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I was asked in an interview today below question. I gave O(nlgn) solution but I was asked to give O(n) solution. I could not come up with O(n) solution. Can you help?
An input array is given like [1,2,4] then every element of it is doubled and
appended into the array. So the array now looks like [1,2,4,2,4,8]. How
this array is randomly shuffled. One possible random arrangement is
[4,8,2,1,2,4]. Now we are given this random shuffled array and we want to
get original array [1,2,4] in O(n) time.
The original array can be returned in any order. How can I do it?
Here's an O(N) Java solution that could be improved by first making sure that the array is of the proper form. For example it shouldn't accept [0] as an input:
import java.util.*;
class Solution {
public static int[] findOriginalArray(int[] changed) {
if (changed.length % 2 != 0)
return new int[] {};
// set Map size to optimal value to avoid rehashes
Map<Integer,Integer> count = new HashMap<>(changed.length*100/75);
int[] original = new int[changed.length/2];
int pos = 0;
// count frequency for each number
for (int n : changed) {
count.put(n, count.getOrDefault(n,0)+1);
}
// now decide which go into the answer
for (int n : changed) {
int smallest = n;
for (int m=n; m > 0 && count.getOrDefault(m,0) > 0; m = m/2) {
//System.out.println(m);
smallest = m;
if (m % 2 != 0) break;
}
// trickle up from smallest to largest while count > 0
for (int m=smallest, mm = 2*m; count.getOrDefault(mm,0) > 0; m = mm, mm=2*mm){
int ct = count.getOrDefault(mm,0);
while (count.get(m) > 0 && ct > 0) {
//System.out.println("adding "+m);
original[pos++] = m;
count.put(mm, ct -1);
count.put(m, count.get(m) - 1);
ct = count.getOrDefault(mm,0);
}
}
}
// check for incorrect format
if (count.values().stream().anyMatch(x -> x > 0)) {
return new int[] {};
}
return original;
}
public static void main(String[] args) {
int[] changed = {1,2,4,2,4,8};
System.out.println(Arrays.toString(changed));
System.out.println(Arrays.toString(findOriginalArray(changed)));
}
}
But I've tried to keep it simple.
The output is NOT guaranteed to be sorted. If you want it sorted it's going to cost O(NlogN) inevitably unless you use a Radix sort or something similar (which would make it O(NlogE) where E is the max value of the numbers you're sorting and logE the number of bits needed).
Runtime
This may not look that it is O(N) but you can see that it is because for every loop it will only find the lowest number in the chain ONCE, then trickle up the chain ONCE. Or said another way, in every iteration it will do O(X) iterations to process X elements. What will remain is O(N-X) elements. Therefore, even though there are for's inside for's it is still O(N).
An example execution can be seen with [64,32,16,8,4,2].
If this where not O(N) if you print out each value that it traverses to find the smallest you'd expect to see the values appear over and over again (for example N*(N+1)/2 times).
But instead you see them only once:
finding smallest 64
finding smallest 32
finding smallest 16
finding smallest 8
finding smallest 4
finding smallest 2
adding 2
adding 8
adding 32
If you're familiar with the Heapify algorithm you'll recognize the approach here.
def findOriginalArray(self, changed: List[int]) -> List[int]:
size = len(changed)
ans = []
left_elements = size//2
#IF SIZE IS ODD THEN RETURN [] NO SOLN. IS POSSIBLE
if(size%2 !=0):
return ans
#FREQUENCY DICTIONARY given array [0,0,2,1] my map will be: {0:2,2:1,1:1}
d = {}
for i in changed:
if(i in d):
d[i]+=1
else:
d[i] = 1
# CHECK THE EDGE CASE OF 0
if(0 in d):
count = d[0]
half = count//2
if((count % 2 != 0) or (half > left_elements)):
return ans
left_elements -= half
ans = [0 for i in range(half)]
#CHECK REST OF THE CASES : considering the values will be 10^5
for i in range(1,50001):
if(i in d):
if(d[i] > 0):
count = d[i]
if(count > left_elements):
ans = []
break
left_elements -= d[i]
for j in range(count):
ans.append(i)
if(2*i in d):
if(d[2*i] < count):
ans = []
break
else:
d[2*i] -= count
else:
ans = []
break
return ans
I have a simple idea which might not be the best, but I could not think of a case where it would not work. Having the array A with the doubled elements and randomly shuffled, keep a helper map. Process each element of the array and, each time you find a new element, add it to the map with the value 0. When an element is processed, increment map[i] and decrement map[2*i]. Next you iterate over the map and print the elements that have a value greater than zero.
