I've been trying to figure out if there would be a way to get the optimal minimum set of coins that would be used to make the change.
The greedy algorithm approach for this has an issue such as if we have the set of coins {1, 5, 6, 9} and we wanted to get the value 11. The greedy algorithm would give us {9,1,1} however the most optimal solution would be {5, 6}
From reading through this site I've found that this method can give us the total minimum number of coins needed. Would there be a way to get the set of coins from that as well?
I'm assuming you already know the Dynamic Programming method to find only the minimum number of coins needed. Let's say that you want to find the minimum number of coins to create a total value K. Then, your code could be
vector <int> min_coins(K + 1);
min_coins[0] = 0; // base case
for(int k = 1; k <= K; ++k) {
min_coins[k] = INF;
for(int c : coins) { // coins[] contains all values of coins
if(k - c >= 0) {
min_coins[k] = min(min_coins[k], min_coins[k - c] + 1);
}
}
}
Answer to your question: In order to find the actual set of coins that is minimal in size, we can simply keep another array last_coin[] where last_coin[k] is equal to the coin that was last added to the optimal set of coins for a sum of k. To illustrate this:
vector <int> min_coins(K + 1), last_coin(K + 1);
min_coins[0] = 0; // base case
for(int k = 1; k <= K; ++k) {
min_coins[k] = INF;
for(int c : coins) {
if(k - c >= 0) {
if(min_coins[k - c] + 1 < min_coins[k]) {
min_coins[k] = min_coins[k - c] + 1;
last_coin[k] = c; // !!!
}
}
}
}
How does this let you find the set of coins? Let's say we wanted to find the best set of coins that sum to K. Then, we know that last_coin[K] holds one of the coins in the set, so we can add last_coin[K] to the set. After that, we subtract last_coin[K] from K and repeat until K = 0. Clearly, this will construct a (not necessarily the) min-size set of coins that sums to K.
Possible implementation:
int value_left = K;
while(value_left > 0) {
last_coin[value_left] is added to the set
value_left -= last_coin[value_left]
}
Related
I have the following problem to solve: given a number N and 1<=k<=N, count the number of possible sums of (1,...,k) which add to N. There may be equal factors (e.g. if N=3 and k=2, (1,1,1) is a valid sum), but permutations must not be counted (e.g., if N=3 and k=2, count (1,2) and (2,1) as a single solution). I have implemented the recursive Python code below but I'd like to find a better solution (maybe with dynamic programming? ). It seems similar to the triple step problem, but with the extra constraint of not counting permutations.
def find_num_sums_aux(n, min_k, max_k):
# base case
if n == 0:
return 1
count = 0
# due to lower bound min_k, we evaluate only ordered solutions and prevent permutations
for i in range(min_k, max_k+1):
if n-i>=0:
count += find_num_sums_aux(n-i, i, max_k)
return count
def find_num_sums(n, k):
count = find_num_sums_aux(n,1,k)
return count
This is a standard problem in dynamic programming (subset sum problem).
Lets define the function f(i,j) which gives the number of ways you can get the sum j using a subset of the numbers (1...i), then the result to your problem will be f(k,n).
for each number x of the range (1...i), x might be a part of the sum j or might not, so we need to count these two possibilities.
Note: f(i,0) = 1 for any i, which means that you can get the sum = 0 in one way and this way is by not taking any number from the range (1...i).
Here is the code written in C++:
int n = 10;
int k = 7;
int f[8][11];
//initializing the array with zeroes
for (int i = 0; i <= k; i++)
for (int j = 0; j <= n; j++)
f[i][j] = 0;
f[0][0] = 1;
for (int i = 1; i <= k; i++) {
for (int j = 0; j <= n; j++) {
if (j == 0)
f[i][j] = 1;
else {
f[i][j] = f[i - 1][j];//without adding i to the sum j
if (j - i >= 0)
f[i][j] = f[i][j] + f[i - 1][j - i];//adding i to the sum j
}
}
}
cout << f[k][n] << endl;//print f(k,n)
Update
To handle the case where we can repeat the elements like (1,1,1) will give you the sum 3, you just need to allow picking the same element multiple times by changing the following line of code:
f[i][j] = f[i][j] + f[i - 1][j - i];//adding i to the sum
To this:
f[i][j] = f[i][j] + f[i][j - i];
Imagine you're given an unsorted array [5,11,7,4,19,8,11,6,17] and a max weight 23. [similar to Two Sum Problem by a bit different]
One needs to find two optimal weights (by which I mean if two weights that are (almost or not) half of the weight you're trying to find) in this case [5,17], [3,19], [11,11] so I need to return [11,11].
