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This is a homework question. They say it takes
O(logN + logM)
where N
and M
are the arrays lengths.Let's name the arrays
First let's compare
a
and b
. Obviously we can ignore all a[i]
and b[i]
where i > k.First let's compare
a[k/2]
and b[k/2]
. Let b[k/2]
>a[k/2]
. Therefore we can discard also all b[i]
, where i > k/2.Now we have all
a[i]
, where i < k and all b[i]
, where i < k/2 to find the answer.What is the next step?
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MichaelMichael
16 Answers
You've got it, just keep going! And be careful with the indexes..
To simplify a bit I'll assume that N and M are > k, so the complexity here is O(log k), which is O(log N + log M).
Pseudo-code:
For the demonstration you can use the loop invariant i + j = k, but I won't do all your homework :)
Jules OlléonJules Olléon
I hope I am not answering your homework as it has been over a year since this question was asked. Here is a tail recursive solution that will take log(len(a)+len(b)) time.
Assumption: The inputs are right. i.e. k is in the range [0, len(a)+len(b)]
Base cases:
- If length of one of the arrays is 0, the answer is kth element of the second array.
Reduction steps:
- If mid index of
a
+ mid index ofb
is less thank
- If mid element of
a
is greater than mid element ofb
, we can ignore the first half ofb
, adjustk
. - else ignore the first half of
a
, adjustk
.
- If mid element of
- Else if
k
is less than sum of mid indices ofa
andb
:- If mid element of
a
is greater than mid element ofb
, we can safely ignore second half ofa
- else we can ignore second half of
b
- If mid element of
Code:
Please note that my solution is creating new copies of smaller arrays in every call, this can be easily eliminated by only passing start and end indices on the original arrays.
lambdapilgrimlambdapilgrim
Many people answered this 'kth smallest element from two sorted array' question, but usually with only general ideas, not a clear working code or boundary conditions analysis.
Here I'd like to elaborate it carefully with the way I went though to help some novices to understand, with my correct working Java code.
A1
and A2
are two sorted ascending arrays, with size1
and size2
as length respectively. We need to find the k-th smallest element from the union of those two arrays. Here we reasonably assume that (k > 0 && k <= size1 + size2)
, which implies that A1
and A2
can't be both empty.First, let's approach this question with a slow O(k) algorithm. The method is to compare the first element of both array,
A1[0]
and A2[0]
. Take the smaller one, say A1[0]
away into our pocket. Then compare A1[1]
with A2[0]
, and so on. Repeat this action until our pocket reached k
elements. Very important: In the first step, we can only commit to A1[0]
in our pocket. We can NOT include or exclude A2[0]
!!!The following O(k) code gives you one element before the correct answer. Here I use it to show my idea, and analysis boundary condition. I have correct code after this one:
The most powerful idea is that in each loop, we always use the base case approach. After committed to the current smallest element, we get one step closer to the target: the k-th smallest element. Never jump into the middle and make yourself confused and lost!
By observing the above code base case
k 1, k size1+size2
, and combine with that A1
and A2
can't both be empty. We can turn the logic into below more concise style.Here is a slow but correct working code:
Now we can try a faster algorithm runs at O(log k). Similarly, compare
A1[k/2]
with A2[k/2]
; if A1[k/2]
is smaller, then all the elements from A1[0]
to A1[k/2]
should be in our pocket. The idea is to not just commit to one element in each loop; the first step contains k/2
elements. Again, we can NOT include or exclude A2[0]
to A2[k/2]
anyway. So in the first step, we can't go more than k/2
elements. For the second step, we can't go more than k/4
elements..After each step, we get much closer to k-th element. At the same time each step get smaller and smaller, until we reach
(step 1)
, which is (k-1 index1+index2)
. Then we can refer to the simple and powerful base case again.Here is the working correct code:
Some people may worry what if
(index1+index2)
jump over k-1? Could we miss the base case (k-1 index1+index2)
? That's impossible. You can add up 0.5+0.25+0.125.., and you will never go beyond 1.Of course, it is very easy to turn the above code into recursive algorithm:
Hope the above analysis and Java code could help you to understand. But never copy my code as your homework! Cheers ;)
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FeiFei
Here's a C++ iterative version of @lambdapilgrim's solution (see the explanation of the algorithm there):
It works for all
0 <= n < (size(a) + size(b))
indexes and has O(log(size(a)) + log(size(b)))
complexity.Example
Community♦
jfsjfs
My attempt for first k numbers, kth number in 2 sorted arrays, and in n sorted arrays:
The complete code with debug utils can be found at: https://github.com/brainclone/teasers/tree/master/kth
Qichao DongQichao Dong
superbsuperb
Here is my implementation in C, you can refer to @Jules Olléon 's explains for the algorithm: the idea behind the algorithm is that we maintain i + j = k, and find such i and j so that a[i-1] < b[j-1] < a[i] (or the other way round). Now since there are i elements in 'a' smaller than b[j-1], and j-1 elements in 'b' smaller than b[j-1], b[j-1] is the i + j-1 + 1 = kth smallest element. To find such i,j the algorithm does a dichotomic search on the arrays.