A simple example, say that the vector is:
[1, 2, 3]
And the doubled/shuffled version is:
A = [3, 2, 1, 4, 2, 6]
When processing 3, first add the keys 3 and 6 to the map with value zero. Increment map[3] and decrement map[6]. This way, map[3] = 1 and map[6] = -1. Then for the next element map[2] = 1 and map[4] = -1 and so forth. The final state of the map in this example would be map[1] = 1, map[2] = 1, map[3] = 1, map[4] = -1, map[6] = 0, map[8] = -1, map[12] = -1.
Then you just process the keys of the map and, for each key with a value greater than zero, add it to the output. There are certainly more efficient solutions, but this one is O(n).
In C++, you can try this.
With time is O(N + KlogK) where N is the length of input, and K is the number of unique elements in input.
class Solution {
public:
vector<int> findOriginalArray(vector<int>& input) {
if (input.size() % 2) return {};
unordered_map<int, int> m;
for (int n : input) m[n]++;
vector<int> nums;
for (auto [n, cnt] : m) nums.push_back(n);
sort(begin(nums), end(nums));
vector<int> out;
for (int n : nums) {
if (m[2 * n] < m[n]) return {};
for (int i = 0; i < m[n]; ++i, --m[2 * n]) out.push_back(n);
}
return out;
}
};
Not so clear about the space complexity required in the question, so this is my top-of-the-mind attempt to this question if this requires O(n) time complexity.
If the length of the input array is not even, then its wrong !!
Create a map, add the elements of the input array to it.
Divide each element in the input array by 2 and check if that value exists in the map. If it exists, add it to the array (slice) orig.
There is a chance we have added duplicate values to this original array, clean it!!
Here is a sample go code:
https://go.dev/play/p/w4mm-rloHyi
I am sure we can optimize this code in a lot of ways for space complexities. But its O(n) time complexity.
You have given a array and You have to give number of continuous subarray which the sum is zero.
example:
1) 0 ,1,-1,0 => 6 {{0},{1,-1},{0,1,-1},{1,-1,0},{0}};
2) 5, 2, -2, 5 ,-5, 9 => 3.
With O(n^2) it can be done.I am trying to find the solution below this complexity.
Consider S[0..N] - prefix sums of your array, i.e. S[k] = A[0] + A[1] + ... + A[k-1] for k from 0 to N.
Now sum of elements from L to R-1 is zero if and only if S[R] = S[L]. It means that you have to find number of indices 0 <= L < R <= N such that S[L] = S[R].
This problem can be solved with a hash table. Iterate over elements of S[] while maintaining for each value X number of times it was met in the already processed part of S[]. These counts should be stored in a hash map, where the number X is a key, and the count H[X] is the value. When you meet a new elements S[i], add H[S[i]] to your answer (these account for substrings ending with (i-1)-st element), then increment H[S[i]] by one.
Note that if sum of absolute values of array elements is small, you can use a simple array instead of hash table. The complexity is linear on average.
Here is the code:
long long CountZeroSubstrings(vector<int> A) {
int n = A.size();
vector<long long> S(n+1, 0);
for (int i = 0; i < n; i++)
S[i+1] = S[i] + A[i];
long long answer = 0;
unordered_map<long long, int> H;
for (int i = 0; i <= n; i++) {
if (H.count(S[i]))
answer += H[S[i]];
H[S[i]]++;
}
return answer;
}
This can be solved in linear time by keeping a hash table of sums reached during the array traversal. The number of subsets can then be directly calculated from the counts of revisited sums.