I was taken back by the problem, and could not solve it. [I was not allowed to use structs]
I tried to sort [3, 5, 6, 7, 8, 11, 11, 17, 19] and search from both ends and store indexes of values that were <= max weight in a vector as a pair (like v[i], v[i+1] and check them later by their pairs) then return a pair with both largest vals, but got confused.
[although, weights were doubles and I did not see duplicates at that set I did not use unsorted_map(hashMap), might it've worked?]
Can anyone suggest how should I go about this problem? is it similar to "knapsack problem"? Thank you
You can use Two Pointer Approach for the problem.
Algorithm:
Sort the array.
Have two pointers startIndex and endIndex to 0 and arraySize-1.
sum = arr[startIndex] + arr[endIndex]
If sum is less than or equal to 23, increment startIndex else decrement endIndex.
keep track of closest value using a variable.
finish when startIndex == endIndex
Code in Java:
public class Solution {
private ArrayList<Integer> twoSumClosest(ArrayList<Integer> a, int s) {
// Sort the arraylist
Collections.sort(a);
// closests sum we got till now
int sumClosest = Integer.MIN_VALUE;
// indexes used to traverse
int startIndex = 0;
int endIndex = a.size() - 1 ;
// answer Indexes
int firstIndex = 1;
int secondIndex = a.size() - 1;
while( startIndex < endIndex ) {
if( a.get(startIndex) + a.get(endIndex) > s) {
endIndex--;
continue;
} else {
if( a.get(startIndex) + a.get(endIndex) > sumClosest ) {
sumClosest = a.get(startIndex) + a.get(endIndex);
firstIndex = startIndex;
secondIndex = endIndex;
}
startIndex++;
}
}
ArrayList<Integer> ans = new ArrayList<>();
ans.add(firstIndex);
ans.add(secondIndex);
return ans;
}
}
Time Complexity: O(nlogn)
O(n) if array was already sorted
Extra Space Needed: O(1)
Given a bag with a maximum of 100 chips,each chip has its value written over it.
Determine the most fair division between two persons. This means that the difference between the amount each person obtains should be minimized. The value of a chips varies from 1 to 1000.
Input: The number of coins m, and the value of each coin.
Output: Minimal positive difference between the amount the two persons obtain when they divide the chips from the corresponding bag.
I am finding it difficult to form a DP solution for it. Please help me.
Initially I had to tried it as a Non DP solution.Actually I havent thought of solving it using DP. I simply sorted the value array. And assigned the largest value to one of the person, and incrementally assigned the other values to one of the two depending upon which creates minimum difference. But that solution actually didnt work.
I am posting my solution here :
bool myfunction(int i, int j)
{
return(i >= j) ;
}
int main()
{
int T, m, sum1, sum2, temp_sum1, temp_sum2,i ;
cin >> T ;
while(T--)
{
cin >> m ;
sum1 = 0 ; sum2 = 0 ; temp_sum1 = 0 ; temp_sum2 = 0 ;
vector<int> arr(m) ;
for(i=0 ; i < m ; i++)
{
cin>>arr[i] ;
}
if(m==1 )
{
if(arr[0]%2==0)
cout<<0<<endl ;
else
cout<<1<<endl ;
}
else {
sort(arr.begin(), arr.end(), myfunction) ;
// vector<int> s1 ;
// vector<int> s2 ;
for(i=0 ; i < m ; i++)
{
temp_sum1 = sum1 + arr[i] ;
temp_sum2 = sum2 + arr[i] ;
if(abs(temp_sum1 - sum2) <= abs(temp_sum2 -sum1))
{
sum1 = sum1 + arr[i] ;
}
else
{
sum2 = sum2 + arr[i] ;
}
temp_sum1 = 0 ;
temp_sum2 = 0 ;
}
cout<<abs(sum1 -sum2)<<endl ;
}
}
return 0 ;
}
what i understand from your question is you want to divide chips in two persons so as to minimize the difference between sum of numbers written on those.
