Given an array **arr[]** of **N** integers and an integer **K**, the task is to find the minimum and maximum of all subarrays of size K.

**Examples:**

Input:arr[] = {2, -2, 3, -9, -5, -8}, K = 4Output:

-9 3

-9 3

-9 3Explanation:

Below are the subarray of size 4 and minimum and maximum value of each subarray:

1. {2, -2, 3, -9}, minValue = -9, maxValue = 3

2. {-2, 3, -9, -5}, minValue = -9, maxValue = 3

3. {3, -9, -5, -8}, minValue = -9, maxValue = 3

Input:arr[] = { 5, 4, 3, 2, 1, 6, 3, 5, 4, 2, 1 }, K = 3Output:

3 5

2 4

1 3

1 6

1 6

3 6

3 5

2 5

1 4

**Approach:**

- Traverse the given array upto K elements and store the count of each element into a map.
- After inserting K elements, for each remaining elements do the following:
- Increase the frequency of current element
**arr[i]**by 1. - Decrease the frequency of
**arr[i – K + 1]**by 1 to store the frequency of current subarray(**arr[i – K + 1, i]**) of size**K**. - Since map stores the key value pair in sorted order. Therefore the iterator at the starting of the map stores the minimum element and at the ending of the map stores the maximum element. Print the minimum and maximum element of the current subarray.

- Increase the frequency of current element
- Repeat the above steps for each subarray formed.

Below is the implementation of the above approach:

`// C++ program for the above approach` ` ` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to find the minimum and` `// maximum element for each subarray` `// of size K` `int` `maxSubarray(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` ` ` `// To store the frequency of element` ` ` `// for every subarray` ` ` `map<` `int` `, ` `int` `> Map;` ` ` ` ` `// To count the subarray array size` ` ` `// while traversing array` ` ` `int` `l = 0;` ` ` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// Increment till we store the` ` ` `// frequency of first K element` ` ` `l++;` ` ` ` ` `// Update the count for current` ` ` `// element` ` ` `Map[arr[i]]++;` ` ` ` ` `// If subarray size is K, then` ` ` `// find the minimum and maximum` ` ` `// for each subarray` ` ` `if` `(l == k) {` ` ` ` ` `// Iterator points to end` ` ` `// of the Map` ` ` `auto` `itMax = Map.end();` ` ` `itMax--;` ` ` ` ` `// Iterator points to start of` ` ` `// the Map` ` ` `auto` `itMin = Map.begin();` ` ` ` ` `// Print the minimum and maximum` ` ` `// element of current sub-array` ` ` `cout << itMin->first << ` `' '` ` ` `<< itMax->first << endl;` ` ` ` ` `// Decrement the frequency of` ` ` `// arr[i - K + 1]` ` ` `Map[arr[i - k + 1]]--;` ` ` ` ` `// if arr[i - K + 1] is zero` ` ` `// remove from the map` ` ` `if` `(Map[arr[i - k + 1]] == 0) {` ` ` `Map.erase(arr[i - k + 1]);` ` ` `}` ` ` ` ` `l--;` ` ` `}` ` ` `}` ` ` `return` `0;` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { 5, 4, 3, 2, 1, 6,` ` ` `3, 5, 4, 2, 1 };` ` ` ` ` `// Subarray size` ` ` `int` `k = 3;` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` ` ` `// Function Call` ` ` `maxSubarray(arr, n, k);` ` ` `return` `0;` `}` |

**Output:**

3 5 2 4 1 3 1 6 1 6 3 6 3 5 2 5 1 4

**Time Complexity:** *O(N*log K)*, where N is the number of element.**Auxiliary Space:** *O(K)*, where K is the size of subarray.

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