] == n[1]
&& j[0] == n[2]
&& j[1] == n[3])
return 1;
if (i[0] == n[1]
&& i[1] == n[0]
&& j[0] == n[2]
&& j[1] == n[3])
return 1;
if (i[0] == n[0]
&& i[1] == n[1]
&& j[0] == n[3]
&& j[1] == n[2])
return 1;
if (i[0] == n[1]
&& i[1] == n[0]
&& j[0] == n[3]
&& j[1] == n[2])
return 1;
// et cetera, the following 20 repetitions are redacted for clarity
// (this really should be a loop, shouldn't it?)
return 0;
}
int main()
{
for (int i = 10; i < 100; i++) {
for (int j = 10; j < 100; j++) {
int n = i * j;
if (n < 1000)
continue;
char ndigits[4];
getdigits(ndigits, n);
char idigits[2];
char jdigits[2];
getdigits(idigits, i);
getdigits(jdigits, j);
if (is_vampire(ndigits, idigits, jdigits))
printf("%d * %d = %d\n", i, j, n);
}
}
return 0;
}
I wouldn't have given up so easily on brute force. You have distinct set of numbers, 1000 to 9999 that you must run through. I would divide up the set into some number of subsets, and then spin up threads to handle each subset.
You could further divide the work be coming up with the various combinations of each number; IIRC my discrete math, you have 4*3*2 or 24 combinations for each number to try.
A producer / consumer approach might be worthwhile.
Iteration seems fine to me, since you only need to do this once to find all the values and you can just cache them afterwards. Python (3) version that takes about 1.5 seconds:
# just some setup
from itertools import product, permutations
dtoi = lambda *digits: int(''.join(str(digit) for digit in digits))
gen = ((dtoi(*digits), digits) for digits in product(range(10), repeat=4) if digits[0] != 0)
l = []
for val, digits in gen:
for check1, check2 in ((dtoi(*order[:2]), dtoi(*order[2:])) for order in permutations(digits) if order[0] > 0 and order[2] > 0):
if check1 * check2 == val:
l.append(val)
break
print(l)
Which will give you [1260, 1395, 1435, 1530, 1827, 2187, 6880]
EDIT: full brute force that weeds out identical X and Y values...
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class Vampire {
public static void main(String[] args) {
for (int x = 10; x < 100; x++) {
String sx = String.valueOf(x);
for (int y = x; y < 100; y++) {
int v = x * y;
String sy = String.valueOf(y);
String sv = String.valueOf(v);
if (sortVampire(sx + sy).equals(sortVampire(sv))) {
System.out.printf("%d * %d = %d%n", x, y, v);
}
}
}
}
private static List<Character> sortVampire(String v) {
List<Character> vc = new ArrayList<Character>();
for (int j = 0; j < v.length(); j++) {
vc.add(v.charAt(j));
}
Collections.sort(vc);
return vc;
}
}
Brute force version in C# with LINQ:
class VampireNumbers
{
static IEnumerable<int> numberToDigits(int number)
{
while(number > 0)
{
yield return number % 10;
number /= 10;
}
}
static bool isVampire(int first, int second, int result)
{
var resultDigits = numberToDigits(result).OrderBy(x => x);
var vampireDigits = numberToDigits(first)
.Concat(numberToDigits(second))
.OrderBy(x => x);
return resultDigits.SequenceEqual(vampireDigits);
}
static void Main(string[] args)
{
var vampires = from fang1 in Enumerable.Range(10, 89)
from fang2 in Enumerable.Range(10, 89)
where fang1 < fang2
&& isVampire(fang1, fang2, fang1 * fang2)
select new { fang1, fang2 };
foreach(var vampire in vampires)
{
Console.WriteLine(vampire.fang1 * vampire.fang2
+ " = "
+ vampire.fang1
+ " * "
+ vampire.fang2);
}
}
}
Similar to someone mentioned above, my method is to first find all permutations of a number, then split them in half to form two 2-digit numbers, and test if their product equal to the original number.
Another interesting discussion above is how many permutations a number can have. Here is my opinion:
(1) a number whose four digitals are the same has 1 permutation;
(2) a number who has only two different digits has 6 permutations (it doesn't matter if it contains zeros, because we don't care after permutation if it is still a 4-digit number);
(3) a number who has three different digits has 12 permutations;
(4) a number with all four different digits has 24 permutations.
public class VampireNumber {
// method to find all permutations of a 4-digit number
public static void permuta(String x, String s, int v)
{for(int i = 0; i < s.length(); i++)
{permuta( x + s.charAt(i), s.substring(0,i) + s.substring(i+1), v);
if (s.length() == 1)
{x = x + s;
int leftpart = Integer.parseInt(x.substring(0,2));
int rightpart = Integer.parseInt(x.substring(2));
if (leftpart*rightpart == v)
{System.out.println("Vampir = " + v);
}
}
}
}
public static void main(String[] args){
for (int i = 1000; i < 10000; i++) {
permuta("", Integer.toString(i), i); //convert the integer to a string
}
}
}
The approach I would try would be to loop through each number in [1000, 9999], and test if any permutation of its digits (split in the middle) multiplied to make it.
This will require (9999 - 1000) * 24 = 215,976 tests, which should execute acceptably fast on a modern machine.
I would definitely store the digits separately, so you can avoid having to do something like a bunch of division to extract the digits from a single integer.
