Given an array of integers, take any two non-intersecting subarrays, A1 and A2. A subarray could be empty too. we have to return the max(sum(A1,A2)). The return value is the maximum possible sum of two subarrays of the given array.
My Thoughts:
Find the max sum subarray. This will give two new subarrays, the left of it and the right of it. Find the max sum subarray left and right and choose the optimal one. I am not sure if this will yield the max answer.
Any ideas or approach?
Find the max subarray ending at index i going left and find the max subarray ending at index i going right. Then for each candidate, update the best result with the larger of the current best or max_subarray_sum_on_the_left[i] + max_subarray_sum_on_the_right[i+1].
In case it was unclear -- when performing Kadane's algorithm for each direction, save for each index the best seen from that direction. This leads to an O(n) overall solution.
Create array left[] such that left[i] = max subarray sum ending at index i (takes O(N) time)
Create array right[] such that right[i] = max subarray sum starting at index i (takes O(N) time)
Iterate over the arrays and keep track of the max value visited such that left[i] = max possible sum over all subarrays ending at i and right[i] = max possible sum over all subarrays starting at i. (takes O(N) time)
Loop i from 1 to N-1, where i is the index about which we'll "partition" our array and choose non-intersecting subarrays. For a given i, the answer will be left[i] + right[i]
The solution has O(N) space and time complexity.
Say I had N arrays. These N arrays are in an array of arrays A. How many N-tuples are there such that for a tuple t,
sum = 0
for i = 0 ... N-1
sum += A[i][t[i]]
sum == k
What is an efficient way to solve this? The best I can come up with is just enumerating all possibilities.
P.S. This isn't a homework question. I saw this on LeetCode and was curious about a solution to the general case.
Conceptual solution (can be improved):
sort the elements in each array
shift the elements in each array by the absolute minimum of all arrays (abs_min - to shift all arrays you'll subtract the abs_min from each element of all arrays) - you now have all arrays with non-negative elements and you are searching for a target_sum = initial_sum - num_arrays*abs_min
set your curr_array as the first one
binary search for the position of target_sum in the curr_array. You will need to consider all the elements in curr_array with indices under this position. Take one such element, subtract it from the target_sum, and recursively repeat the search with the next array.
I believe the (amortised) complexity will be somewhere of O(num_arrays*N*log(N)) where N is the (maximum) number of elements in the arrays.
Opportunities for improvement:
I kinda feel that shifting all arrays by abs_min is unnecessary (just an artifice that helps the thinking). Maybe before going one step deeper in recursion in step 4, the target_sum may be shifted by the min of current array?
reordering the arrays so that the shorter ones are considered first will perhaps improve the performance (lower number of elements in the upper levels of the recursion to consider) [Edit] or maybe reordering the arrays in the descending order of their min value (take out from the target_sum in the most aggressive way possible)?
adopting a scheme which eliminate/multiplexes the duplicates inside the initial arrays may help - i.e a map with the index_key=unique_value and the map-value the set of indexes). If the specific tuples are not required, then a map of unique-value->occurrence_count would be enough. (this may be useful if one can be sure that duplicate exist - e.g. the values in the arrays are within tight ranges and arrays are pretty long - pigeonhole principle)
[Edited to show how it works in the example of {{1, 2, 3}, {42,43, 44, 45, 46, 47}}]
Upper limit = index of the element strictly greater than the provided value. If you want values lesser or equal, take values strictly below that index!!
Zero-based index convention
49 target sum in the first array gets the upper limit of index=3 (so all indexes under 3 need to to be considered)
first array - start index=2 / value=3 in the first array, you will be looking for a target_sum of 46 in the second. Upper limit by binary search in the second is index=5 (and will be looking strictly under), so start with index=4/value=46 (the algo cuts out the value of 47). 46 is good and retained, index=3/value=45 is not enough and (not having a 3-rd array to recurse into it) the algo won't even consider under index=3/value=45.
first array, index=1/value=2, looking for a target_sum of 47 in the second array. Get an upper limit (binary search) affords index=7 (to search strictly under it) so index=6/value=47. 47 is retained, 46 and below and the algo cut out
down in the first array, index=0/value=1, looking for a target_sum of 48 in the second array. Upper limit is again 7, at index=6/value=47 we get an insufficient value and terminate.
