{"id":1612,"date":"2020-07-01T09:30:23","date_gmt":"2020-07-01T09:30:23","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=1612"},"modified":"2022-03-22T07:48:51","modified_gmt":"2022-03-22T07:48:51","slug":"kth-smallest-number","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/kth-smallest-number\/","title":{"rendered":"Kth Smallest Number"},"content":{"rendered":"

\"\"<\/p>\n

Concepts Used<\/h3>\n
\n

Heap<\/p>\n<\/blockquote>\n

Difficulty Level<\/h3>\n
\n

Easy<\/p>\n<\/blockquote>\n

Problem Statement :<\/h3>\n
\n

Given an array arr[ ] and a number K<\/code> where K<\/code> is smaller than the size of the array, the task is to find the Kth<\/code> smallest element in the given array. It is given that all array elements are distinct.<\/p>\n<\/blockquote>\n

<\/strong><\/u><\/a><\/p>\n

Solution Approach :<\/h3>\n

Introduction :<\/h4>\n
\n

Idea is to create a min-heap<\/strong> then extract the first (k-1) elements.
\nNow print the first element of the heap.<\/p>\n<\/blockquote>\n

Method 1 :<\/h4>\n
\n

Sort the array in increasing order using any sorting algorithm. Now print the kth<\/sup> value in the array which is our kth<\/code> smallest value.<\/p>\n<\/blockquote>\n

Method 2 :<\/h4>\n
\n

We will create a min-heap of size n<\/code>. In a min-heap<\/strong>, the root always stores the smaller value as compared to its left<\/strong> & right<\/strong> subtree, this condition needs to be true for every node. We need to insert each item one by one such that it’s parent is always smaller than the item itself. If the parent is greater, then we need to swap the current item with its parent. <\/p>\n

In order to print the kth<\/sup> smallest element we need to extract()<\/strong> the first k-1<\/code> elements and now print the first element of our heap.<\/p>\n<\/blockquote>\n

extract()<\/strong>: Removes the minimum element from Min-Heap<\/strong>. Time Complexity of this Operation is O(Logn) as this operation needs to maintain the heap property (by calling heapify()<\/strong>) after removing root.<\/p>\n

heapify()<\/strong>: Maintains the heap property for each node. If any node does not follow heap property it swaps the node with the node which is smaller ,or greater (in case of max-heap<\/strong>), than the node.<\/p>\n

\n

Below is the algorithm of above approach.<\/p>\n<\/blockquote>\n

Algorithms :<\/h3>\n

Insert() :<\/h4>\n
\n
    \n
  1. Insert the item at the last index, and increment the size by 1.<\/li>\n
  2. Then, check if the inserted item is smaller than its parent,<\/li>\n
  3. If yes, then swap the inserted item with its parent.<\/li>\n
  4. If no, then do nothing.<\/li>\n
  5. Now, go to step 2<\/code> and repeat untill we reach root (first element).<\/li>\n<\/ol>\n<\/blockquote>\n

    Extract() :<\/h4>\n
    \n
      \n
    1. Store the value of the first node of our heap ( temp = heap[0]<\/code> ).<\/li>\n
    2. Replace the root node with the farthest right node (last element).<\/li>\n
    3. Decrease the size by 1<\/code>. (heap[0] = heap[size-1])<\/code><\/li>\n
    4. Perform heapify<\/strong> starting from the new root. <\/li>\n
    5. Return the stored value.(temp<\/strong>)<\/li>\n<\/ol>\n<\/blockquote>\n

      Heapify ():<\/h4>\n
      \n
        \n
      1. if the heap property holds true then you are done.<\/li>\n
      2. else if<\/li>\n
      3. the replacement node value <= its parent nodes value then swap them, and repeat step 3.<\/li>\n
      4. else<\/li>\n
      5. swap the replacement node with the smallest child node, and repeat step 3.<\/li>\n<\/ol>\n<\/blockquote>\n

        Example:<\/h4>\n

        \"\"<\/p>\n

        Solutions:<\/h3>\n

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