If you write your code such that you're only ever doing integer addition and multiplication (and maybe the occasional division to carry), it should be pretty fast. You could further increase the speed by skipping two-digit pairs which "obviously" won't work - e.g., ones with leading zeros (note that the largest product than can be produced by a one digit number and a two digit number is 9 * 99, or 891).
Also note that this approach is embarassingly parallel (http://en.wikipedia.org/wiki/Embarrassingly_parallel), so if you really need to speed it up even more then you should look into testing the numbers in separate threads.
<?php
for ($i = 10; $i <= 99; $j++) {
// Extract digits
$digits = str_split($i);
// Loop through 2nd number
for ($j = 10; $j <= 99; $j++) {
// Extract digits
$j_digits = str_split($j);
$digits[2] = $j_digits[0];
$digits[3] = $j_digits[1];
$product = $i * $j;
$product_digits = str_split($product);
// check if fangs
$inc = 0;
while (in_array($digits[$inc], $product_digits)) {
// Remove digit from product table
/// So AAAA -> doesnt match ABCD
unset($product_digits[$array_serach($digits[$inc], $product_digits)]);
$inc++;
// If reached 4 -> vampire number
if ($inc == 4) {
$vampire[] = $product;
break;
}
}
}
}
// Print results
print_r($vampire);
?>
Took less than a second on PHP. couldn't even tell it had to run 8100 computations... computers are fast!
Results:
Gives you all the 4 digits plus some are repeated. You can further process the data and remove duplicates.
It seems to me that to perform the fewest possible tests without relying on any particularly abstract insights, you probably want to iterate over the fangs and cull any obviously pointless candidates.
For example, since x*y == y*x about half your search space can be eliminated by only evaluating cases where y > x. If the largest two-digit fang is 99 then the smallest which can make a four-digit number is 11, so don't start lower than 11.
EDIT:
OK, throwing everything I thought of into the mix (even though it looks silly against the leading solution).
for (x = 11; x < 100; x++)
{
/* start y either at x, or if x is too small then 1000 / x */
for (y = (x * x < 1000 ? 1000 / x : x); y < 100; y++)
{
int p = x * y;
/* if sum of digits in product is != sum of digits in x+y, then skip */
if ((p - (x + y)) % 9 != 0)
continue;
if (is_vampire(p, x, y))
printf("%d\n", p);
}
}
and the test, since I haven't seen anyone use a histogram, yet:
int is_vampire(int p, int x, int y)
{
int h[10] = { 0 };
int i;
for (i = 0; i < 4; i++)
{
h[p % 10]++;
p /= 10;
}
for (i = 0; i < 2; i++)
{
h[x % 10]--;
h[y % 10]--;
x /= 10;
y /= 10;
}
for (i = 0; i < 10; i++)
if (h[i] != 0)
return 0;
return 1;
}
1260 1395 1435 1530 1827 2187 6880 is vampire
I am new to programming... But there are only 12 combinations in finding all 4-digit vampire numbers. My poor answer is:
public class VampNo {
public static void main(String[] args) {
for(int i = 1000; i < 10000; i++) {
int a = i/1000;
int b = i/100%10;
int c = i/10%10;
int d = i%10;
if((a * 10 + b) * (c * 10 + d) == i || (b * 10 + a) * (d * 10 + c) == i ||
(a * 10 + d) * (b * 10 + c) == i || (d * 10 + a) * (c * 10 + b) == i ||
(a * 10 + c) * (b * 10 + d) == i || (c * 10 + a) * (d * 10 + b) == i ||
(a * 10 + b) * (d * 10 + c) == i || (b * 10 + a) * (c * 10 + d) == i ||
(b * 10 + c) * (d * 10 + a) == i || (c * 10 + b) * (a * 10 + d) == i ||
(a * 10 + c) * (d * 10 + b) == i || (c * 10 + a) * (b * 10 + d) == i)
System.out.println(i + " is vampire");
}
}
}
The main task now is to simplify boolean expression in If() block
I've edited Owlstead's algorithm a bit to make it more understandable to Java beginners/learners.