Zhiwen FangZhiwen Fang
Here's my solution. The C++ code prints the kth smallest value as well as the number of iterations to get the kth smallest value using a loop, which in my opinion is in the order of log(k). The code however requires k to be smaller than the length of the first array which is a limitation.
Karthikeyan SvKarthikeyan Sv
The first pseudo code provided above, does not work for many values. For example,here are two arrays. int[] a = { 1, 5, 6, 8, 9, 11, 15, 17, 19 }; int[] b = { 4, 7, 8, 13, 15, 18, 20, 24, 26 };
It did not work for k=3 and k=9 in it. I have another solution. It is given below.
But.. it is also not working for k=5. There is this even/odd catch of k which is not letting it to be simple.
sn.anuragsn.anurag
Hrishikesh MishraHrishikesh Mishra
Here is mine solution in java . Will try to further optimize it
This is inspired from Algo at wonderful youtube video
M SachM Sach
Link to code complexity (log(n)+log(m))
Link to Code (log(n)*log(m))
Implementation of (log(n)+log(m)) solution
Wilkos
I would like to add my explanation to the problem. This is a classic problem where we have to use the fact that the two arrays are sorted .we have been given two sorted arrays arr1 of size sz1 and arr2 of size sz2
a)Lets suppose if
Checking If k is valid
k is > (sz1+sz2)
then we cannot find kth smallest element in union of both sorted arrays ryt So return Invalid data.b)Now if above condition holds false and we have valid and feasible value of k,
Managing Edge Cases
We will append both the arrays by -infinity values at front and +infinity values at end to cover the edge cases of k = 1,2 and k = (sz1+sz2-1),(sz1+sz2)etc.
B N M Poles
![Life Life](/uploads/1/2/6/3/126358215/553622034.png)
Now both the arrays have size (sz1+2) and (sz2+2) respectively
Main Algorithm
Now,we will do binary search on arr1 .We will do binary search on arr1 looking for an index i , startIndex <= i <= endIndex
such that if we find corresponding index j in arr2 using constraint {(i+j) = k},then if
if (arr2[j-1] < arr1[i] < arr2[j]),then arr1[i] is the kth smallest (Case 1)
else if (arr1[i-1] < arr2[j] < arr1[i]) ,then arr2[i] is the kth smallest (Case 2)
else signifies either arr1[i] < arr2[j-1] < arr2[j] (Case3)
or arr2[j-1] < arr2[j] < arr1[i] (Case4)
Since we know that the kth smallest element has (k-1) elements smaller than it in union of both the arrays ryt? So,
In Case1, what we did , we ensured that there are a total of (k-1) smaller elements to arr1[i] because elements smaller than arr1[i] in arr1 array are i-1 in number than we know (arr2[j-1] < arr1[i] < arr2[j]) and number of elements smaller than arr1[i] in arr2 is j-1 because j is found using (i-1)+(j-1) = (k-1) So kth smallest element will be arr1[i]
But answer may not always come from the first array ie arr1 so we checked for case2 which also satisfies similarly like case 1 because (i-1)+(j-1) = (k-1) . Now if we have (arr1[i-1] < arr2[j] < arr1[i]) we have a total of k-1 elements smaller than arr2[j] in union of both the arrays so its the kth smallest element.
In case3 , to form it to any of case 1 or case 2, we need to increment i and j will be found according using constraint {(i+j) = k} ie in binary search move to right part ie make startIndex = middleIndex
In case4, to form it to any of case 1 or case 2, we need to decrement i and j will be found according using constraint {(i+j) = k} ie in binary search move to left part ie make endIndex = middleIndex.
Now how to decide startIndex and endIndex at beginning of binary search over arr1 with startindex = 1 and endIndex = ??.We need to decide.
If k > sz1,endIndex = (sz1+1) , else endIndex = k;
Because if k is greater than the size of the first array we may have to do binary search over the entire array arr1 else we only need to take first k elements of it because sz1-k elements can never contribute in calculating kth smallest.
CODE Shown Below
For Solution of complexity (log(n)*log(m))
Just i missed using advantage of the fact that for each i the j can be found using constraint {(i-1)+(j-1)=(k-1)} So for each i i was further applying binary search on second array to find j such that arr2[j] <= arr1[i].So this solution can be optimized further
Vinayak SangarVinayak Sangar
Basically, via this approach you can discard k/2 elements at each step.The K will recursively change from k => k/2 => k/4 => .. till it reaches 1.So, Time Complexity is O(logk)
At k=1 , we get the lowest of the two arrays.
The following code is in JAVA. Please note that the we are subtracting 1 (-1) in the code from the indices because Java array's index starts from 0 and not 1, eg. k=3 is represented by the element in 2nd index of an array.
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HeadAndTailHeadAndTail
Anantha KrishnanAnantha Krishnan
Below C# code to Find the k-th Smallest Element in the Union of Two Sorted Arrays. Time Complexity : O(logk)
Piyush PatelPiyush Patel