If understanding is correct, then potentially you can follow below approach to arrive at solution.
Sort the values array i.e. int values[100]
Start adding elements from both ends of array in for loop i.e. for(i=0; j=values.length;i<j;i++,j--)
Odd numbered iteration sum belongs to one person & even numbered sum to other person
run the loop till i < j
now, the difference between two sums obtained in odd & even iterations should be minimum as array was sorted earlier.
If my understanding of the question is correct, then this solution should resolve your problem.
Reflect as appropriate.
Thanks
Ravindra
Based on How to determine if a list of polygon points are in clockwise order?
I've come up with the following code:
bool PointsClockwise(const std::vector<MyPoint>& points)
{
double sum = 0.0;
for(size_t i = 0; i < points.size() - 1; ++i)
sum += (points[i+1].x()-points[i].x()) * (points[i+1].y()+points[i].y());
return sum > 0.0;
}
However, this seems to have wrong result in certain cases. Take for example the following ring:
LINESTRING(0 119,0 60,694 70,704 72,712 77,719 83,723 92,725 102,723 111,719 120,712 126,703 130)
It is in counter-clockwise order, but the function returns true.
Thanks!
You missed one of the line segments from your summation - namely the one connecting the last point with the first.
Try that:
bool PointsClockwise(const std::vector<MyPoint>& points)
{
double sum = 0.0;
for(size_t i = 0; i < points.size() - 1; ++i)
sum += (points[i+1].x()-points[i].x()) * (points[i+1].y()+points[i].y());
sum += (points[0].x()-points[points.size()-1].x()) * (points[0].y()+points[points.size()-1].y());
return sum > 0.0;
}
You need to include the case i == points.size() - 1, but to do that, you need to do some modular arithmetic in the loop, or else separate out the last iteration. Actually, just initialize sum to the last iteration:
double sum = (points[0].x() - points[points.size() - 1].x())
* (points[0].y() + points[points.size() - 1].y());
and end the iteration at i < points.size() - 1
Project Euler problem 14:
The following iterative sequence is
defined for the set of positive
integers:
n → n/2 (n is even) n → 3n + 1 (n is
odd)
Using the rule above and starting with
13, we generate the following
sequence: 13 → 40 → 20 → 10 → 5 → 16 →
8 → 4 → 2 → 1
It can be seen that this sequence
(starting at 13 and finishing at 1)
contains 10 terms. Although it has not
been proved yet (Collatz Problem), it
is thought that all starting numbers
finish at 1.
Which starting number, under one
million, produces the longest chain?
My first instinct is to create a function to calculate the chains, and run it with every number between 1 and 1 million. Obviously, that takes a long time. Way longer than solving this should take, according to Project Euler's "About" page. I've found several problems on Project Euler that involve large groups of numbers that a program running for hours didn't finish. Clearly, I'm doing something wrong.
How can I handle large groups of numbers quickly?
What am I missing here?
Have a read about memoization. The key insight is that if you've got a sequence starting A that has length 1001, and then you get a sequence B that produces an A, you don't to repeat all that work again.
This is the code in Mathematica, using memoization and recursion. Just four lines :)
f[x_] := f[x] = If[x == 1, 1, 1 + f[If[EvenQ[x], x/2, (3 x + 1)]]];
Block[{$RecursionLimit = 1000, a = 0, j},
Do[If[a < f[i], a = f[i]; j = i], {i, Reverse#Range#10^6}];
Print#a; Print[j];
]
Output .... chain length´525´ and the number is ... ohhhh ... font too small ! :)
BTW, here you can see a plot of the frequency for each chain length
Starting with 1,000,000, generate the chain. Keep track of each number that was generated in the chain, as you know for sure that their chain is smaller than the chain for the starting number. Once you reach 1, store the starting number along with its chain length. Take the next biggest number that has not being generated before, and repeat the process.
This will give you the list of numbers and chain length. Take the greatest chain length, and that's your answer.