import java.util.Arrays;
public class Vampire {
public static void main(String[] args) {
for (int x = 10; x < 100; x++) {
String sx = Integer.toString(x);
for (int y = x; y < 100; y++) {
int v = x * y;
String sy = Integer.toString(y);
String sv = Integer.toString(v);
if( Arrays.equals(sortVampire(sx + sy), sortVampire(sv)))
System.out.printf("%d * %d = %d%n", x, y, v);
}
}
}
private static char[] sortVampire (String v){
char[] sortedArray = v.toCharArray();
Arrays.sort(sortedArray);
return sortedArray;
}
}
This python code run very fast (O(n2))
result = []
for i in range(10,100):
for j in range(10, 100):
list1 = []
list2 = []
k = i * j
if k < 1000 or k > 10000:
continue
else:
for item in str(i):
list1.append(item)
for item in str(j):
list1.append(item)
for item in str(k):
list2.append(item)
flag = 1
for each in list1:
if each not in list2:
flag = 0
else:
list2.remove(each)
for each in list2:
if each not in list1:
flag = 0
if flag == 1:
if k not in result:
result.append(k)
for each in result:
print(each)
And here is my code. To generate zombie numbers we need to use Random class :)
import java.io.PrintStream;
import java.util.Set;
import java.util.HashSet;
import java.util.Iterator;
class VampireNumbers {
static PrintStream p = System.out;
private static Set<Integer> findVampireNumber() {
Set<Integer> vampireSet = new HashSet<Integer>();
for (int y = 1000; y <= 9999; y++) {
char[] numbersSeparately = ("" + y).toCharArray();
int numberOfDigits = numbersSeparately.length;
for (int i = 0; i < numberOfDigits; i++) {
for (int j = 0; j < numberOfDigits; j++) {
if (i != j) {
int value1 = Integer.valueOf("" + numbersSeparately[i] + numbersSeparately[j]);
int ki = -1;
for (int k = 0; k < numberOfDigits; k++) {
if (k != i && k != j) {
ki = k;
}
}
int kj = -1;
for (int t = 0; t < numberOfDigits; t++) {
if (t != i && t != j && t != ki) {
kj = t;
}
}
int value21 = Integer.valueOf("" + numbersSeparately[ki] + numbersSeparately[kj]);
int value22 = Integer.valueOf("" + numbersSeparately[kj] + numbersSeparately[ki]);
if (value1 * value21 == y && !(numbersSeparately[j] == 0 && numbersSeparately[kj] == 0)
|| value1 * value22 == y
&& !(numbersSeparately[j] == 0 && numbersSeparately[ki] == 0)) {
vampireSet.add(y);
}
}
}
}
}
return vampireSet;
}
public static void main(String[] args) {
Set<Integer> vampireSet = findVampireNumber();
Iterator<Integer> i = vampireSet.iterator();
int number = 1;
while (i.hasNext()) {
p.println(number + ": " + i.next());
number++;
}
}
}
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I found a piece of code that I was writing for interview prep few months ago.
According to the comment I had, it was trying to solve this problem:
Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars), find all the combinations of coins that make up the dollar value.
There are only pennies (1¢), nickels (5¢), dimes (10¢), and quarters (25¢) allowed.
For example, if 100 was given, the answer should be:
4 quarter(s) 0 dime(s) 0 nickel(s) 0 pennies
3 quarter(s) 1 dime(s) 0 nickel(s) 15 pennies
etc.
I believe that this can be solved in both iterative and recursive ways. My recursive solution is quite buggy, and I was wondering how other people would solve this problem. The difficult part of this problem was making it as efficient as possible.
I looked into this once a long time ago, and you can read my little write-up on it. Here’s the Mathematica source.
By using generating functions, you can get a closed-form constant-time solution to the problem. Graham, Knuth, and Patashnik’s Concrete Mathematics is the book for this, and contains a fairly extensive discussion of the problem. Essentially you define a polynomial where the nth coefficient is the number of ways of making change for n dollars.
Pages 4-5 of the writeup show how you can use Mathematica (or any other convenient computer algebra system) to compute the answer for 10^10^6 dollars in a couple seconds in three lines of code.
(And this was long enough ago that that’s a couple of seconds on a 75Mhz Pentium...)
Note: This only shows the number of ways.
Scala function:
def countChange(money: Int, coins: List[Int]): Int =
if (money == 0) 1
else if (coins.isEmpty || money < 0) 0
else countChange(money - coins.head, coins) + countChange(money, coins.tail)
I would favor a recursive solution. You have some list of denominations, if the smallest one can evenly divide any remaining currency amount, this should work fine.
Basically, you move from largest to smallest denominations.
Recursively,
You have a current total to fill, and a largest denomination (with more than 1 left).
If there is only 1 denomination left, there is only one way to fill the total. You can use 0 to k copies of your current denomination such that k * cur denomination <= total.
For 0 to k, call the function with the modified total and new largest denomination.
Add up the results from 0 to k. That's how many ways you can fill your total from the current denomination on down. Return this number.
Here's my python version of your stated problem, for 200 cents. I get 1463 ways. This version prints all the combinations and the final count total.
#!/usr/bin/python
# find the number of ways to reach a total with the given number of combinations
cents = 200
denominations = [25, 10, 5, 1]
names = {25: "quarter(s)", 10: "dime(s)", 5 : "nickel(s)", 1 : "pennies"}
def count_combs(left, i, comb, add):
if add: comb.append(add)
if left == 0 or (i+1) == len(denominations):
if (i+1) == len(denominations) and left > 0:
if left % denominations[i]:
return 0
comb.append( (left/denominations[i], demoninations[i]) )
i += 1
while i < len(denominations):
comb.append( (0, denominations[i]) )
i += 1
print(" ".join("%d %s" % (n,names[c]) for (n,c) in comb))
return 1
cur = denominations[i]
return sum(count_combs(left-x*cur, i+1, comb[:], (x,cur)) for x in range(0, int(left/cur)+1))
count_combs(cents, 0, [], None)
Scala function :
def countChange(money: Int, coins: List[Int]): Int = {
def loop(money: Int, lcoins: List[Int], count: Int): Int = {
// if there are no more coins or if we run out of money ... return 0
if ( lcoins.isEmpty || money < 0) 0
else{
if (money == 0 ) count + 1
/* if the recursive subtraction leads to 0 money left - a prefect division hence return count +1 */
else
/* keep iterating ... sum over money and the rest of the coins and money - the first item and the full set of coins left*/
loop(money, lcoins.tail,count) + loop(money - lcoins.head,lcoins, count)
}
}
val x = loop(money, coins, 0)
Console println x
x
}
Here's some absolutely straightforward C++ code to solve the problem which did ask for all the combinations to be shown.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
if (argc != 2)
{
printf("usage: change amount-in-cents\n");
return 1;
}
int total = atoi(argv[1]);
printf("quarter\tdime\tnickle\tpenny\tto make %d\n", total);
int combos = 0;
for (int q = 0; q <= total / 25; q++)
{
int total_less_q = total - q * 25;
for (int d = 0; d <= total_less_q / 10; d++)
{
int total_less_q_d = total_less_q - d * 10;
for (int n = 0; n <= total_less_q_d / 5; n++)
{
int p = total_less_q_d - n * 5;
printf("%d\t%d\t%d\t%d\n", q, d, n, p);
combos++;
}
}
}
printf("%d combinations\n", combos);
return 0;
}
But I'm quite intrigued about the sub problem of just calculating the number of combinations. I suspect there's a closed-form equation for it.