So, grand totals:
Total binary searches: 1-in the first array, 3 in the second.
Total successful equalities tested=2 (two tuples found).
Total unsuccesful equalities tested=3 (until the second array does no longer offer a satisfactory answer).
Total additions/subtraction performed=3 (one for each value in the first array)
By contrast, the exhaustive scanning would get:
no binary searches
total additions = 3*6=18
total successful equality tested = 2
total un-successful equality tested = 16
Language: C++
Constraints: A[i][j] >= 0
Complexity: O(N * k)
int A [MAX_N][MAX_N], memo[MAX_N][MAX_K+1];
int countWays2(int m, int sum, int N){
if(memo [m][sum] != -1)
return memo [m][sum];
if(m == 0)
return memo[m][sum] = count(A[0], A[0]+N, sum);
int ways = 0;
for(int i = 0; i < N; ++i)
if(sum >= A[m][i])
ways += countWays2(m-1, sum - A[m][i], N);
return memo[m][sum] = ways;
}
int countWays(int N, int k){
if(k < 0) return 0;
for(int i = 0; i < N; ++i)
fill(memo[i], memo[i] + k + 1, -1); //Initialize memoization table
return countWays2(N-1, k, N);
}
The answer is countWays(N, k)
How do I find/store maximum/minimum of all possible non-empty sub-arrays of an array of length n?
I generated the segment tree of the array and the for each possible sub array if did query into segment tree but that's not efficient. How do I do it in O(n)?
P.S n <= 10 ^7
For eg. arr[]= { 1, 2, 3 }; // the array need not to be sorted
sub-array min max
{1} 1 1
{2} 2 2
{3} 3 3
{1,2} 1 2
{2,3} 2 3
{1,2,3} 1 3
I don't think it is possible to store all those values in O(n). But it is pretty easy to create, in O(n), a structure that makes possible to answer, in O(1) the query "how many subsets are there where A[i] is the maximum element".
Naïve version:
Think about the naïve strategy: to know how many such subsets are there for some A[i], you could employ a simple O(n) algorithm that counts how many elements to the left and to the right of the array that are less than A[i]. Let's say:
A = [... 10 1 1 1 5 1 1 10 ...]
This 5 up has 3 elements to the left and 2 to the right lesser than it. From this we know there are 4*3=12 subarrays for which that very 5 is the maximum. 4*3 because there are 0..3 subarrays to the left and 0..2 to the right.
Optimized version:
This naïve version of the check would take O(n) operations for each element, so O(n^2) after all. Wouldn't it be nice if we could compute all these lengths in O(n) in a single pass?
Luckily there is a simple algorithm for that. Just use a stack. Traverse the array normally (from left to right). Put every element index in the stack. But before putting it, remove all the indexes whose value are lesser than the current value. The remaining index before the current one is the nearest larger element.
To find the same values at the right, just traverse the array backwards.
Here's a sample Python proof-of-concept that shows this algorithm in action. I implemented also the naïve version so we can cross-check the result from the optimized version:
from random import choice
from collections import defaultdict, deque
def make_bounds(A, fallback, arange, op):
stack = deque()
bound = [fallback] * len(A)
for i in arange:
while stack and op(A[stack[-1]], A[i]):
stack.pop()
if stack:
bound[i] = stack[-1]
stack.append(i)
return bound
def optimized_version(A):
T = zip(make_bounds(A, -1, xrange(len(A)), lambda x, y: x<=y),
make_bounds(A, len(A), reversed(xrange(len(A))), lambda x, y: x<y))
answer = defaultdict(lambda: 0)
for i, x in enumerate(A):
left, right = T[i]
answer[x] += (i-left) * (right-i)
return dict(answer)
def naive_version(A):
answer = defaultdict(lambda: 0)
for i, x in enumerate(A):
left = next((j for j in range(i-1, -1, -1) if A[j]>A[i]), -1)
right = next((j for j in range(i+1, len(A)) if A[j]>=A[i]), len(A))
answer[x] += (i-left) * (right-i)
return dict(answer)
A = [choice(xrange(32)) for i in xrange(8)]
MA1 = naive_version(A)
MA2 = optimized_version(A)
print 'Array: ', A
print 'Naive: ', MA1
print 'Optimized:', MA2
print 'OK: ', MA1 == MA2
I don't think it is possible to it directly in O(n) time: you need to iterate over all the elements of the subarrays, and you have n of them. Unless the subarrays are sorted.
