Save my name, email, and website in this browser for the next time I comment. So the subarrays become. 1. two sorted array and find k'th smallest number. Ukkonen's suffix tree algorithm in plain English. rev2023.7.24.43543. By the end of merging all lists, we should have one final sorted linked list.The time complexity of the solution will be big O(N * log(K)) where N is the number of nodes we must merge together and K is the number of recursive calls we make to divide the arrays. The dumping elements in A and B are all smaller than this element in A+B. If elements are leftover in either the first or second array, we finally copy those elements in the new merged array. Can a simply connected manifold satisfy ? K-th smallest element of two sorted arrays - OpenGenus IQ 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, Below C# code to Find the k-th Smallest Element in the Union of Two Sorted Arrays. Steps: Firstly we take the given arrays and form a new merged array. if a>b, ia_right = ia_right (ia_right ia_left)/2, ib_right = ib_left + (ib_right ib_left)/2 Recursively compare medians a and b of each array. In the end, we'll present a comparison between both approaches and when to use them. Array A[m], B[n]. Write a pseudocode/describe your strategy for a function kthelement (Arr1, Arr2, k) that uses the concepts mentioned in the divide and conquer technique. Great explanation. If aFind kthsmallest element from two sorted arrays So let's focus just on the the first k elements of array A. You can add up 0.5+0.25+0.125, and you will never go beyond 1. What its like to be on the Python Steering Council (Ep. The runtime for this algorithm is as follows. By clicking Post Your Answer, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct. In case of small array one or 2 binary searches in larger array is enough to find needed element. or slowly? The idea of Merge Sort is to break up the array into two subarrays of halfthe size (1), recursively sort each of them, and then merge the sorted subarrays to create theentire sorted list. PDF Lecture 4 Divide and Conquer while it's good but it's harder to implement as there are a lot of edge cases that we need to take care of. Keep track of count while comparing elements of two, [pastacode lang=c manual=%2F%2F%20A%20Simple%20Merge%20based%20O(n)%20solution%20to%20find%20median%20of%0A%2F%2F%20two%20sorted%20arrays%0A%23include%20%3Cstdio.h%3E%0A%20%0A%2F*%20This%20function%20returns%20median%20of%20ar1%5B%5D%20and%20ar2%5B%5D.%0A%20%20%20Assumptions%20in%20this%20function%3A%0A%20%20%20Both%20ar1%5B%5D%20and%20ar2%5B%5D%20are%20sorted%20arrays%0A%20%20%20Both%20have%20n%20elements%20*%2F%0Aint%20getMedian(int%20ar1%5B%5D%2C%20int%20ar2%5B%5D%2C%20int%20n)%0A%7B%0A%20%20%20%20int%20i%20%3D%200%3B%20%20%2F*%20Current%20index%20of%20i%2Fp%20array%20ar1%5B%5D%20*%2F%0A%20%20%20%20int%20j%20%3D%200%3B%20%2F*%20Current%20index%20of%20i%2Fp%20array%20ar2%5B%5D%20*%2F%0A%20%20%20%20int%20count%3B%0A%20%20%20%20int%20m1%20%3D%20-1%2C%20m2%20%3D%20-1%3B%0A%20%0A%20%20%20%20%2F*%20Since%20there%20are%202n%20elements%2C%20median%20will%20be%20average%0A%20%20%20%20%20of%20elements%20at%20index%20n-1%20and%20n%20in%20the%20array%20obtained%20after%0A%20%20%20%20%20merging%20ar1%20and%20ar2%20*%2F%0A%20%20%20%20for%20(count%20%3D%200%3B%20count%20%3C%3D%20n%3B%20count%2B%2B)%0A%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20%2F*Below%20is%20to%20handle%20case%20where%20all%20elements%20of%20ar1%5B%5D%20are%0A%20%20%20%20%20%20%20%20%20%20smaller%20than%20smallest(or%20first)%20element%20of%20ar2%5B%5D*%2F%0A%20%20%20%20%20%20%20%20if%20(i%20%3D%3D%20n)%0A%20%20%20%20%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20%20%20%20%20m1%20%3D%20m2%3B%0A%20%20%20%20%20%20%20%20%20%20%20%20m2%20%3D%20ar2%5B0%5D%3B%0A%20%20%20%20%20%20%20%20%20%20%20%20break%3B%0A%20%20%20%20%20%20%20%20%7D%0A%20%0A%20%20%20%20%20%20%20%20%2F*Below%20is%20to%20handle%20case%20where%20all%20elements%20of%20ar2%5B%5D%20are%0A%20%20%20%20%20%20%20%20%20%20smaller%20than%20smallest(or%20first)%20element%20of%20ar1%5B%5D*%2F%0A%20%20%20%20%20%20%20%20else%20if%20(j%20%3D%3D%20n)%0A%20%20%20%20%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20%20%20%20%20m1%20%3D%20m2%3B%0A%20%20%20%20%20%20%20%20%20%20%20%20m2%20%3D%20ar1%5B0%5D%3B%0A%20%20%20%20%20%20%20%20%20%20%20%20break%3B%0A%20%20%20%20%20%20%20%20%7D%0A%20%0A%20%20%20%20%20%20%20%20if%20(ar1%5Bi%5D%20%3C%20ar2%5Bj%5D)%0A%20%20%20%20%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20%20%20%20%20m1%20%3D%20m2%3B%20%20%2F*%20Store%20the%20prev%20median%20*%2F%0A%20%20%20%20%20%20%20%20%20%20%20%20m2%20%3D%20ar1%5Bi%5D%3B%0A%20%20%20%20%20%20%20%20%20%20%20%20i%2B%2B%3B%0A%20%20%20%20%20%20%20%20%7D%0A%20%20%20%20%20%20%20%20else%0A%20%20%20%20%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20%20%20%20%