Given a sequence a=(a1,a2....an) from n postive integers. We call Disorder D(ak) of ak=(a1,a2...ak) the diference between ak's max from ak's min. We call Total Disorder the sum all D(ak) for all subsequences from k=2 to k=n. We are looking for a dp algorithm with a recursive solution for b*, witch is a permutation of a,and it achieves minimum D(ak) from k=2 to k=n.
Exmples:
a=(6, 2, 3, 1, 3, 3) then b*=(3, 3, 3, 2, 1, 6)[with D(b*) = 0 + 0 + 1 + 2 + 5 = 8]
a=(1, 3, 3, 3, 6, 6) then b*=(3, 3, 3, 6, 6, 1)[with D(b*) = 0 + 0 + 3 + 3 + 5 = 11]
The only thing i was able to prove was that at the end of b* the number will be either the max of a or min of a.
Pls help.
First sort the input array, and then consider building the result permutation backwards from the end towards the start.
For every element you will either remove the first or last element of the sorted array. Also, for every position k, the disorder of the subarray ending at that position is known -- it's just the difference between the two ends of the remaining element array.
To find the optimal selection, then, you can use DP[k,n] = the minimum disorder so far if we've chosen n elements from the front of the sorted array (with the remainder chosen from the back).
DP[k,n] is easily calculated from DP[k+1,n] and DP[k+1,n-1], and the minimum DP[0,?] is the minimum total disorder.
Related
I came across an algorithm question recently and I still haven't been able to come up with a way to solve it. Can anyone help with pseudocode or logic?
Here is the question:
There are n elements in the array. N is an odd number. When we exclude 1 element from array, we are finding the minimum pairing cost.
Rules:
1 <= n <= 1001
n % 2 = 1
Example:
Given array is [4, 2, 1, 7, 8]
When we pair items with the closest ones [[1,2], [7,8]] and "4" is excluded.
So the minimum cost is |1 - 2| + |7 - 8| = 2;
What i tried:
Sort array first: [1,2,4,7,8]
Remove the middle element: 4
Pair items with the next ones: [[1, 2], [7, 8]]
According the example it works but what if the given array is [1, 7, 8, 16, 17]?
Sort array first: [1, 7, 8, 16, 17]
Remove the middle element: 8
Pair items with the next ones: [[1, 7], [16, 17]] Wrong Answer
"1" must be excluded and the pairs must be [[7, 8], [16, 17]]
Once the array is sorted, you can pair all elements from left to right, keep track of the total sum, and replace the last pairing with one starting from the right, updating the total sum if it's smaller.
In pseudo-code (all zero-based indexing):
let S be the sum of all pairing costs of
elements 2i and 2i+1 for i from 0 to (n-3)/2
(that is all pairings when you exclude the very last element)
let j = (n-1)/2
for i from (n-3)/2 to 0 (included):
let L be the pairing cost of elements 2i and 2i+1
let R be the pairing cost of elements 2i+1 and 2i+2
let S' = S - L + R
if S' < S
replace S with S'
replace j with i
2j is the element to exclude
Sorting the array first is a good start. Once you've done that, you have a choice of removing any value from index 1..N. A brute-force approach would be to calculate the pairing cost of omitting index 1, then recalculate omitting only index 2, and so on until you reach index N.
You'd be calculating many of the pairs over and over. To avoid that, consider that all the pairs to the left of your omitted index are paired odd-even (from the perspective of starting at element 1) and to the right of the omitted index will be even-odd. If you precalculate the sums of the left pairings and the sums of the right pairings into two arrays, you could determine the minimum cost at each position as the minimum sum of both values at each position of those two arrays.
I have a list of integers, and I need to find a way to get the maximum sum of a subset of them, adding elements to the total until the sum is equal to (or greater than) a fixed cutoff. I know this seems similar to the knapsack, but I was unsure whether it was equivalent.
Sorting the array and adding the maximum element until sum <= cutoff does not work. Observe the following list:
list = [6, 5, 4, 4, 4, 3, 2, 2, 1]
cutoff = 15
For this list, doing it the naive way results in a sum of 15, which is very sub-optimal. As far as I can see, the maximum you could arrive at using this list is 20, by adding 4 + 4 + 4 + 2 + 6. If this is just a different version of knapsack, I can just implement a knapsack solution, as I probably have small enough lists to get away with this, but I'd prefer to do something more efficient.