You could, on the other hand, when initialising the subarrays, instead of making them normal arrays, you could build heaps, specifically min heaps when you want to find the minimum and max heaps when you want to find the maximum.
Building a heap is a linear time operation, and retrieving the maximum and minimum respectively for a max heap and min heap is a constant time operation, since those elements are found at the first place of the heap.
Heaps can be easily implemented just using a normal array.
Check this article on Wikipedia about binary heaps: https://en.wikipedia.org/wiki/Binary_heap.
I do not understand what exactly you mean by maximum of sub-arrays, so I will assume you are asking for one of the following
The subarray of maximum/minimum length or some other criteria (in which case the problem will reduce to finding max element in a 1 dimensional array)
The maximum elements of all your sub-arrays either in the context of one sub-array or in the context of the entire super-array
Problem 1 can be solved by simply iterating your super-array and storing a reference to the largest element. Or building a heap as nbro had said. Problem 2 also has a similar solution. However a linear scan is through n arrays of length m is not going to be linear. So you will have to keep your class invariants such that the maximum/minimum is known after every operation. Maybe with the help of some data structure like a heap.
Assuming you mean contiguous sub-arrays, create the array of partial sums where Yi = SUM(i=0..i)Xi, so from 1,4,2,3 create 0,1,1+4=5,1+4+2=7,1+4+2+3=10. You can create this from left to right in linear time, and the value of any contiguous subarray is one partial sum subtracted from another, so 4+2+3 = 1+4+2+3 - 1= 9.
Then scan through the partial sums from left to right, keeping track of the smallest value seen so far (including the initial zero). At each point subtract this from the current value and keep track of the highest value produced in this way. This should give you the value of the contiguous sub-array with largest sum, and you can keep index information, too, to find where this sub-array starts and ends.
To find the minimum, either change the above slightly or just reverse the sign of all the numbers and do exactly the same thing again: min(a, b) = -max(-a, -b)
I think the question you are asking is to find the Maximum of a subarry.
bleow is the code that cand do that in O(n) time.
int maxSumSubArr(vector<int> a)
{
int maxsum = *max_element(a.begin(), a.end());
if(maxsum < 0) return maxsum;
int sum = 0;
for(int i = 0; i< a.size; i++)
{
sum += a[i];
if(sum > maxsum)maxsum = sum;
if(sum < 0) sum = 0;
}
return maxsum;
}
Note: This code is not tested please add comments if found some issues.
Given two sorted arrays of numbers, we want to find the pair with the kth largest possible sum. (A pair is one element from the first array and one element from the second array). For example, with arrays
[2, 3, 5, 8, 13]
[4, 8, 12, 16]
The pairs with largest sums are
13 + 16 = 29
13 + 12 = 25
8 + 16 = 24
13 + 8 = 21
8 + 12 = 20
So the pair with the 4th largest sum is (13, 8). How to find the pair with the kth largest possible sum?
Also, what is the fastest algorithm? The arrays are already sorted and sizes M and N.
I am already aware of the O(Klogk) solution , using Max-Heap given here .
It also is one of the favorite Google interview question , and they demand a O(k) solution .
I've also read somewhere that there exists a O(k) solution, which i am unable to figure out .
Can someone explain the correct solution with a pseudocode .
P.S.
Please DON'T post this link as answer/comment.It DOESN'T contain the answer.