20m1%20%3D%20m2%3B%20%20%2F*%20Store%20the%20prev%20median%20*%2F%0A%20%20%20%20%20%20%20%20%20%20%20%20m2%20%3D%20ar2%5Bj%5D%3B%0A%20%20%20%20%20%20%20%20%20%20%20%20j%2B%2B%3B%0A%20%20%20%20%20%20%20%20%7D%0A%20%20%20%20%7D%0A%20%0A%20%20%20%20return%20(m1%20%2B%20m2)%2F2%3B%0A%7D%0A%20%0A%2F*%20Driver%20program%20to%20test%20above%20function%20*%2F%0Aint%20main()%0A%7B%0A%20%20%20%20int%20ar1%5B%5D%20%3D%20%7B1%2C%2012%2C%2015%2C%2026%2C%2038%7D%3B%0A%20%20%20%20int%20ar2%5B%5D%20%3D%20%7B2%2C%2013%2C%2017%2C%2030%2C%2045%7D%3B%0A%20%0A%20%20%20%20int%20n1%20%3D%20sizeof(ar1)%2Fsizeof(ar1%5B0%5D)%3B%0A%20%20%20%20int%20n2%20%3D%20sizeof(ar2)%2Fsizeof(ar2%5B0%5D)%3B%0A%20%20%20%20if%20(n1%20%3D%3D%20n2)%0A%20%20%20%20%20%20%20%20printf(%22Median%20is%20%25d%22%2C%20getMedian(ar1%2C%20ar2%2C%20n1))%3B%0A%20%20%20%20else%0A%20%20%20%20%20%20%20%20printf(%22Doesnt%20work%20for%20arrays%20of%20unequal%20size%22)%3B%0A%20%20%20%20getchar()%3B%0A%20%20%20%20return%200%3B%0A%7D message= highlight= provider=manual/]. I have correct code after this one: The most powerful idea is that in each loop, we always use the base case approach. How can I animate a list of vectors, which have entries either 1 or 0? You can't have an algorithm that doesn't check at least n elements, even verifying a solution would require checking that many. 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. 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. kth element of two sorted arrays | My Blog Why does CNN's gravity hole in the Indian Ocean dip the sea level instead of raising it? We know for a fact that the kth smallest value can't appear in the array in array A after position k (assuming all the elements are distinct). 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. Since a is the median of A, a is also the median of b. How to find the $k$th smallest item among the union of $C$ disjoint, sorted arrays? here are two arrays. (Bathroom Shower Ceiling). The K will recursively change from k => k/2 => k/4 => till it reaches 1. Dumping smaller parts of A and B and cut both sizes to m+n-k+1. Cold water swimming - go in quickly? Find centralized, trusted content and collaborate around the technologies you use most. PDF Divide and Conquer - UW Computer Sciences User Pages Am I in trouble? Do I have a misconception about probability? Take the smaller one, say A1[0] away into our pocket. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. The index of the array will go out of boundary. 2) if a == b, return a. But the size of the problem here should surely mean the size of the arrays, so an algorithm that only checks n elements is sublinear. The (mid1 - L1) + (mid2 - L2) + 1 is actually meaning there should be two pointers in two arrays: one is the kth and another one is the one which makes the recursion end. 1 Given two sorted array of size M and N. I was trying to implement an algorithm with time complexity O (logM+logN). If we add up the position of the elements in A and B, we have the total number of elements in the two arrays smaller than the current element. Could we miss the base case (k-1 == index1+index2)? (Bathroom Shower Ceiling). For the demonstration you can use the loop invariant i + j = k, but I won't do all your homework :). Thus no dumping element exist between front and the kth element in A+B or A+B. Find the Kth Smallest Element in Two Sorted Arrays in Java I believe that you can solve this problem using a variant on binary search. Wikitechy Founder, Author, International Speaker, and Job Consultant. Shortest Paths C/C++ Dijkstras shortest path algorithm Given a graph and a source vertex in graph, find shortest paths from source to all vertices, Java programming-Median of two sorted arrays-Divide and Conquer-There are 2 sorted arrays A and B of size n each. Cold water swimming - go in quickly? (log n)^2 isn't little-o of log n - it grows asymptotically faster - but it is little-o of n. -1 Binary search doesn't read every element. 1. Given an integer array nums and an integer k, return the k th largest element in the array. If this is less than k, then we recurse in the upper half of the first k elements of A, and if this is greater than k we recurse in the lower half of the first elements of k, etc. Implement an algorithm using divide and conquer technique: Given two sorted arrays of size m and n respectively, find the element that would be at the kth position in combined sorted array. TheMergesortalgorithm is a divide-and-conquer algorithm for sorting an array of comparableelements. Why do capacitors have less energy density than batteries? I think that the answer is yes, using this algorithm. Since size of the set for which we are looking for median is even (2n), we take average of middle two numbers in all below solutions and return floor of the average. A question on Demailly's