The sub problem is a typical Dynamic Programming problem.
/* Q: Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars),
find the number of combinations of coins that make up the dollar value.
There are only penny, nickel, dime, and quarter.
(quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */
/* A:
Reference: http://andrew.neitsch.ca/publications/m496pres1.nb.pdf
f(n, k): number of ways of making change for n cents, using only the first
k+1 types of coins.
+- 0, n < 0 || k < 0
f(n, k) = |- 1, n == 0
+- f(n, k-1) + f(n-C[k], k), else
*/
#include <iostream>
#include <vector>
using namespace std;
int C[] = {1, 5, 10, 25};
// Recursive: very slow, O(2^n)
int f(int n, int k)
{
if (n < 0 || k < 0)
return 0;
if (n == 0)
return 1;
return f(n, k-1) + f(n-C[k], k);
}
// Non-recursive: fast, but still O(nk)
int f_NonRec(int n, int k)
{
vector<vector<int> > table(n+1, vector<int>(k+1, 1));
for (int i = 0; i <= n; ++i)
{
for (int j = 0; j <= k; ++j)
{
if (i < 0 || j < 0) // Impossible, for illustration purpose
{
table[i][j] = 0;
}
else if (i == 0 || j == 0) // Very Important
{
table[i][j] = 1;
}
else
{
// The recursion. Be careful with the vector boundary
table[i][j] = table[i][j-1] +
(i < C[j] ? 0 : table[i-C[j]][j]);
}
}
}
return table[n][k];
}
int main()
{
cout << f(100, 3) << ", " << f_NonRec(100, 3) << endl;
cout << f(200, 3) << ", " << f_NonRec(200, 3) << endl;
cout << f(1000, 3) << ", " << f_NonRec(1000, 3) << endl;
return 0;
}
The code is using Java to solve this problem and it also works... This method may not be a good idea because of too many loops, but it's really a straight forward way.
public class RepresentCents {
public static int sum(int n) {
int count = 0;
for (int i = 0; i <= n / 25; i++) {
for (int j = 0; j <= n / 10; j++) {
for (int k = 0; k <= n / 5; k++) {
for (int l = 0; l <= n; l++) {
int v = i * 25 + j * 10 + k * 5 + l;
if (v == n) {
count++;
} else if (v > n) {
break;
}
}
}
}
}
return count;
}
public static void main(String[] args) {
System.out.println(sum(100));
}
}
This is a really old question, but I came up with a recursive solution in java that seemed smaller than all the others, so here goes -
public static void printAll(int ind, int[] denom,int N,int[] vals){
if(N==0){
System.out.println(Arrays.toString(vals));
return;
}
if(ind == (denom.length))return;
int currdenom = denom[ind];
for(int i=0;i<=(N/currdenom);i++){
vals[ind] = i;
printAll(ind+1,denom,N-i*currdenom,vals);
}
}
Improvements:
public static void printAllCents(int ind, int[] denom,int N,int[] vals){
if(N==0){
if(ind < denom.length) {
for(int i=ind;i<denom.length;i++)
vals[i] = 0;
}
System.out.println(Arrays.toString(vals));
return;
}
if(ind == (denom.length)) {
vals[ind-1] = 0;
return;
}
int currdenom = denom[ind];
for(int i=0;i<=(N/currdenom);i++){
vals[ind] = i;
printAllCents(ind+1,denom,N-i*currdenom,vals);
}
}
Let C(i,J) the set of combinations of making i cents using the values in the set J.
You can define C as that:
(first(J) takes in a deterministic way an element of a set)
It turns out a pretty recursive function... and reasonably efficient if you use memoization ;)
semi-hack to get around the unique combination problem - force descending order:
$denoms = [1,5,10,25]
def all_combs(sum,last)
return 1 if sum == 0
return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom|
total+all_combs(sum-denom,denom)}
end
This will run slow since it won't be memoized, but you get the idea.