First of all in any sum, you won't have produced a worse result by adding the largest element last. So there is no harm in assuming that the elements are sorted from smallest to largest as a first step.
And now you use a dynamic programming approach similar to the usual subset sum.
def best_cutoff_sum (cutoff, elements):
elements = sorted(elements)
sums = {0: None}
for e in elements:
next_sums = {}
for v, path in sums.iteritems():
next_sums[v] = path
if v < cutoff:
next_sums[v + e] = [e, path]
sums = next_sums
best = max(sums.keys())
return (best, sums[best])
print(best_cutoff_sum(15, [6, 5, 4, 4, 4, 3, 2, 2, 1]))
With a little work you can turn the path from the nested array it currently is to whatever format you want.
If your list of non-negative elements has n elements, your cutoff is c and your maximum value is v, then this algorithm will take time O(n * (k + v))
I faced this problem in placement exam of SAP labs:
It's your birthday, so you are given a bag with a fixed space 'S'. You can go to a store and pick as many items you like which can be accommodated inside your bag. The store has 'n' items and each item occupies a space s[i]. You have to find out the maximum space in bag which you can fill.
For example, say the limit of you bag is S = 15 and the store has 10 items of sizes [1, 7, 3, 5, 4, 10, 6, 15, 20, 8]. Now you can fill 15 space by various ways such as [1, 7, 3, 4], [7, 3, 5], [15], [5, 10] and many more. So you return 15.
Note: There is quirk in the sizes of items. All of the items but at most 15 follow the following rule: *for all i, j, either size[i]>=2*size[j]+1 or size[j] >= 2*size[i] +1 if i ≠ j.*
Constraints:
1<= n <= 60.
1<= size[i] <= 10^17.
1<= S <= 10^18.
Example: S = 9, n = 5, sizes = [1, 7, 4, 4, 10].
Output: 8. You can't fill exactly 9 space in any way. You can fill 8 space either by using [1, 7] or [4, 4].
Let´s call x the elements that follow that rule. Note that for this set of elements, we have some nice properties:
Given x in sorted ascending order, sum(x[i..j]) < x[j + 1]
To solve maximum sum <= k, just iterate in sorted descending order and substract from k x[i] whenever possible. original k - remaining k is the solution. Assuming elements were already sorted, this is O(|x|).
One way to obtain this set is to iterate items sorted by size in ascending order and add to set if :
set has no elements or
current element >= 2 * size[lastElementAdded] + 1
Now we are left with at most 15 items that do not follow this rule. So we can´t use the efficient solving like before. For each item, we can consider to put it or not in the bag. This leads to 2^15 possible sums. For each of those sums, we can run our method for the elements that follow the rule.
Overall complexity: 2^15 * (n - 15). For n = 60, this should be solved in less than a second.
As an exercise: by using accumulated sums and binary search, it can be brought down to 2^15 * log2(n - 15).
Problem is simple -
Suppose I have an array of following numbers -
4,1,4,5,7,4,3,1,5
I have to find number of sets of k elements each that can be created from above numbers having largest sum. Two sets are considered to be different if they have at least one different element.
e.g.
if k = 2, then there can be two sets - {7,5} and {7,5}. Note: 5 appears twice in above array.
I think I can start with something like-
1. Sort array
2. Create two arrays. One for different number and an other in parallel for number's occurence.
But I am stuck now. Any suggestions?
The algorithm is as follows:
1) Sort elements in descending order.
2) Look at this array. It may look something like this:
a ... a b ... b c ... c d ...
| <- k -> |
Now obviously all elements a and b will be in the sets with the largest sum. You can't replace any of them with a smaller element, because then the sum wouldn't be the largest possible. So you have no choice here, you have to choose all a and b for any of the sets.
On the other hand only some of the elements c will be in those sets. So the answer is just the number of possibilities, to choose c's to fill the positions left in the sets, after you have taken all larger elements. That is the binomial coefficient:
count of c's choose (k - (count of elements larger than c))
For example for an array (already sorted here)
[9, 8, 7, 7, 5, 5, 5, 5, 4, 4, 2, 2, 1, 1, 1]
and k = 6, you must choose 9, 8 and both 7's for every set with the largest sum (which is 41). And then you can choose any two out of the four 5's. So the result will be 4 choose 2 = 6.