proof to the cannonical isomorphism of tangent bundle of Grassmannian. Was the release of "Barbie" intentionally coordinated to be on the same day as "Oppenheimer"? That is, can we determine the values in these arrays without necessarily constructing each element? This solution involves the knowledge of the merge sort algorithm where we must divide our array recursively and merge the lists to sort them at each recursive call. find kth smallest element in two sorted arrays, median of two sorted arrays binary search, median of two sorted arrays of different sizes, merge two sorted arrays into one sorted array in java, write a program to merge two sorted arrays, Shortest Paths C/C++ Dijkstras shortest path algorithm, Java programming Median of two sorted arrays, The median of a finite list of numbers can be found by arranging all the. Can you solve it without sorting? (LogOut/ 1. Given 2 sorted arrays of integers, find the nth largest number in sublinear time. Can I spin 3753 Cruithne and keep it spinning? That sounded interesting enough I had to try it out: Just a small addition, you don't have to search through the entire B array either, once the binary search takes you strictly above or under index k/2, you already know the answer to whether the sum is greater than k. Your last line should be: O(lg k lg n) = O(lg2 n) = o(logn), which is sublinear. Finding the Kth Smallest Element in the Union of Two Sorted Arrays For the second step, we can't go more than k/4 elements After each step, we get much closer to k-th element. If not, each turn throw half of the array based on the comparison result of the medians. Suppose 3 4 b a <. 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. Following is the complete algorithm. Cheers ;). By the following implementation, we have the complexity of O(log(min(n,m)) with O(1) space complexity. Do a binary move in corresponding B places. Median is the middle element of the sorted array i.e. Required fields are marked *. I have another solution. 1) Let ia_left = 0, ia_right = k-1; ib_left = 0, ib_right = k-1, ..// other is same as 1 Draw the recursive Q1) Using Divide and conquer approach, solve the kth element in 2 sorted arrays problem. Start at position k/2; this is the k/2th smallest element in array A. #100daysofcodewithGFGSubmit your solutions here-: https://practice.geeksforgeeks.org/problems/k-th-element-of-two-sorted-array1317/1Free resources that can n. Let us repeat the process for above two subarrays: m1 is greater than m2. ..b) Insert next element from the array from which the element is extracted. The k/2'th element of what remains is guaranteed to be bigger than the bottom half of A, so it's guaranteed to be the k'th element of the original. The kth smallest number in two arrays, one sorted the other unsorted. now its working one. I'll get back to you if this works. Keep track of count while comparing elements of two arrays. Besides that most of the implementations are recursive which adds the space complexity of recursion stack. So 0 <= x <= len (A) and 0<= k - x <= len (B) --> max {0, k - len (B)} <= x <= min {len (A), k} It sounds like using some divide-and-conquer algorithms but I'm not sure. @user2612743: Might be an indexing-from-0 vs indexing-from-1 issue. Unfortunately approach I is incorrect starting from step 2. use routine of finding the median of two sorted arrays. Here is a tail recursive solution that will take log(len(a)+len(b)) time. . Median of two sorted arrays of same size - GeeksforGeeks Vice versa. Divide-and-conqueralgorithms Chapter2 Divide-and-conqueralgorithms Thedivide-and-conquerstrategysolvesaproblemby: Breakingitintoproblem subproblemsthatarethemselvessmallerinstancesofthesametypeof Recursivelysolvingthesesubproblems Appropriatelycombiningtheiranswers 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. Assumption: The inputs are correct, i.e., k is in the range [0,len(a)+len(b)]. if we stuck in the same position it means that we found the solution and we will return the max of values in the position of sizeA-1 from a and sizeB-1 from b. Find kth smallest element from two sorted arrays In the reduction steps, it is important to get rid of a number of elements in one of the arrays proportional to its length in order to make the run-time logarithmic. What is a time efficient algorithm to find the k th smallest element in the union of both arrays? I was asked this question in an interview. This is a question one of my friends told me he was asked while interviewing, I've been thinking about a solution. Closest Pair of Points Divide and Conquer We are given an array of n points in the plane, and the problem is to find out the closes together. For the above ar1[] and ar2[], m1 is smaller than m2. Implement an algorithm using divide and conquer technique: Given two sorted arrays of size m and n respectively, find