# short and sweet with O(n) table memory
#include <iostream>
#include <vector>
int count( std::vector<int> s, int n )
{
std::vector<int> table(n+1,0);
table[0] = 1;
for ( auto& k : s )
for(int j=k; j<=n; ++j)
table[j] += table[j-k];
return table[n];
}
int main()
{
std::cout << count({25, 10, 5, 1}, 100) << std::endl;
return 0;
}
This is my answer in Python. It does not use recursion:
def crossprod (list1, list2):
output = 0
for i in range(0,len(list1)):
output += list1[i]*list2[i]
return output
def breakit(target, coins):
coinslimit = [(target / coins[i]) for i in range(0,len(coins))]
count = 0
temp = []
for i in range(0,len(coins)):
temp.append([j for j in range(0,coinslimit[i]+1)])
r=[[]]
for x in temp:
t = []
for y in x:
for i in r:
t.append(i+[y])
r = t
for targets in r:
if crossprod(targets, coins) == target:
print targets
count +=1
return count
if __name__ == "__main__":
coins = [25,10,5,1]
target = 78
print breakit(target, coins)
Example output
...
1 ( 10 cents) 2 ( 5 cents) 58 ( 1 cents)
4 ( 5 cents) 58 ( 1 cents)
1 ( 10 cents) 1 ( 5 cents) 63 ( 1 cents)
3 ( 5 cents) 63 ( 1 cents)
1 ( 10 cents) 68 ( 1 cents)
2 ( 5 cents) 68 ( 1 cents)
1 ( 5 cents) 73 ( 1 cents)
78 ( 1 cents)
Number of solutions = 121
var countChange = function (money,coins) {
function countChangeSub(money,coins,n) {
if(money==0) return 1;
if(money<0 || coins.length ==n) return 0;
return countChangeSub(money-coins[n],coins,n) + countChangeSub(money,coins,n+1);
}
return countChangeSub(money,coins,0);
}
Both: iterate through all denominations from high to low, take one of denomination, subtract from requried total, then recurse on remainder (constraining avilable denominations to be equal or lower to current iteration value.)
If the currency system allows it, a simple greedy algorithm that takes as many of each coin as possible, starting with the highest value currency.
Otherwise, dynamic programming is required to find an optimal solution quickly since this problem is essentially the knapsack problem.
For example, if a currency system has the coins: {13, 8, 1}, the greedy solution would make change for 24 as {13, 8, 1, 1, 1}, but the true optimal solution is {8, 8, 8}
Edit: I thought we were making change optimally, not listing all the ways to make change for a dollar. My recent interview asked how to make change so I jumped ahead before finishing to read the question.
I know this is a very old question. I was searching through the proper answer and couldn't find anything that is simple and satisfactory. Took me some time but was able to jot down something.
function denomination(coins, original_amount){
var original_amount = original_amount;
var original_best = [ ];
for(var i=0;i<coins.length; i++){
var amount = original_amount;
var best = [ ];
var tempBest = [ ]
while(coins[i]<=amount){
amount = amount - coins[i];
best.push(coins[i]);
}
if(amount>0 && coins.length>1){
tempBest = denomination(coins.slice(0,i).concat(coins.slice(i+1,coins.length)), amount);
//best = best.concat(denomination(coins.splice(i,1), amount));
}
if(tempBest.length!=0 || (best.length!=0 && amount==0)){
best = best.concat(tempBest);
if(original_best.length==0 ){
original_best = best
}else if(original_best.length > best.length ){
original_best = best;
}
}
}
return original_best;
}
denomination( [1,10,3,9] , 19 );
This is a javascript solution and uses recursion.
In Scala Programming language i would do it like this:
def countChange(money: Int, coins: List[Int]): Int = {
money match {
case 0 => 1
case x if x < 0 => 0
case x if x >= 1 && coins.isEmpty => 0
case _ => countChange(money, coins.tail) + countChange(money - coins.head, coins)
}
}
This is a simple recursive algorithm that takes a bill, then takes a smaller bill recursively until it reaches the sum, it then takes another bill of same denomination, and recurses again. See sample output below for illustration.
var bills = new int[] { 100, 50, 20, 10, 5, 1 };
void PrintAllWaysToMakeChange(int sumSoFar, int minBill, string changeSoFar)
{
for (int i = minBill; i < bills.Length; i++)
{
var change = changeSoFar;
var sum = sumSoFar;
while (sum > 0)
{
if (!string.IsNullOrEmpty(change)) change += " + ";
change += bills[i];
sum -= bills[i];
if (sum > 0)
{
PrintAllWaysToMakeChange(sum, i + 1, change);
}
}
if (sum == 0)
{
Console.WriteLine(change);
}
}
}
PrintAllWaysToMakeChange(15, 0, "");
Prints the following:
10 + 5
10 + 1 + 1 + 1 + 1 + 1
5 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
5 + 5 + 1 + 1 + 1 + 1 + 1
5 + 5 + 5
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
Duh, I feel stupid right now. Below there is an overly complicated solution, which I'll preserve because it is a solution, after all. A simple solution would be this:
// Generate a pretty string
val coinNames = List(("quarter", "quarters"),
("dime", "dimes"),
("nickel", "nickels"),
("penny", "pennies"))
def coinsString =
Function.tupled((quarters: Int, dimes: Int, nickels:Int, pennies: Int) => (
List(quarters, dimes, nickels, pennies)
zip coinNames // join with names
map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
map (t => t._1 + " " + t._2) // qty name
mkString " "
))
def allCombinations(amount: Int) =
(for{quarters <- 0 to (amount / 25)
dimes <- 0 to ((amount - 25*quarters) / 10)
nickels <- 0 to ((amount - 25*quarters - 10*dimes) / 5)
} yield (quarters, dimes, nickels, amount - 25*quarters - 10*dimes - 5*nickels)
) map coinsString mkString "\n"
Here is the other solution. This solution is based on the observation that each coin is a multiple of the others, so they can be represented in terms of them.