I'll make some code to clarify.
public static long nextInChain(long n) {
if (n==1) return 1;
if (n%2==0) {
return n/2;
} else {
return (3 * n) + 1;
}
}
public static void main(String[] args) {
long iniTime=System.currentTimeMillis();
HashSet<Long> numbers=new HashSet<Long>();
HashMap<Long,Long> lenghts=new HashMap<Long, Long>();
long currentTry=1000000l;
int i=0;
do {
doTry(currentTry,numbers, lenghts);
currentTry=findNext(currentTry,numbers);
i++;
} while (currentTry!=0);
Set<Long> longs = lenghts.keySet();
long max=0;
long key=0;
for (Long aLong : longs) {
if (max < lenghts.get(aLong)) {
key = aLong;
max = lenghts.get(aLong);
}
}
System.out.println("number = " + key);
System.out.println("chain lenght = " + max);
System.out.println("Elapsed = " + ((System.currentTimeMillis()-iniTime)/1000));
}
private static long findNext(long currentTry, HashSet<Long> numbers) {
for(currentTry=currentTry-1;currentTry>=0;currentTry--) {
if (!numbers.contains(currentTry)) return currentTry;
}
return 0;
}
private static void doTry(Long tryNumber,HashSet<Long> numbers, HashMap<Long, Long> lenghts) {
long i=1;
long n=tryNumber;
do {
numbers.add(n);
n=nextInChain(n);
i++;
} while (n!=1);
lenghts.put(tryNumber,i);
}
Suppose you have a function CalcDistance(i) that calculates the "distance" to 1. For instance, CalcDistance(1) == 0 and CalcDistance(13) == 9. Here is a naive recursive implementation of this function (in C#):
public static int CalcDistance(long i)
{
if (i == 1)
return 0;
return (i % 2 == 0) ? CalcDistance(i / 2) + 1 : CalcDistance(3 * i + 1) + 1;
}
The problem is that this function has to calculate the distance of many numbers over and over again. You can make it a little bit smarter (and a lot faster) by giving it a memory. For instance, lets create a static array that can store the distance for the first million numbers:
static int[] list = new int[1000000];
We prefill each value in the list with -1 to indicate that the value for that position is not yet calculated. After this, we can optimize the CalcDistance() function:
public static int CalcDistance(long i)
{
if (i == 1)
return 0;
if (i >= 1000000)
return (i % 2 == 0) ? CalcDistance(i / 2) + 1 : CalcDistance(3 * i + 1) + 1;
if (list[i] == -1)
list[i] = (i % 2 == 0) ? CalcDistance(i / 2) + 1: CalcDistance(3 * i + 1) + 1;
return list[i];
}
If i >= 1000000, then we cannot use our list, so we must always calculate it. If i < 1000000, then we check if the value is in the list. If not, we calculate it first and store it in the list. Otherwise we just return the value from the list. With this code, it took about ~120ms to process all million numbers.
This is a very simple example of memoization. I use a simple list to store intermediate values in this example. You can use more advanced data structures like hashtables, vectors or graphs when appropriate.
Minimize how many levels deep your loops are, and use an efficient data structure such as IList or IDictionary, that can auto-resize itself when it needs to expand. If you use plain arrays they need to be copied to larger arrays as they expand - not nearly as efficient.
This variant doesn't use an HashMap but tries only to not repeat the first 1000000 numbers. I don't use an hashmap because the biggest number found is around 56 billions, and an hash map could crash.
I have already done some premature optimization. Instead of / I use >>, instead of % I use &. Instead of * I use some +.
void Main()
{
var elements = new bool[1000000];
int longestStart = -1;
int longestRun = -1;
long biggest = 0;
for (int i = elements.Length - 1; i >= 1; i--) {
if (elements[i]) {
continue;
}
elements[i] = true;
int currentStart = i;
int currentRun = 1;
long current = i;
while (current != 1) {
if (current > biggest) {
biggest = current;
}
if ((current & 1) == 0) {
current = current >> 1;
} else {
current = current + current + current + 1;
}
currentRun++;
if (current < elements.Length) {
elements[current] = true;
}
}
if (currentRun > longestRun) {
longestStart = i;
longestRun = currentRun;
}
}
Console.WriteLine("Longest Start: {0}, Run {1}", longestStart, longestRun);
Console.WriteLine("Biggest number: {0}", biggest);
}