the element that would be at the kth position in combined sorted array. We continue this finding, comparing, dumping until we have a==b. I have 2 sorted arrays of integers, how do I find the kth biggest item in O(logn) time? I. Throw away the bottom half of A and the top half of B, then recursively find the k/2'th element of what remains. What its like to be on the Python Steering Council (Ep. To find such i,j the algorithm does a dichotomic search on the arrays. .. If so, thentheais sorted in place by making at most one swap. Let b [k/2] > a [k/2]. I think this is two concurrent binary searches on the subarrays A[0..n-1] and B[0..n-1], which is O(log n). It's astonishing how many wrong answers are circulating over the internet regarding this problem. Connect and share knowledge within a single location that is structured and easy to search. Take the average of the elements at indexes n-1 and n in the merged array. Since a is the median of 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. Then use routine 1 to find the median of A + B. Is this mold/mildew? To give a robust definition: the kth order statistic is the element that would appear at position k if the data were sorted. The following code is in JAVA. 708. You should be removing the left half of A and the right half of B, rather than the other way around. Then, compare the k/2'th elements of each array. This videos goes through the O(log(n) + log(m)) time algorithm to solve the \"Kth element of two sorted arrays\" question, and applies the algorithm to solve the \"Median of two sorted arrays\" problem.0:00 Problem description1:09 Conceptual overview of log time solution14:41 Solution implementation23:00 Application of the solution to find the median of two sorted arraysProblem links:\"Kth element of two sorted arrays\": https://practice.geeksforgeeks.org/problems/k-th-element-of-two-sorted-array1317/1\"Median of two sorted arrays\": https://leetcode.com/problems/median-of-two-sorted-arrays/ Otherwise, repeat this process on array B. By confidently eliminating the half we know for sure is not going to have the kth element. By observing the above code base case k == 1, k == size1+size2, and combine with that A1 and A2 can't both be empty. 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. What are the pitfalls of indirect implicit casting? The time complexity is O(log(m+n-k+1)) < O(log(2k-k)) = O(log k). They say it takes O (logN + logM) where N and M are the arrays lengths. PDF Part Four - Stanford University To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Input: The idea is to not just commit to one element in each loop; the first step contains k/2 elements. Using robocopy on windows led to infinite subfolder duplication via a stray shortcut file. How can I avoid this? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. For deciding which half of which array to eliminate, we take advantage of the fact that both arrays are sorted, so if k is larger than sum of half-lengths, we can eliminate first half of the array whose middle element is the smaller of the two middle elements. then we cannot find kth smallest element in union of both sorted arrays ryt So return Invalid data. Entries (RSS) and Comments (RSS). sizeB is the same definition except we calculate the value such a way that sizeA+sizeB=k. The idea behind mergesort is totake a list, divideit into two smaller sublists, conquereach sublist by sorting it, and thencombinethe two solutionsfor the subproblems into a single solution. Here's my code based on Jules Olleon's solution: Here is my implementation in C, you can refer to @Jules Ollon '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). Your email address will not be published. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. I think what you probably want to do is just a standard binary search over both ranges starting in the middle of each and maintain an a_low, a_high, and a_mid value (plus one of each for b). So start solving today and subscribe to our channel for the free coding lessons.Getting bored just watching? Google | Phone Screen | Kth Largest Element of Two Sorted Arrays Medium. Describe an algorithm to find the kth smallest element in A B C in O (log n) time using divide and conquer (not heap) Expert Answer Solution: Using the divide and conquer approach, below solution time complexity is: O (log n) Check whether kth element lies in first half of either of the 2 arrays. Merge Two Lists Video: https://www.youtube.com/watch?v=c86I16V8P58 Join the community Discord: https://discord.gg/aVWsAaaCtT Support me on Patreon: htt. 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). This shouldn't work. Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer. Using robocopy on windows led to infinite subfolder duplication via a stray shortcut file. How can I avoid this? Sublinear time implies logarithmic to me, so perhaps some kind of divide and conquer method. If it is not greater than A[0], it is the result, else dump kth element in B. (A modification to) Jon Prez Laraudogoitas "Beautiful Supertask" What assumptions of Noether's theorem fail? Obviously we can ignore all a[i] and b[i] where i > k. Order Statistics Given a collection of data, the kth order statistic is the kth smallest value in the data set. Not the answer you're looking for? 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). To learn more, see our tips on writing great answers. Of course not, but to select the nth element from a list of any size, you have to read at least n elements. 3. kth element of two arrays, m<=n ?.We need to decide. 3. (LogOut/ Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. This means that the total time for this search is O(lg k lg n). Solution pseudocode 2) Sort the above created n/5 groups and find median of all groups. Instead of counting from the smallest to largest, we think from larger side. But I'll give you +1 for an interesting question nonetheless. Recursively compare medians a and b of each array. The following O(k) code gives you one element before the correct answer. 14.5K. there is no bug, i have tested my code before i posted to SO, Thanks sammy333, I have updated the code. Methinks that you don't want to scale up by 3/2. 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. Opposite if k is smaller. Therefore we can discard also all b[i], where i > k/2. So the median is the kth element of original two arrays. A car dealership sent a 8300 form after I paid $10k in cash for a car. Then we can refer to the simple and powerful base case again. ))) (max((log in time runs algorithm best The sequences. Median. The kth element in A+B is the same as the kth element in A+B. The time complexity is O(log m) . The space complexity will be big O(log(K)) where K is the number of recursive calls made that fill up our stack space. Merge Two Lists Video: https://www.youtube.com/watch?v=c86I16V8P58 Join the community Discord: https://discord.gg/aVWsAaaCtT Support me on Patreon: https://www.patreon.com/michaelmuinosFollow me on LinkedIn: https://www.linkedin.com/in/michael-muinosFollow me on Github: https://github.com/MichaelMuinosCheck out my interview prep platform for learning the patterns! Interview Prep Platform: https://algoswithmichael.comIn this video I go over the popular Leetcode problem Merge K Sorted Lists involving the divide and conquer approach. The first pseudo code provided above, does not work for many values. 593), Stack Overflow at WeAreDevelopers World Congress in Berlin, Temporary policy: Generative AI (e.g., ChatGPT) is banned. Who counts as pupils or as a student in Germany? Level up your coding skills and quickly land a job. How might I find the largest number contained in a JavaScript array? As we know, all elements in the array are distinct. A very concise approach can be found here too with each time adding k/2, k/4, k/8 to approach k: http://stackoverflow.com/questions/4607945/how-to-find-the-kth-smallest-element-in-the-union-of-two-sorted-arrays. 0. k'th highest of two sorted arrays. minimalistic ext4 filesystem without journal and other advanced features, Best estimator of the mean of a normal distribution based only on box-plot statistics. Let us see different methods to get the median of two sorted arrays of size n each. The based on the values on those two borders with conclude that we have to extend to the right side in array a or shrink to the left side. The intuition behind this algorithm is as follows. 4. D&C.pdf - Discussion Divide-and-Conquer 1. K-th Element of Two A1 and A2 are two sorted ascending arrays, with size1 and size2 as length respectively. 1. K-th Element of Two Sorted Arrays - GeeksforGeeks (Here we are getting rid of half). The search of array A does a binary search over k elements, which takes O(lg k) iterations. Sublinear time implies logarithmic to me, so perhaps some kind of divide and conquer method. I'm a frequent speaker at tech conferences and events. K-th Element of two sorted arrays - Arrays - Tutorial - takeuforward Find order statistic in union of 2 sorted lists on logarithmic time, How to find the kth largest number in two sorted arrays of different sizes, Finding kth smallest number from n sorted arrays, An algorithm to find the nth largest number in two arrays of size n, Find kthsmallest element from two sorted arrays, Efficient algorithm to compute the median of pariwise absolute sums of a sorted array, Finding the kth-smallest element in union of sorted arrays, Finding kth smallest element in union of 2 sorted array, k-th Smallest Element in the Union of Two Sorted Arrays, kth smallest element in two sorted array - O(log n) Solution.
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