// Just to make things a bit more readable, as these routines will access
// arrays a lot
val coinValues = List(25, 10, 5, 1)
val coinNames = List(("quarter", "quarters"),
("dime", "dimes"),
("nickel", "nickels"),
("penny", "pennies"))
val List(quarter, dime, nickel, penny) = coinValues.indices.toList
// Find the combination that uses the least amount of coins
def leastCoins(amount: Int): Array[Int] =
((List(amount) /: coinValues) {(list, coinValue) =>
val currentAmount = list.head
val numberOfCoins = currentAmount / coinValue
val remainingAmount = currentAmount % coinValue
remainingAmount :: numberOfCoins :: list.tail
}).tail.reverse.toArray
// Helper function. Adjust a certain amount of coins by
// adding or subtracting coins of each type; this could
// be made to receive a list of adjustments, but for so
// few types of coins, it's not worth it.
def adjust(base: Array[Int],
quarters: Int,
dimes: Int,
nickels: Int,
pennies: Int): Array[Int] =
Array(base(quarter) + quarters,
base(dime) + dimes,
base(nickel) + nickels,
base(penny) + pennies)
// We decrease the amount of quarters by one this way
def decreaseQuarter(base: Array[Int]): Array[Int] =
adjust(base, -1, +2, +1, 0)
// Dimes are decreased this way
def decreaseDime(base: Array[Int]): Array[Int] =
adjust(base, 0, -1, +2, 0)
// And here is how we decrease Nickels
def decreaseNickel(base: Array[Int]): Array[Int] =
adjust(base, 0, 0, -1, +5)
// This will help us find the proper decrease function
val decrease = Map(quarter -> decreaseQuarter _,
dime -> decreaseDime _,
nickel -> decreaseNickel _)
// Given a base amount of coins of each type, and the type of coin,
// we'll produce a list of coin amounts for each quantity of that particular
// coin type, up to the "base" amount
def coinSpan(base: Array[Int], whichCoin: Int) =
(List(base) /: (0 until base(whichCoin)).toList) { (list, _) =>
decrease(whichCoin)(list.head) :: list
}
// Generate a pretty string
def coinsString(base: Array[Int]) = (
base
zip coinNames // join with names
map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
map (t => t._1 + " " + t._2)
mkString " "
)
// So, get a base amount, compute a list for all quarters variations of that base,
// then, for each combination, compute all variations of dimes, and then repeat
// for all variations of nickels.
def allCombinations(amount: Int) = {
val base = leastCoins(amount)
val allQuarters = coinSpan(base, quarter)
val allDimes = allQuarters flatMap (base => coinSpan(base, dime))
val allNickels = allDimes flatMap (base => coinSpan(base, nickel))
allNickels map coinsString mkString "\n"
}
So, for 37 coins, for example:
scala> println(allCombinations(37))
0 quarter 0 dimes 0 nickels 37 pennies
0 quarter 0 dimes 1 nickel 32 pennies
0 quarter 0 dimes 2 nickels 27 pennies
0 quarter 0 dimes 3 nickels 22 pennies
0 quarter 0 dimes 4 nickels 17 pennies
0 quarter 0 dimes 5 nickels 12 pennies
0 quarter 0 dimes 6 nickels 7 pennies
0 quarter 0 dimes 7 nickels 2 pennies
0 quarter 1 dime 0 nickels 27 pennies
0 quarter 1 dime 1 nickel 22 pennies
0 quarter 1 dime 2 nickels 17 pennies
0 quarter 1 dime 3 nickels 12 pennies
0 quarter 1 dime 4 nickels 7 pennies
0 quarter 1 dime 5 nickels 2 pennies
0 quarter 2 dimes 0 nickels 17 pennies
0 quarter 2 dimes 1 nickel 12 pennies
0 quarter 2 dimes 2 nickels 7 pennies
0 quarter 2 dimes 3 nickels 2 pennies
0 quarter 3 dimes 0 nickels 7 pennies
0 quarter 3 dimes 1 nickel 2 pennies
1 quarter 0 dimes 0 nickels 12 pennies
1 quarter 0 dimes 1 nickel 7 pennies
1 quarter 0 dimes 2 nickels 2 pennies
1 quarter 1 dime 0 nickels 2 pennies
This blog entry of mine solves this knapsack like problem for the figures from an XKCD comic. A simple change to the items dict and the exactcost value will yield all solutions for your problem too.
If the problem were to find the change that used the least cost, then a naive greedy algorithm that used as much of the highest value coin might well fail for some combinations of coins and target amount. For example if there are coins with values 1, 3, and 4; and the target amount is 6 then the greedy algorithm might suggest three coins of value 4, 1, and 1 when it is easy to see that you could use two coins each of value 3.
