{"id":27393,"date":"2018-02-02T21:12:47","date_gmt":"2018-02-02T15:42:47","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=27393"},"modified":"2018-02-02T21:12:47","modified_gmt":"2018-02-02T15:42:47","slug":"cc-programming-maximum-of-all-subarrays-of-size-k","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/cc-programming-maximum-of-all-subarrays-of-size-k\/","title":{"rendered":"C\/C++ Programming-Maximum of all sub arrays of size k"},"content":{"rendered":"

Given an array and an integer k, find the maximum for each and every contiguous subarray of size k.<\/span><\/p>\n

Examples:<\/p>\n

Input :
\narr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}
\nk = 3
\nOutput :
\n3 3 4 5 5 5 6<\/p>\n

Input :
\narr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}
\nk = 4
\nOutput :
\n10 10 10 15 15 90 90<\/p>\n

\u00a0Method 1 (Simple)<\/strong>
\nRun two loops. In the outer loop, take all subarrays of size k. In the inner loop, get the maximum of the current subarray<\/div>\n
<\/div>\n
C Programming<\/strong><\/div>\n
\n
<\/div> 
#include<stdio.h> void printKMax(int arr[], int n, int k){    int j, max;     for (int i = 0; i <= n-k; i++)    {        max = arr[i];         for (j = 1; j < k; j++)        {            if (arr[i+j] > max)               max = arr[i+j];        }        printf("%d ", max);    }}  int main(){    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};    int n = sizeof(arr)\/sizeof(arr[0]);    int k = 3;    printKMax(arr, n, k);    return 0;}<\/code><\/pre> <\/div>\n
Time Complexity: The outer loop runs n-k+1 times and the inner loop runs k times for every iteration of outer loop. So time complexity is O((n-k+1)*k) which can also be written as O(nk).<\/p>\n[ad type=”banner”]\n
Method 2 (Use Self-Balancing BST)<\/strong>
\n1) Pick first k elements and create a Self-Balancing Binary Search Tree (BST) of size k.
\n2) Run a loop for i = 0 to n \u2013 k
\na) Get the maximum element from the BST, and print it.
\nb) Search for arr[i] in the BST and delete it from the BST.
\nc) Insert arr[i+k] into the BST.<\/p>\n
Time Complexity: Time Complexity of step 1 is O(kLogk). Time Complexity of steps 2(a), 2(b) and 2(c) is O(Logk). Since steps 2(a), 2(b) and 2(c) are in a loop that runs n-k+1 times, time complexity of the complete algorithm is O(kLogk + (n-k+1)*Logk) which can also be written as O(nLogk).<\/p>\n[ad type=”banner”]\n
Method 3 (A O(n) method: use Dequeue)<\/strong>
\nWe create a Dequeue<\/a>, Qi <\/em>of capacity k, that stores only useful elements of current window of k elements. An element is useful if it is in current window and is greater than all other elements on left side of it in current window. We process all array elements one by one and maintain Qi <\/em>to contain useful elements of current window and these useful elements are maintained in sorted order. The element at front of the Qi <\/em>is the largest and element at rear of Qi <\/em>is the smallest of current window.<\/p>\n[ad type=”banner”]\n
Following is C++ implementation of this method.<\/strong><\/p>\n
 
 <\/div> 
#include <iostream>#include <deque> using namespace std; \/\/ A Dequeue (Double ended queue) based method for printing maixmum element of\/\/ all subarrays of size kvoid printKMax(int arr[], int n, int k){    \/\/ Create a Double Ended Queue, Qi that will store indexes of array elements    \/\/ The queue will store indexes of useful elements in every window and it will    \/\/ maintain decreasing order of values from front to rear in Qi, i.e.,     \/\/ arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order    std::deque<int>  Qi(k);     \/* Process first k (or first window) elements of array *\/    int i;    for (i = 0; i < k; ++i)    {        \/\/ For very element, the previous smaller elements are useless so        \/\/ remove them from Qi        while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()])            Qi.pop_back();  \/\/ Remove from rear         \/\/ Add new element at rear of queue        Qi.push_back(i);    }     \/\/ Process rest of the elements, i.e., from arr[k] to arr[n-1]    for ( ; i < n; ++i)    {        \/\/ The element at the front of the queue is the largest element of        \/\/ previous window, so print it        cout << arr[Qi.front()] << " ";         \/\/ Remove the elements which are out of this window        while ( (!Qi.empty()) && Qi.front() <= i - k)            Qi.pop_front();  \/\/ Remove from front of queue         \/\/ Remove all elements smaller than the currently        \/\/ being added element (remove useless elements)        while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()])            Qi.pop_back();          \/\/ Add current element at the rear of Qi        Qi.push_back(i);    }     \/\/ Print the maximum element of last window    cout << arr[Qi.front()];} \/\/ Driver program to test above functionsint main(){    int arr[] = {12, 1, 78, 90, 57, 89, 56};    int n = sizeof(arr)\/sizeof(arr[0]);    int k = 3;    printKMax(arr, n, k);    return 0;}<\/code><\/pre> <\/div>\n
Output:<\/p>\n
78 90 90 90 89<\/pre>\n
Time Complexity: O(n). It seems more than O(n) at first look. If we take a closer look, we can observe that every element of array is added and removed at most once. So there are total 2n operations.
\nAuxiliary Space: O(k)<\/p>\n[ad type=”banner”]\n
<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
C\/C++ Programming Maximum of all sub arrays of size k Given an array and an integer k, find the maximum for each and every contiguous sub array of size k.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[69866,1,73012,80124],"tags":[81630,81623,81626,83872,83874,83873,81625,83875],"class_list":["post-27393","post","type-post","status-publish","format-standard","hentry","category-c-programming","category-coding","category-data-structures","category-queue","tag-find-maximum-of-minimum-for-every-window-size-in-a-given-array","tag-maximum-of-all-subarrays-of-size-k-java","tag-maximum-sum-of-all-subarrays-of-size-k","tag-reverse-array-in-groups","tag-sliding-window-maximum","tag-sliding-window-maximum-maximum-of-all-subarrays-of-size-k","tag-sliding-window-maximum-java","tag-sliding-window-maximum-python"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/27393","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=27393"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/27393\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=27393"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=27393"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=27393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}