Haskell version:
import qualified Data.Map as M
import Data.List (foldl')
f = foldl' (\b a -> b + div (a * (a + 1)) 2) 0 . M.elems . snd
. foldl' (\(s,m) x -> let s' = s + x in case M.lookup s' m of
Nothing -> (s',M.insert s' 0 m)
otherwise -> (s',M.adjust (+1) s' m)) (0,M.fromList[(0,0)])
Output:
*Main> f [0,1,-1,0]
6
*Main> f [5,2,-2,5,-5,9]
3
*Main> f [0,0,0,0]
10
*Main> f [0,1,0,0]
4
*Main> f [0,1,0,0,2,3,-3]
5
*Main> f [0,1,-1,0,0,2,3,-3]
11
C# version of #stgatilov answer https://stackoverflow.com/a/31489960/3087417 with readable variables:
int[] sums = new int[arr.Count() + 1];
for (int i = 0; i < arr.Count(); i++)
sums[i + 1] = sums[i] + arr[i];
int numberOfFragments = 0;
Dictionary<int, int> sumToNumberOfRepetitions = new Dictionary<int, int>();
foreach (int item in sums)
{
if (sumToNumberOfRepetitions.ContainsKey(item))
numberOfFragments += sumToNumberOfRepetitions[item];
else
sumToNumberOfRepetitions.Add(item, 0);
sumToNumberOfRepetitions[item]++;
}
return numberOfFragments;
If you want to have sum not only zero but any number k, here is the hint:
int numToFind = currentSum - k;
if (sumToNumberOfRepetitions.ContainsKey(numToFind))
numberOfFragments += sumToNumberOfRepetitions[numToFind];
I feel it can be solved using DP:
Let the state be :
DP[i][j] represents the number of ways j can be formed using all the subarrays ending at i!
Transitions:
for every element in the initial step ,
Increase the number of ways to form Element[i] using i elements by 1 i.e. using the subarray of length 1 starting from i and ending with i i.e
DP[i][Element[i]]++;
then for every j in Range [ -Mod(highest Magnitude of any element ) , Mod(highest Magnitude of any element) ]
DP[i][j]+=DP[i-1][j-Element[i]];
Then your answer will be the sum of all the DP[i][0] (Number of ways to form 0 using subarrays ending at i ) where i varies from 1 to Number of elements
Complexity is O(MOD highest magnitude of any element * Number of Elements)
https://www.techiedelight.com/find-sub-array-with-0-sum/
This would be an exact solution.
# Utility function to insert <key, value> into the dict
def insert(dict, key, value):
# if the key is seen for the first time, initialize the list
dict.setdefault(key, []).append(value)
# Function to print all sub-lists with 0 sum present
# in the given list
def printallSublists(A):
# create an empty -dict to store ending index of all
# sub-lists having same sum
dict = {}
# insert (0, -1) pair into the dict to handle the case when
# sub-list with 0 sum starts from index 0
insert(dict, 0, -1)
result = 0
sum = 0
# traverse the given list
for i in range(len(A)):
# sum of elements so far
sum += A[i]
# if sum is seen before, there exists at-least one
# sub-list with 0 sum
if sum in dict:
list = dict.get(sum)
result += len(list)
# find all sub-lists with same sum
for value in list:
print("Sublist is", (value + 1, i))
# insert (sum so far, current index) pair into the -dict
insert(dict, sum, i)
print("length :", result)
if __name__ == '__main__':
A = [0, 1, 2, -3, 0, 2, -2]
printallSublists(A)
I don't know what the complexity of my suggestion would be but i have an idea :)
What you can do is try to reduce element from main array which are not able to contribute for you solution
suppose elements are -10, 5, 2, -2, 5,7 ,-5, 9,11,19
so you can see that -10,9,11 and 19 are element
that are never gone be useful to make sum 0 in your case
so try to remove -10,9,11, and 19 from your main array
to do this what you can do is
1) create two sub array from your main array
`positive {5,7,2,9,11,19}` and `negative {-10,-2,-5}`
2) remove element from positive array which does not satisfy condition
condition -> value should be construct from negative arrays element
or sum of its elements
ie.