Paddy.
public class Coins {
static int ac = 421;
static int bc = 311;
static int cc = 11;
static int target = 4000;
public static void main(String[] args) {
method2();
}
public static void method2(){
//running time n^2
int da = target/ac;
int db = target/bc;
for(int i=0;i<=da;i++){
for(int j=0;j<=db;j++){
int rem = target-(i*ac+j*bc);
if(rem < 0){
break;
}else{
if(rem%cc==0){
System.out.format("\n%d, %d, %d ---- %d + %d + %d = %d \n", i, j, rem/cc, i*ac, j*bc, (rem/cc)*cc, target);
}
}
}
}
}
}
I found this neat piece of code in the book "Python For Data Analysis" by O'reily. It uses lazy implementation and int comparison and i presume it can be modified for other denominations using decimals. Let me know how it works for you!
def make_change(amount, coins=[1, 5, 10, 25], hand=None):
hand = [] if hand is None else hand
if amount == 0:
yield hand
for coin in coins:
# ensures we don't give too much change, and combinations are unique
if coin > amount or (len(hand) > 0 and hand[-1] < coin):
continue
for result in make_change(amount - coin, coins=coins,
hand=hand + [coin]):
yield result
This is the improvement of Zihan's answer. The great deal of unnecessary loops comes when the denomination is just 1 cent.
It's intuitive and non-recursive.
public static int Ways2PayNCents(int n)
{
int numberOfWays=0;
int cent, nickel, dime, quarter;
for (quarter = 0; quarter <= n/25; quarter++)
{
for (dime = 0; dime <= n/10; dime++)
{
for (nickel = 0; nickel <= n/5; nickel++)
{
cent = n - (quarter * 25 + dime * 10 + nickel * 5);
if (cent >= 0)
{
numberOfWays += 1;
Console.WriteLine("{0},{1},{2},{3}", quarter, dime, nickel, cent);
}
}
}
}
return numberOfWays;
}
Straightforward java solution:
public static void main(String[] args)
{
int[] denoms = {4,2,3,1};
int[] vals = new int[denoms.length];
int target = 6;
printCombinations(0, denoms, target, vals);
}
public static void printCombinations(int index, int[] denom,int target, int[] vals)
{
if(target==0)
{
System.out.println(Arrays.toString(vals));
return;
}
if(index == denom.length) return;
int currDenom = denom[index];
for(int i = 0; i*currDenom <= target;i++)
{
vals[index] = i;
printCombinations(index+1, denom, target - i*currDenom, vals);
vals[index] = 0;
}
}
Lots of variations here but couldn't find a PHP solution for the number of combinations anywhere so I'll add one in.
/**
* #param int $money The total value
* #param array $coins The coin denominations
* #param int $sum The countable sum
* #return int
*/
function getTotalCombinations($money, $coins, &$sum = 0){
if ($money == 0){
return $sum++;
} else if (empty($coins) || $money < 0){
return $sum;
} else {
$firstCoin = array_pop(array_reverse($coins));
getTotalCombinations($money - $firstCoin, $coins, $sum) + getTotalCombinations($money, array_diff($coins, [$firstCoin]), $sum);
}
return $sum;
}
$totalCombinations = getTotalCombinations($money, $coins);
/*
* make a list of all distinct sets of coins of from the set of coins to
* sum up to the given target amount.
* Here the input set of coins is assumed yo be {1, 2, 4}, this set MUST
* have the coins sorted in ascending order.
* Outline of the algorithm:
*
* Keep track of what the current coin is, say ccn; current number of coins
* in the partial solution, say k; current sum, say sum, obtained by adding
* ccn; sum sofar, say accsum:
* 1) Use ccn as long as it can be added without exceeding the target
* a) if current sum equals target, add cc to solution coin set, increase
* coin coin in the solution by 1, and print it and return
* b) if current sum exceeds target, ccn can't be in the solution, so
* return
* c) if neither of the above, add current coin to partial solution,
* increase k by 1 (number of coins in partial solution), and recuse
* 2) When current denomination can no longer be used, start using the
* next higher denomination coins, just like in (1)
* 3) When all denominations have been used, we are done
*/
#include <iostream>
#include <cstdlib>
using namespace std;
// int num_calls = 0;
// int num_ways = 0;
void print(const int coins[], int n);
void combine_coins(
const int denoms[], // coins sorted in ascending order
int n, // number of denominations
int target, // target sum
int accsum, // accumulated sum
int coins[], // solution set, MUST equal
// target / lowest denom coin
int k // number of coins in coins[]
)
{
int ccn; // current coin
int sum; // current sum
// ++num_calls;
for (int i = 0; i < n; ++i) {
/*
* skip coins of lesser denomination: This is to be efficient
* and also avoid generating duplicate sequences. What we need
* is combinations and without this check we will generate
* permutations.