I start with a simple but not quite linear-time algorithm. We choose some value between array1[0]+array2[0] and array1[N-1]+array2[N-1]. Then we determine how many pair sums are greater than this value and how many of them are less. This may be done by iterating the arrays with two pointers: pointer to the first array incremented when sum is too large and pointer to the second array decremented when sum is too small. Repeating this procedure for different values and using binary search (or one-sided binary search) we could find Kth largest sum in O(N log R) time, where N is size of the largest array and R is number of possible values between array1[N-1]+array2[N-1] and array1[0]+array2[0]. This algorithm has linear time complexity only when the array elements are integers bounded by small constant.
Previous algorithm may be improved if we stop binary search as soon as number of pair sums in binary search range decreases from O(N2) to O(N). Then we fill auxiliary array with these pair sums (this may be done with slightly modified two-pointers algorithm). And then we use quickselect algorithm to find Kth largest sum in this auxiliary array. All this does not improve worst-case complexity because we still need O(log R) binary search steps. What if we keep the quickselect part of this algorithm but (to get proper value range) we use something better than binary search?
We could estimate value range with the following trick: get every second element from each array and try to find the pair sum with rank k/4 for these half-arrays (using the same algorithm recursively). Obviously this should give some approximation for needed value range. And in fact slightly improved variant of this trick gives range containing only O(N) elements. This is proven in following paper: "Selection in X + Y and matrices with sorted rows and columns" by A. Mirzaian and E. Arjomandi. This paper contains detailed explanation of the algorithm, proof, complexity analysis, and pseudo-code for all parts of the algorithm except Quickselect. If linear worst-case complexity is required, Quickselect may be augmented with Median of medians algorithm.
This algorithm has complexity O(N). If one of the arrays is shorter than other array (M < N) we could assume that this shorter array is extended to size N with some very small elements so that all calculations in the algorithm use size of the largest array. We don't actually need to extract pairs with these "added" elements and feed them to quickselect, which makes algorithm a little bit faster but does not improve asymptotic complexity.
If k < N we could ignore all the array elements with index greater than k. In this case complexity is equal to O(k). If N < k < N(N-1) we just have better complexity than requested in OP. If k > N(N-1), we'd better solve the opposite problem: k'th smallest sum.
I uploaded simple C++11 implementation to ideone. Code is not optimized and not thoroughly tested. I tried to make it as close as possible to pseudo-code in linked paper. This implementation uses std::nth_element, which allows linear complexity only on average (not worst-case).
A completely different approach to find K'th sum in linear time is based on priority queue (PQ). One variation is to insert largest pair to PQ, then repeatedly remove top of PQ and instead insert up to two pairs (one with decremented index in one array, other with decremented index in other array). And take some measures to prevent inserting duplicate pairs. Other variation is to insert all possible pairs containing largest element of first array, then repeatedly remove top of PQ and instead insert pair with decremented index in first array and same index in second array. In this case there is no need to bother about duplicates.
OP mentions O(K log K) solution where PQ is implemented as max-heap. But in some cases (when array elements are evenly distributed integers with limited range and linear complexity is needed only on average, not worst-case) we could use O(1) time priority queue, for example, as described in this paper: "A Complexity O(1) Priority Queue for Event Driven Molecular Dynamics Simulations" by Gerald Paul. This allows O(K) expected time complexity.
Advantage of this approach is a possibility to provide first K elements in sorted order. Disadvantages are limited choice of array element type, more complex and slower algorithm, worse asymptotic complexity: O(K) > O(N).
EDIT: This does not work. I leave the answer, since apparently I am not the only one who could have this kind of idea; see the discussion below.
A counter-example is x = (2, 3, 6), y = (1, 4, 5) and k=3, where the algorithm gives 7 (3+4) instead of 8 (3+5).
Let x and y be the two arrays, sorted in decreasing order; we want to construct the K-th largest sum.