5 = -5 //so keep it //don't consider the sign
7 = (-5 + -2 ) // keep
2 = -2 // keep
9 // cannot be construct using -10,-2,-5
same for all 11 and 19
3) remove element form negative array which does not satisfy condition
condition -> value should be construct from positive arrays element
or sum of its elements
i.e. -10 // cannot be construct so discard
-2 = 2 // keep
-5 = 5 // keep
so finally you got an array which contains -2,-5,5,7,2 create all possible sub array form it and check for sum = 0
(Note if your input array contains 0 add all 0's in final array)
Given a array with +ve and -ve integer , find the maximum sum such that you are not allowed to skip 2 contiguous elements ( i.e you have to select at least one of them to move forward).
eg :-
10 , 20 , 30, -10 , -50 , 40 , -50, -1, -3
Output : 10+20+30-10+40-1 = 89
This problem can be solved using Dynamic Programming approach.
Let arr be the given array and opt be the array to store the optimal solutions.
opt[i] is the maximum sum that can be obtained starting from element i, inclusive.
opt[i] = arr[i] + (some other elements after i)
Now to solve the problem we iterate the array arr backwards, each time storing the answer opt[i].
Since we cannot skip 2 contiguous elements, either element i+1 or element i+2 has to be included
in opt[i].
So for each i, opt[i] = arr[i] + max(opt[i+1], opt[i+2])
See this code to understand:
int arr[n]; // array of given numbers. array size = n.
nput(arr, n); // input the array elements (given numbers)
int opt[n+2]; // optimal solutions.
memset(opt, 0, sizeof(opt)); // Initially set all optimal solutions to 0.
for(int i = n-1; i >= 0; i--) {
opt[i] = arr[i] + max(opt[i+1], opt[i+2]);
}
ans = max(opt[0], opt[1]) // final answer.
Observe that opt array has n+2 elements. This is to avoid getting illegal memory access exception (memory out of bounds) when we try to access opt[i+1] and opt[i+2] for the last element (n-1).
See the working implementation of the algorithm given above
Use a recurrence that accounts for that:
dp[i] = max(dp[i - 1] + a[i], <- take two consecutives
dp[i - 2] + a[i], <- skip a[i - 1])
Base cases left as an exercise.
If you see a +ve integer add it to the sum. If you see a negative integer, then inspect the next integer pick which ever is maximum and add it to the sum.
10 , 20 , 30, -10 , -50 , 40 , -50, -1, -3
For this add 10, 20, 30, max(-10, -50), 40 max(-50, -1) and since there is no element next to -3 discard it.
The last element will go to sum if it was +ve.
Answer:
I think this algorithm will help.
1. Create a method which gives output the maximum sum of particular user input array say T[n], where n denotes the total no. of elements.
2. Now this method will keep on adding array elements till they are positive. As we want to maximize the sum and there is no point in dropping positive elements behind.
3. As soon as our method encounters a negative element, it will transfer all consecutive negative elements to another method which create a new array say N[i] such that this array will contain all the consecutive negative elements that we encountered in T[n] and returns N[i]'s max output.
In this way our main method is not affected and its keep on adding positive elements and whenever it encounters negative element, it instead of adding their real values adds the net max output of that consecutive array of negative elements.
for example: T[n] = 29,34,55,-6,-5,-4,6,43,-8,-9,-4,-3,2,78 //here n=14
Main Method Working:
29+34+55+(sends data & gets value from Secondary method of array [-6,-5,-4])+6+43+(sends data & gets value from Secondary method of array [-8,-9,-4,-3])+2+78
Process Terminates with max output.
Secondary Method Working:
{
N[i] = gets array from Main method or itself as and when required.
This is basically a recursive method.
say N[i] has elements like N1, N2, N3, N4, etc.
for i>=3:
Now choice goes like this.
1. If we take N1 then we can recurse the left off array i.e. N[i-1] which has all elements except N1 in same order. Such that the net max output will be
N1+(sends data & gets value from Secondary method of array N[i-1] recursively)
2. If we doesn't take N1, then we cannot skip N2. So, Now algorithm is like 1st choice but starting with N2. So max output in this case will be
N2+(sends data & gets value from Secondary method of array N[i-2] recursively).
Here N[i-2] is an array containing all N[i] elements except N1 & N2 in same order.
Termination: When we are left with the array of size one ( for N[i-2] ) then we have to choose that particular value as no option.