*/
if (k > 0 && denoms[i] < coins[k - 1])
continue; // skip coins of lesser denomination
ccn = denoms[i];
if ((sum = accsum + ccn) > target)
return; // no point trying higher denominations now
if (sum == target) {
// found yet another solution
coins[k] = ccn;
print(coins, k + 1);
// ++num_ways;
return;
}
coins[k] = ccn;
combine_coins(denoms, n, target, sum, coins, k + 1);
}
}
void print(const int coins[], int n)
{
int s = 0;
for (int i = 0; i < n; ++i) {
cout << coins[i] << " ";
s += coins[i];
}
cout << "\t = \t" << s << "\n";
}
int main(int argc, const char *argv[])
{
int denoms[] = {1, 2, 4};
int dsize = sizeof(denoms) / sizeof(denoms[0]);
int target;
if (argv[1])
target = atoi(argv[1]);
else
target = 8;
int *coins = new int[target];
combine_coins(denoms, dsize, target, 0, coins, 0);
// cout << "num calls = " << num_calls << ", num ways = " << num_ways << "\n";
return 0;
}
Here's a C# function:
public static void change(int money, List<int> coins, List<int> combination)
{
if(money < 0 || coins.Count == 0) return;
if (money == 0)
{
Console.WriteLine((String.Join("; ", combination)));
return;
}
List<int> copy = new List<int>(coins);
copy.RemoveAt(0);
change(money, copy, combination);
combination = new List<int>(combination) { coins[0] };
change(money - coins[0], coins, new List<int>(combination));
}
Use it like this:
change(100, new List<int>() {5, 10, 25}, new List<int>());
It prints:
25; 25; 25; 25
10; 10; 10; 10; 10; 25; 25
10; 10; 10; 10; 10; 10; 10; 10; 10; 10
5; 10; 10; 25; 25; 25
5; 10; 10; 10; 10; 10; 10; 10; 25
5; 5; 10; 10; 10; 10; 25; 25
5; 5; 10; 10; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 10; 25; 25; 25
5; 5; 5; 10; 10; 10; 10; 10; 10; 25
5; 5; 5; 5; 10; 10; 10; 25; 25
5; 5; 5; 5; 10; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 25; 25; 25
5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 10; 10; 25; 25
5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 10; 25; 25
5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 25; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5
Below is a python program to find all combinations of money. This is a dynamic programming solution with order(n) time.
Money is 1,5,10,25
We traverse from row money 1 to row money 25 (4 rows). Row money 1 contains the count if we only consider money 1 in
calculating the number of combinations. Row money 5 produces each column by taking the count in row money r for the
same final money plus the previous 5 count in its own row (current position minus 5). Row money 10 uses row money 5,
which contains counts for both 1,5 and adds in the previous 10 count (current position minus 10). Row money 25 uses row
money 10, which contains counts for row money 1,5,10 plus the previous 25 count.
For example, numbers[1][12] = numbers[0][12] + numbers[1][7] (7 = 12-5) which results in 3 = 1 + 2; numbers[3][12] =
numbers[2][12] + numbers[3][9] (-13 = 12-25) which results in 4 = 0 + 4, since -13 is less than 0.
def cntMoney(num):
mSz = len(money)
numbers = [[0]*(1+num) for _ in range(mSz)]
for mI in range(mSz): numbers[mI][0] = 1
for mI,m in enumerate(money):
for i in range(1,num+1):
numbers[mI][i] = numbers[mI][i-m] if i >= m else 0
if mI != 0: numbers[mI][i] += numbers[mI-1][i]
print('m,numbers',m,numbers[mI])
return numbers[mSz-1][num]
money = [1,5,10,25]
num = 12
print('money,combinations',num,cntMoney(num))
output:
('m,numbers', 1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
('m,numbers', 5, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3])
('m,numbers', 10, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('m,numbers', 25, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('money,combinations', 12, 4)
Java solution
import java.util.Arrays;
import java.util.Scanner;
public class nCents {
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
int cents=input.nextInt();
int num_ways [][] =new int [5][cents+1];
//putting in zeroes to offset
int getCents[]={0 , 0 , 5 , 10 , 25};
Arrays.fill(num_ways[0], 0);
Arrays.fill(num_ways[1], 1);
int current_cent=0;
for(int i=2;i<num_ways.length;i++){
current_cent=getCents[i];
for(int j=1;j<num_ways[0].length;j++){
if(j-current_cent>=0){
if(j-current_cent==0){
num_ways[i][j]=num_ways[i-1][j]+1;
}else{
num_ways[i][j]=num_ways[i][j-current_cent]+num_ways[i-1][j];
}
}else{
num_ways[i][j]=num_ways[i-1][j];
}
}
}
System.out.println(num_ways[num_ways.length-1][num_ways[0].length-1]);
}
}
The below java solution which will print the different combinations as well. Easy to understand. Idea is
for sum 5
The solution is
5 - 5(i) times 1 = 0
if(sum = 0)
print i times 1
5 - 4(i) times 1 = 1
5 - 3 times 1 = 2
2 - 1(j) times 2 = 0
if(sum = 0)
print i times 1 and j times 2
and so on......
If the remaining sum in each loop is lesser than the denomination ie
if remaining sum 1 is lesser than 2, then just break the loop
The complete code below
Please correct me in case of any mistakes
public class CoinCombinbationSimple {
public static void main(String[] args) {
int sum = 100000;
printCombination(sum);
}
static void printCombination(int sum) {
for (int i = sum; i >= 0; i--) {
int sumCopy1 = sum - i * 1;
if (sumCopy1 == 0) {
System.out.println(i + " 1 coins");
}
for (int j = sumCopy1 / 2; j >= 0; j--) {
int sumCopy2 = sumCopy1;
if (sumCopy2 < 2) {
break;
}
sumCopy2 = sumCopy1 - 2 * j;
if (sumCopy2 == 0) {
System.out.println(i + " 1 coins " + j + " 2 coins ");
}
for (int k = sumCopy2 / 5; k >= 0; k--) {
int sumCopy3 = sumCopy2;
if (sumCopy2 < 5) {
break;
}
sumCopy3 = sumCopy2 - 5 * k;
if (sumCopy3 == 0) {
System.out.println(i + " 1 coins " + j + " 2 coins "
+ k + " 5 coins");
}
}
}
}
}
}