![]() ![]() You understand the solution to the question.But most of the time, the company is less interested in the solution and more interested in how you arrived at it. This question is credited to PRAMP, with the heap constructor benefited from the priority queue REACTO problem from Fullstack Academy.When you're taking online assessments or doing live coding interviews, you might think that the only thing that matters is whether or not the test cases pass. The space complexity involved requires another array of length k + 1, or O(k) space complexity. Having to run through the array entirely requires O(n) time, and within each for-loop, inserting and deleting from a heap both take O(log(k+1)) time, or O(log(k)), resulting in an O(nlog(k)) time complexity. The time complexity taken here is O(nlog(k)) time. ![]() Solution Sort Functionįunction MinHeap ( ) Complexity To get the last elements out of the heap, the heap itself needs to be traversed to sort the remaining k + 1 elements into the result array. Repeat the process until you have traversed the entire array. Slide your sub-array one index over, add that new value to the heap, and again remove the new minimum to place back into the array. the node at the top of the heap) out and place that back into the original array. Once the heap is built, take the minimum (i.e. ![]() To apply this to the sliding window, we can create a minimum heap of size k + 1, and fill it with the first k + 1 elements of the array. In a minimum heap, the node parent is always smaller than its children, resulting in the smallest value being at the top of the heap. Heaps are a data structure that, like regular binary trees, will take in data and insert the node in the tree based on comparisons to the other node’s values. Binary HeapĪ data structure that takes advantage of this insertion method is the minimum heap. In order to assume that this window will always be able to sort the smallest value, insertion into this array also needs to be ‘sorted’ from the beginning. This problem requires the output to be a sorted array from smallest to largest. Window => //repeat until the full array is traversed Window = //Now slide the window down 1 index to the right. We can then ‘slide’ the sub-array one index over and repeat to find the smallest element until the entire array has been traversed.Ĭonst arr = const k = 2 const window = //sub-array from index 0 of length K+1 //Sort the first index You can sort the value into the first index correctly, since you know each element is within k indices of its final sorted index. The 1st number in the array can be in any position from 0 to k, since there are no negative indices.Ĭonsider a sub-array of that length, k + 1, filled with the first k + 1 values in the array. How can we guarantee that? ‘Sliding Window’ Sub-Array However, if we can guarantee that the first index’s value is sorted, then this becomes k + 1 options for the next element. K provides a plus/minus deviation from the actual index, providing 2k + 1 options for every element. This would be done in O(nk) time, since for every element in the n length array, you need to loop through k values to obtain the minimum. However, this would ignore that fact that the array is already mostly sorted (k-sorted), resulting in a less optimal solution.Īnother approach utilizes the k values, by finding the minimum of the k values (a window of k values) in front of an element, and comparing its minimum to the element, swapping places if needed. Let arr = let k = 2 sortKMessedArray (arr, k ) //output: ApproachĪn initial approach would be to sort the array using mergeSort or JavaScript’s sort method, in nlog(n) time. Exampleįor an input array of size 10 and k = 2, an element belonging to index 6 in the sorted array will be located at either index 4, 5, 6, 7 or 8 in the input array. Given an array of integers arr where each element is at most k places away from its sorted position, write a function sortKMessedArray that sorts arr. We hope you enjoy! K-Messed Array Sort See Repl Prompt This series of posts will give you a problem to solve and an explanation of both the brute force & optimized solutions. Optimize: Optimize while thinking about Big O.Approach: Talk through your approach to solving the problem.Repeat: Repeat the question back to the interviewer.REACTO is an approach to successfully conquer these interviews. These interviews often require knowledge of Big O (time and space complexity), Data Structures, and Algorithms. ![]() The interview process for software engineers often time requires the interviewee to stand in front of a white board and code out a solution to a prompt that is given to them by the interviewer. ![]()
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