The recursions will finally yield the max outputs and we have to finally choose the output of that choice which is more.
and redirect the max output to wherever required.
for i=2:
we have to choose the value which is bigger
for i=1:
We can surely skip that value.
So max output in this case will be 0.
}
I think this answer will help to you.
Given array:
Given:- 10 20 30 -10 -50 40 -50 -1 -3
Array1:-10 30 60 50 10 90 40 89 86
Array2:-10 20 50 40 0 80 30 79 76
Take the max value of array1[n-1],array1[n],array2[n-1],array2[n] i.e 89(array1[n-1])
Algorithm:-
For the array1 value assign array1[0]=a[0],array1=a[0]+a[1] and array2[0]=a[0],array2[1]=a[1].
calculate the array1 value from 2 to n is max of sum of array1[i-1]+a[i] or array1[i-2]+a[i].
for loop from 2 to n{
array1[i]=max(array1[i-1]+a[i],array1[i-2]+a[i]);
}
similarly for array2 value from 2 to n is max of sum of array2[i-1]+a[i] or array2[i-2]+a[i].
for loop from 2 to n{
array2[i]=max(array2[i-1]+a[i],array2[i-2]+a[i]);
}
Finally find the max value of array1[n-1],array[n],array2[n-1],array2[n];
int max(int a,int b){
return a>b?a:b;
}
int main(){
int a[]={10,20,30,-10,-50,40,-50,-1,-3};
int i,n,max_sum;
n=sizeof(a)/sizeof(a[0]);
int array1[n],array2[n];
array1[0]=a[0];
array1[1]=a[0]+a[1];
array2[0]=a[0];
array2[1]=a[1];
for loop from 2 to n{
array1[i]=max(array1[i-1]+a[i],array1[i-2]+a[i]);
array2[i]=max(array2[i-1]+a[i],array2[i-2]+a[i]);
}
--i;
max_sum=max(array1[i],array1[i-1]);
max_sum=max(max_sum,array2[i-1]);
max_sum=max(max_sum,array2[i]);
printf("The max_sum is %d",max_sum);
return 0;
}
Ans: The max_sum is 89
public static void countSum(int[] a) {
int count = 0;
int skip = 0;
int newCount = 0;
if(a.length==1)
{
count = a[0];
}
else
{
for(int i:a)
{
newCount = count + i;
if(newCount>=skip)
{
count = newCount;
skip = newCount;
}
else
{
count = skip;
skip = newCount;
}
}
}
System.out.println(count);
}
}
Let the array be of size N, indexed as 1...N
Let f(n) be the function, that provides the answer for max sum of sub array (1...n), such that no two left over elements are consecutive.
f(n) = max (a[n-1] + f(n-2), a(n) + f(n-1))
In first option, which is - {a[n-1] + f(n-2)}, we are leaving the last element, and due to condition given in question selecting the second last element.
In the second option, which is - {a(n) + f(n-1)} we are selecting the last element of the subarray, so we have an option to select/deselect the second last element.
Now starting from the base case :
f(0) = 0 [Subarray (1..0) doesn't exist]
f(1) = (a[1] > 0 ? a[1] : 0); [Subarray (1..1)]
f(2) = max( a(2) + 0, a[1] + f(1)) [Choosing atleast one of them]
Moving forward we can calculate any f(n), where n = 1...N, and store them to calculate next results. And yes, obviously, the case f(N) will give us the answer.
Time complexity o(n)
Space complexity o(n)
n = arr.length().
Append a 0 at the end of the array to handle boundary case.
ans: int array of size n+1.
ans[i] will store the answer for array a[0...i] which includes a[i] in the answer sum.
Now,
ans[0] = a[0]
ans[1] = max(a[1], a[1] + ans[0])
for i in [2,n-1]:
ans[i] = max(ans[i-1] , ans[i-2]) + a[i]
Final answer would be a[n]
If you want to avoid using Dynamic Programming
To find the maximum sum, first, you've to add all the positive
numbers.
We'll be skipping only negative elements. Since we're not
allowed to skip 2 contiguous elements, we will put all contiguous
negative elements in a temp array, and can figure out the maximum sum
of alternate elements using sum_odd_even function as defined below.
Then we can add the maximum of all such temp arrays to our sum of all
positive numbers. And the final sum will give us the desired output.