The variables are: i the index in the first array (element x[i]), j the index in the second array (element y[j]), and k the "order" of the sum (k in 1..K), in the sense that S(k)=x[i]+y[j] will be the k-th greater sum satisfying your conditions (this is the loop invariant).
Start from (i, j) equal to (0, 0): clearly, S(1) = x[0]+y[0].
for k from 1 to K-1, do:
if x[i+1]+ y[j] > x[i] + y[j+1], then i := i+1 (and j does not change) ; else j:=j+1
To see that it works, consider you have S(k) = x[i] + y[j]. Then, S(k+1) is the greatest sum which is lower (or equal) to S(k), and such as at least one element (i or j) changes. It is not difficult to see that exactly one of i or j should change.
If i changes, the greater sum you can construct which is lower than S(k) is by setting i=i+1, because x is decreasing and all the x[i'] + y[j] with i' < i are greater than S(k). The same holds for j, showing that S(k+1) is either x[i+1] + y[j] or x[i] + y[j+1].
Therefore, at the end of the loop you found the K-th greater sum.
tl;dr: If you look ahead and look behind at each iteration, you can start with the end (which is highest) and work back in O(K) time.
Although the insight underlying this approach is, I believe, sound, the code below is not quite correct at present (see comments).
Let's see: first of all, the arrays are sorted. So, if the arrays are a and b with lengths M and N, and as you have arranged them, the largest items are in slots M and N respectively, the largest pair will always be a[M]+b[N].
Now, what's the second largest pair? It's going to have perhaps one of {a[M],b[N]} (it can't have both, because that's just the largest pair again), and at least one of {a[M-1],b[N-1]}. BUT, we also know that if we choose a[M-1]+b[N-1], we can make one of the operands larger by choosing the higher number from the same list, so it will have exactly one number from the last column, and one from the penultimate column.
Consider the following two arrays: a = [1, 2, 53]; b = [66, 67, 68]. Our highest pair is 53+68. If we lose the smaller of those two, our pair is 68+2; if we lose the larger, it's 53+67. So, we have to look ahead to decide what our next pair will be. The simplest lookahead strategy is simply to calculate the sum of both possible pairs. That will always cost two additions, and two comparisons for each transition (three because we need to deal with the case where the sums are equal);let's call that cost Q).
At first, I was tempted to repeat that K-1 times. BUT there's a hitch: the next largest pair might actually be the other pair we can validly make from {{a[M],b[N]}, {a[M-1],b[N-1]}. So, we also need to look behind.
So, let's code (python, should be 2/3 compatible):
def kth(a,b,k):
M = len(a)
N = len(b)
if k > M*N:
raise ValueError("There are only %s possible pairs; you asked for the %sth largest, which is impossible" % M*N,k)
(ia,ib) = M-1,N-1 #0 based arrays
# we need this for lookback
nottakenindices = (0,0) # could be any value
nottakensum = float('-inf')
for i in range(k-1):
optionone = a[ia]+b[ib-1]
optiontwo = a[ia-1]+b[ib]
biggest = max((optionone,optiontwo))
#first deal with look behind
if nottakensum > biggest:
if optionone == biggest:
newnottakenindices = (ia,ib-1)
else: newnottakenindices = (ia-1,ib)
ia,ib = nottakenindices
nottakensum = biggest
nottakenindices = newnottakenindices
#deal with case where indices hit 0
elif ia <= 0 and ib <= 0:
ia = ib = 0
elif ia <= 0:
ib-=1
ia = 0
nottakensum = float('-inf')
elif ib <= 0:
ia-=1
ib = 0
nottakensum = float('-inf')
#lookahead cases
elif optionone > optiontwo:
#then choose the first option as our next pair
nottakensum,nottakenindices = optiontwo,(ia-1,ib)
ib-=1
elif optionone < optiontwo: # choose the second
nottakensum,nottakenindices = optionone,(ia,ib-1)
ia-=1
#next two cases apply if options are equal
elif a[ia] > b[ib]:# drop the smallest
nottakensum,nottakenindices = optiontwo,(ia-1,ib)
ib-=1
else: # might be equal or not - we can choose arbitrarily if equal
nottakensum,nottakenindices = optionone,(ia,ib-1)
ia-=1
#+2 - one for zero-based, one for skipping the 1st largest
data = (i+2,a[ia],b[ib],a[ia]+b[ib],ia,ib)
narrative = "%sth largest pair is %s+%s=%s, with indices (%s,%s)" % data
print (narrative) #this will work in both versions of python
if ia <= 0 and ib <= 0:
raise ValueError("Both arrays exhausted before Kth (%sth) pair reached"%data[0])
return data, narrative
For those without python, here's an ideone: http://ideone.com/tfm2MA
At worst, we have 5 comparisons in each iteration, and K-1 iterations, which means that this is an O(K) algorithm.