With the same array and k = 4, the result would be x choose 0 = 1 (that unique set is {9, 8, 7, 7}), with k = 7 the result would be 4 choose 3 = 4, and with k = 9: 2 choose 1 = 2 (choosing any 4 for the set with the largest sum).
EDIT: I edited the answer, because we figured it out that OP needs to count multisets.
First, find the largest k numbers in the array. This is of course easy, and if k is very small, you can do it in O(k) by performing k linear scans. If k is not so small, you can use a binary heap, or a priority queue or just sort the array to do that which is respectively O(n * log(k)) or O(n * log(n)) when using sorting.
Let assume that you have computed k largest numbers. Of course all sets of size k with the largest sum have to contain exactly these k largest numbers and no more other numbers. On the other hand, any different set doesn't have the largest sum.
Let count[i] be the number of occurrences of number i in the input sequence.
Let occ[i] be the number of occurrences of number i in the largest k numbers.
We can compute these both tables in very different ways, for example using a hash table or if input numbers are small, you can use an array indexed by these numbers.
Let B be the array of distinct numbers from the largest k numbers.
Let m be the size of B.
Now let's compute the answer. We will do it in m steps. After i-th step we will have computed the number of different multisets consisting of the first i numbers from B. At the beginning the result is 1 since there is only one empty multiset. In the i-th step, we will multiply the current result by the number of possible chooses of occ[B[i]] elements from count[B[i]] elements, which is equal to binomial(occ[i], count[i])
For example, let's consider your instance with added one more 7 at the end and k set to 3:
k = 3
A = [4, 1, 4, 5, 7, 4, 3, 1, 5, 7]
The largest three numbers in A are 7, 7, 5
At the beginning we have:
count[7] = 2
count[5] = 2
occ[7] = 2
occ[5] = 1
result = 1
B = [7, 5]
We start with the first element in B which is 7. Its count is 2 and its occ is also 2, so we do:
// binomial(2, 2) is 1
result = result * binomial(2, 2)
Next element in B is 5, its count is 2 and its occ is 1, so we do:
// binomial(2, 1) is 2
result = result * binomial(2, 1)
And the final result is 2, since there are two different multisets [7, 7, 5]
I'd create a sorted dictionary of the frequencies of occurrence of the numbers in the input. Then take the two largest numbers and multiply the number of times they occur.
In C++, it could look something like this:
std::vector<int> inputs { 4, 1, 4, 5, 7, 3, 1, 5};
std::map<int, int> counts;
for (auto i : inputs)
++counts[i];
auto last = counts.rbegin();
int largest_count = *last;
int second_count = *++last;
int set_count = largeest_count * second_count;
You can do the following:
1) Sort the elements in descending order;
2) define variable answer=1;
3) Start from the beginning of the array and for each new value you see, count the number of its occurrence (lets call this variable count). every time do: answer = answer * count. The pseudo-code should look like this.
find_count(Array A, K)
{
sort(A,'descending);
int answer=1;
int count=1;
for (int i=1,j=1; i<K && j<A.length;j++)
{
if(A[i] != A[i-1])
{
answer = answer *count;
i++;
count=1;
}
else
count++;
}
return answer;
}
what is a way to add elements of array thats sum would equal the largest element in the array?
example for this array [4, 6, 23, 10, 1, 3] I have sorted the array first resulting in [1, 3, 4, 6, 10, 23] then I pop the last digit or the last element max = 23. I'm left with [1, 3, 4, 6, 10] and need a way to find a way to find the elements that add up to 23 which are 3 + 4 + 6 + 10 = 23. The elements don't have to be subsequent they can be at random points of the array but they must add up to max.
I can find the permutations of the sorted array from 2 elements to n-1 elements and sum them and compare them to max but that seems inefficient. plz help
This is exactly the subset sum problem, which is NP-Complete, but if your numbers are relatively small integers, there is an efficient pseudo-polynomial solution using Dynamic Programming:
D(i,0) = TRUE
D(0,x) = FALSE x>0
D(i,x) = D(i-1,x) OR D(i-1,x-arr[i])
If there is a solution, you need to step back in the matrix created by the DP solution, and "record" each choice you have made along the way, to get the elements used for the summation. This thread deals with how to find the actual elements in a very similar problem (known as knapsack problem), which is solved similarly: How to find which elements are in the bag, using Knapsack Algorithm [and not only the bag's value]?