Code:
def sum_odd_even(arr):
sum1 = sum2 = 0
for i in range(len(arr)):
if i%2 == 0:
sum1 += arr[i]
else:
sum2 += arr[i]
return max(sum1,sum2)
input = [10, 20, 30, -10, -50, 40, -50, -1, -3]
result = 0
temp = []
for i in range(len(input)):
if input[i] > 0:
result += input[i]
if input[i] < 0 and i != len(input)-1:
temp.append(input[i])
elif input[i] < 0:
temp.append(input[i])
result += sum_odd_even(temp)
temp = []
else:
result += sum_odd_even(temp)
temp = []
print result
Simple Solution: Skip with twist :). Just skip the smallest number in i & i+1 if consecutive -ve. Have if conditions to check that till n-2 elements and check for the last element in the end.
int getMaxSum(int[] a) {
int sum = 0;
for (int i = 0; i <= a.length-2; i++) {
if (a[i]>0){
sum +=a[i];
continue;
} else if (a[i+1] > 0){
i++;
continue;
} else {
sum += Math.max(a[i],a[i+1]);
i++;
}
}
if (a[a.length-1] > 0){
sum+=a[a.length-1];
}
return sum;
}
The correct recurrence is as follow:
dp[i] = max(dp[i - 1] + a[i], dp[i - 2] + a[i - 1])
The first case is the one we pick the i-th element. The second case is the one we skip the i-th element. In the second case, we must pick the (i-1)th element.
The problem of IVlad's answer is that it always pick i-th element, which can lead to incorrect answer.
This question can be solved using include,exclude approach.
For first element, include = arr[0], exclude = 0.
For rest of the elements:
nextInclude = arr[i]+max(include, exclude)
nextExclude = include
include = nextInclude
exclude = nextExclude
Finally, ans = Math.max(include,exclude).
Similar questions can be referred at (Not the same)=> https://www.youtube.com/watch?v=VT4bZV24QNo&t=675s&ab_channel=Pepcoding.
Suppose you are a thief and you invaded a house. Inside you found the following items:
A vase that weights 3 pounds and is worth 50 dollars.
A silver nugget that weights 6 pounds and is worth 30 dollars.
A painting that weights 4 pounds and is worth 40 dollars.
A mirror that weights 5 pounds and is worth 10 dollars.
Solution to this Knapsack problem of size 10 pounds is 90 dollars .
Table made from dynamic programming is :-
Now i want to know which elements i put in my sack using this table then how to back track ??
From your DP table we know f[i][w] = the maximum total value of a subset of items 1..i that has total weight less than or equal to w.
We can use the table itself to restore the optimal packing:
def reconstruct(i, w): # reconstruct subset of items 1..i with weight <= w
# and value f[i][w]
if i == 0:
# base case
return {}
if f[i][w] > f[i-1][w]:
# we have to take item i
return {i} UNION reconstruct(i-1, w - weight_of_item(i))
else:
# we don't need item i
return reconstruct(i-1, w)
I have an iterative algorithm inspired by #NiklasB. that works when a recursive algorithm would hit some kind of recursion limit.
def reconstruct(i, w, kp_soln, weight_of_item):
"""
Reconstruct subset of items i with weights w. The two inputs
i and w are taken at the point of optimality in the knapsack soln
In this case I just assume that i is some number from a range
0,1,2,...n
"""
recon = set()
# assuming our kp soln converged, we stopped at the ith item, so
# start here and work our way backwards through all the items in
# the list of kp solns. If an item was deemed optimal by kp, then
# put it in our bag, otherwise skip it.
for j in range(0, i+1)[::-1]:
cur_val = kp_soln[j][w]
prev_val = kp_soln[j-1][w]
if cur_val > prev_val:
recon.add(j)
w = w - weight_of_item[j]
return recon
Using a loop :
for (int n = N, w = W; n > 0; n--)
{
if (sol[n][w] != 0)
{
selected[n] = 1;
w = w - wt[n];
}
else
selected[n] = 0;
}
System.out.print("\nItems with weight ");
for (int i = 1; i < N + 1; i++)
if (selected[i] == 1)
System.out.print(val[i] +" ");
I got asked this question on a interview for Google a couple of weeks ago, I didn't quite get the answer and I was wondering if anyone here could help me out.