Now, it might be possible to exploit information about differences between values to optimise this a little bit, but this accomplishes the goal.
Here's a reference implementation (not O(K), but will always work, unless there's a corner case with cases where pairs have equal sums):
import itertools
def refkth(a,b,k):
(rightia,righta),(rightib,rightb) = sorted(itertools.product(enumerate(a),enumerate(b)), key=lamba((ia,ea),(ib,eb):ea+eb)[k-1]
data = k,righta,rightb,righta+rightb,rightia,rightib
narrative = "%sth largest pair is %s+%s=%s, with indices (%s,%s)" % data
print (narrative) #this will work in both versions of python
return data, narrative
This calculates the cartesian product of the two arrays (i.e. all possible pairs), sorts them by sum, and takes the kth element. The enumerate function decorates each item with its index.
The max-heap algorithm in the other question is simple, fast and correct. Don't knock it. It's really well explained too. https://stackoverflow.com/a/5212618/284795
Might be there isn't any O(k) algorithm. That's okay, O(k log k) is almost as fast.
If the last two solutions were at (a1, b1), (a2, b2), then it seems to me there are only four candidate solutions (a1-1, b1) (a1, b1-1) (a2-1, b2) (a2, b2-1). This intuition could be wrong. Surely there are at most four candidates for each coordinate, and the next highest is among the 16 pairs (a in {a1,a2,a1-1,a2-1}, b in {b1,b2,b1-1,b2-1}). That's O(k).
(No it's not, still not sure whether that's possible.)
[2, 3, 5, 8, 13]
[4, 8, 12, 16]
Merge the 2 arrays and note down the indexes in the sorted array. Here is the index array looks like (starting from 1 not 0)
[1, 2, 4, 6, 8]
[3, 5, 7, 9]
Now start from end and make tuples. sum the elements in the tuple and pick the kth largest sum.
public static List<List<Integer>> optimization(int[] nums1, int[] nums2, int k) {
// 2 * O(n log(n))
Arrays.sort(nums1);
Arrays.sort(nums2);
List<List<Integer>> results = new ArrayList<>(k);
int endIndex = 0;
// Find the number whose square is the first one bigger than k
for (int i = 1; i <= k; i++) {
if (i * i >= k) {
endIndex = i;
break;
}
}
// The following Iteration provides at most endIndex^2 elements, and both arrays are in ascending order,
// so k smallest pairs must can be found in this iteration. To flatten the nested loop, refer
// 'https://stackoverflow.com/questions/7457879/algorithm-to-optimize-nested-loops'
for (int i = 0; i < endIndex * endIndex; i++) {
int m = i / endIndex;
int n = i % endIndex;
List<Integer> item = new ArrayList<>(2);
item.add(nums1[m]);
item.add(nums2[n]);
results.add(item);
}
results.sort(Comparator.comparing(pair->pair.get(0) + pair.get(1)));
return results.stream().limit(k).collect(Collectors.toList());
}
Key to eliminate O(n^2):
Avoid cartesian product(or 'cross join' like operation) of both arrays, which means flattening the nested loop.
Downsize iteration over the 2 arrays.
So:
Sort both arrays (Arrays.sort offers O(n log(n)) performance according to Java doc)
Limit the iteration range to the size which is just big enough to support k smallest pairs searching.