How to find pair with kth largest sum?

Question

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.

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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.

Answer 1

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).

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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).

Answer 2

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).

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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.

Answer 3

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).

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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.

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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.

Answer 4

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.)

Answer 5

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.

Answer 6

[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.

Answer 7

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):

1. Avoid cartesian product(or 'cross join' like operation) of both arrays, which means flattening the nested loop.

2. Downsize iteration over the 2 arrays.

So:

1. Sort both arrays (Arrays.sort offers O(n log(n)) performance according to Java doc)

2. Limit the iteration range to the size which is just big enough to support k smallest pairs searching.