You have an array with n elements. The elements are either 0 or 1.
You want to split the array into k contiguous subarrays. The size of each subarray can vary between ceil(n/2k) and floor(3n/2k). You can assume that k << n.
After you split the array into k subarrays. One element of each subarray will be randomly selected.
Devise an algorithm for maximizing the sum of the randomly selected elements from the k subarrays.
Basically means that we will want to split the array in such way such that the sum of all the expected values for the elements selected from each subarray is maximum.
You can assume that n is a power of 2.
Example:
Array: [0,0,1,1,0,0,1,1,0,1,1,0]
n = 12
k = 3
Size of subarrays can be: 2,3,4,5,6
Possible subarrays [0,0,1] [1,0,0,1] [1,0,1,1,0]
Expected Value of the sum of the elements randomly selected from the subarrays: 1/3 + 2/4 + 3/5 = 43/30 ~ 1.4333333
Optimal split: [0,0,1,1,0,0][1,1][0,1,1,0]
Expected value of optimal split: 1/3 + 1 + 1/2 = 11/6 ~ 1.83333333
I think we can solve this problem using dynamic programming.
Basically, we have:
f(i,j) is defined as the maximum sum of all expected values chosen from an array of size i and split into j subarrays. Therefore the solution should be f(n,k).
The recursive equation is:
f(i,j) = f(i-x,j-1) + sum(i-x+1,i)/x where (n/2k) <= x <= (3n/2k)
I don't know if this is still an open question or not, but it seems like the OP has managed to add enough clarifications that this should be straightforward to solve. At any rate, if I am understanding what you are saying this seems like a fair thing to ask in an interview environment for a software development position.
Here is the basic O(n^2 * k) solution, which should be adequate for small k (as the interviewer specified):
def best_val(arr, K):
n = len(arr)
psum = [ 0.0 ]
for x in arr:
psum.append(psum[-1] + x)
tab = [ -100000 for i in range(n) ]
tab.append(0)
for k in range(K):
for s in range(n - (k+1) * ceil(n/(2*K))):
terms = range(s + ceil(n/(2*K)), min(s + floor((3*n)/(2*K)) + 1, n+1))
tab[s] = max( [ (psum[t] - psum[s]) / (t - s) + tab[t] for t in terms ])
return tab[0]
I used the numpy ceil/floor functions but you basically get the idea. The only `tricks' in this version is that it does windowing to reduce the memory overhead to just O(n) instead of O(n * k), and that it precalculates the partial sums to make computing the expected value for a box a constant time operation (thus saving a factor of O(n) from the inner loop).
I don't know if anyone is still interested to see the solution for this problem. Just stumbled upon this question half an hour ago and thought of posting my solution(Java). The complexity for this is O(n*K^log10). The proof is a little convoluted so I would rather provide runtime numbers:
n k time(ms)
48 4 25
48 8 265
24 4 20
24 8 33
96 4 51
192 4 143
192 8 343919
The solution is the same old recursive one where given an array, choose the first partition of size ceil(n/2k) and find the best solution recursively for the rest with number of partitions = k -1, then take ceil(n/2k) + 1 and so on.
Code:
public class PartitionOptimization {
public static void main(String[] args) {
PartitionOptimization p = new PartitionOptimization();
int[] input = { 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0};
int splitNum = 3;
int lowerLim = (int) Math.ceil(input.length / (2.0 * splitNum));
int upperLim = (int) Math.floor((3.0 * input.length) / (2.0 * splitNum));
System.out.println(input.length + " " + lowerLim + " " + upperLim + " " +
splitNum);
Date currDate = new Date();
System.out.println(currDate);
System.out.println(p.getMaxPartExpt(input, lowerLim, upperLim,
splitNum, 0));
System.out.println(new Date().getTime() - currDate.getTime());
}
public double getMaxPartExpt(int[] input, int lowerLim, int upperLim,
int splitNum, int startIndex) {
if (splitNum <= 1 && startIndex<=(input.length -lowerLim+1)){
double expt = findExpectation(input, startIndex, input.length-1);
return expt;
}
if (!((input.length - startIndex) / lowerLim >= splitNum))
return -1;
double maxExpt = 0;
double curMax = 0;
int bestI=0;
for (int i = startIndex + lowerLim - 1; i < Math.min(startIndex
+ upperLim, input.length); i++) {
double curExpect = findExpectation(input, startIndex, i);
double splitExpect = getMaxPartExpt(input, lowerLim, upperLim,
splitNum - 1, i + 1);
if (splitExpect>=0 && (curExpect + splitExpect > maxExpt)){
bestI = i;
curMax = curExpect;
maxExpt = curExpect + splitExpect;
}
}
return maxExpt;
}
public double findExpectation(int[] input, int startIndex, int endIndex) {
double expectation = 0;
for (int i = startIndex; i <= endIndex; i++) {
expectation = expectation + input[i];
}
expectation = (expectation / (endIndex - startIndex + 1));
return expectation;
}
}
Not sure I understand, the algorithm is to split the array in groups, right? The maximum value the sum can have is the number of ones. So split the array in "n" groups of 1 element each and the addition will be the maximum value possible. But it must be something else and I did not understand the problem, that seems too silly.
I think this can be solved with dynamic programming. At each possible split location, get the maximum sum if you split at that location and if you don't split at that point. A recursive function and a table to store history might be useful.
sum_i = max{ NumOnesNewPart/NumZerosNewPart * sum(NewPart) + sum(A_i+1, A_end),
sum(A_0,A_i+1) + sum(A_i+1, A_end)
}
This might lead to something...
I think its a bad interview question, but it is also an easy problem to solve.
Every integer contributes to the expected value with weight 1/s where s is the size of the set where it has been placed. Therefore, if you guess the sizes of the sets in your partition, you just need to fill the sets with ones starting from the smallest set, and then fill the remaining largest set with zeroes.
You can easily see then that if you have a partition, filled as above, where the sizes of the sets are S_1, ..., S_k and you do a transformation where you remove one item from set S_i and move it to set S_i+1, you have the following cases:
Both S_i and S_i+1 were filled with ones; then the expected value does not change
Both them were filled with zeroes; then the expected value does not change
S_i contained both 1's and 0's and S_i+1 contains only zeroes; moving 0 to S_i+1 increases the expected value because the expected value of S_i increases
S_i contained 1's and S_i+1 contains both 1's and 0's; moving 1 to S_i+1 increases the expected value because the expected value of S_i+1 increases and S_i remains intact
In all these cases, you can shift an element from S_i to S_i+1, maintaining the filling rule of filling smallest sets with 1's, so that the expected value increases. This leads to the simple algorithm:
Create a partitioning where there is a maximal number of maximum-size arrays and maximal number of minimum-size arrays
Fill the arrays starting from smallest one with 1's
Fill the remaining slots with 0's
How about a recursive function:
int BestValue(Array A, int numSplits)
// Returns the best value that would be obtained by splitting
// into numSplits partitions.
This in turn uses a helper:
// The additional argument is an array of the valid split sizes which
// is the same for each call.
int BestValueHelper(Array A, int numSplits, Array splitSizes)
{
int result = 0;
for splitSize in splitSizes
int splitResult = ExpectedValue(A, 0, splitSize) +
BestValueHelper(A+splitSize, numSplits-1, splitSizes);
if splitResult > result
result = splitResult;
}
ExpectedValue(Array A, int l, int m) computes the expected value of a split of A that goes from l to m i.e. (A[l] + A[l+1] + ... A[m]) / (m-l+1).
BestValue calls BestValueHelper after computing the array of valid split sizes between ceil(n/2k) and floor(3n/2k).
I have omitted error handling and some end conditions but those should not be too difficult to add.
Let
a[] = given array of length n
from = inclusive index of array a
k = number of required splits
minSize = minimum size of a split
maxSize = maximum size of a split
d = maxSize - minSize
expectation(a, from, to) = average of all element of array a from "from" to "to"
Optimal(a[], from, k) = MAX[ for(j>=minSize-1 to <=maxSize-1) { expectation(a, from, from+j) + Optimal(a, j+1, k-1)} ]
Runtime (assuming memoization or dp) = O(n*k*d)