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Minimum and Maximum of all subarrays of size K using Map

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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 = 4 Output: -9 3 -9 3 -9 3 Explanation: 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 = 3 Output: 3 5 2 4 1 3 1 6 1 6 3 6 3 5 2 5 1 4

Approach:

  1. Traverse the given array upto K elements and store the count of each element into a map.
  2. 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.
  3. Repeat the above steps for each subarray formed.

Below is the implementation of the above approach: 

CPP

// 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;
}

                    

Java

/*package whatever //do not write package name here */
import java.util.*;
 
class GFG {
 
  // Function to find the minimum and
  // maximum element for each subarray
  // of size K
  static int maxSubarray(int arr[], int n, int k)
  {
 
    // To store the frequency of element
    // for every subarray
    TreeMap<Integer, Integer> Map = new TreeMap<>();
 
    // 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.put(arr[i],Map.getOrDefault(arr[i],0)+1);
 
      // If subarray size is K, then
      // find the minimum and maximum
      // for each subarray
      if (l == k) {
 
        // Iterator points to end
        // of the Map
        var itMax = getLast(Map);
 
 
        // Iterator points to start of
        // the Map
        var itMin = getFirst(Map);
 
        // Print the minimum and maximum
        // element of current sub-array
        System.out.println(itMin.getKey() + " " + itMax.getKey());
 
        // Decrement the frequency of
        // arr[i - K + 1]
        Map.put(arr[i - k + 1],Map.getOrDefault(arr[i - k + 1],0)-1);
 
        // if arr[i - K + 1] is zero
        // remove from the map
        if (Map.get(arr[i - k + 1]) == 0) {
          Map.remove(arr[i - k + 1]);
        }
 
        l--;
      }
    }
    return 0;
  }
 
  static Map.Entry<Integer, Integer> getFirst(TreeMap<Integer, Integer> lhm){
    int count = 1;
    for (Map.Entry<Integer, Integer> it :
         lhm.entrySet()) {
      if (count == 1) {
        return it;
      }
      count++;
    }
 
    return null;
  }
 
  static Map.Entry<Integer, Integer> getLast(TreeMap<Integer, Integer> lhm) {
    int count = 1;
 
    for (Map.Entry<Integer, Integer> it : lhm.entrySet()) {
 
      if (count == lhm.size()) {
        return it;
      }
      count++;
    }
 
    return null;
  }
 
  public static void main (String[] args) {
 
    // Given array arr[]
    int arr[] = { 5, 4, 3, 2, 1, 6,3, 5, 4, 2, 1 };
 
    // Subarray size
    int k = 3;
    int n = arr.length;
 
    // Function Call
    maxSubarray(arr, n, k);
  }
}
 
// This code is contributed by aadityaburujwale.

                    

Python3

# Function to find the minimum and
# maximum element for each subarray
# of size K
def maxSubarray(arr, n, k):
    # To store the frequency of element
    # for every subarray
    map = dict()
 
    # To count the subarray array size
    # while traversing array
    l = 0
 
    # Traverse the array
    for i in range(n):
        # Increment till we store the
        # frequency of first K element
        l += 1
 
        # Update the count for current
        # element
        if arr[i] in map:
            map[arr[i]] += 1
        else:
            map[arr[i]] = 1
 
        # If subarray size is K, then
        # find the minimum and maximum
        # for each subarray
        if l == k:
            # Iterator points to end
            # of the Map
            itMax = max(map.keys())
 
            # Iterator points to start of
            # the Map
            itMin = min(map.keys())
 
            # Print the minimum and maximum
            # element of current sub-array
            print(itMin, itMax)
 
            # Decrement the frequency of
            # arr[i - K + 1]
            map[arr[i - k + 1]] -= 1
 
            # if arr[i - K + 1] is zero
            # remove from the map
            if map[arr[i - k + 1]] == 0:
                del map[arr[i - k + 1]]
            l -= 1
    return 0
 
# Given array arr[]
arr = [5, 4, 3, 2, 1, 6, 3, 5, 4, 2, 1]
 
# Subarray size
k = 3
 
n = len(arr)
 
# Function Call
maxSubarray(arr, n, k)
 
# This code is contributed by akashish__

                    

C#

using System;
using System.Collections.Generic;
 
class GFG
{
  // Function to find the minimum and
  // maximum element for each subarray
  // of size K
  static int maxSubarray(int[] arr, int n, int k)
  {
    // To store the frequency of element
    // for every subarray
    SortedDictionary<int, int> Map = new SortedDictionary<int, int>();
 
    // 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
      if (!Map.ContainsKey(arr[i]))
      {
        Map.Add(arr[i], 1);
      }
      else
      {
        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
        var itMax = getLast(Map);
 
        // Iterator points to start of
        // the Map
        var itMin = getFirst(Map);
 
        // Print the minimum and maximum
        // element of current sub-array
        Console.WriteLine(itMin.Key + " " + itMax.Key);
 
        // 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.Remove(arr[i - k + 1]);
        }
 
        l--;
      }
    }
    return 0;
  }
 
  static KeyValuePair<int, int> getFirst(SortedDictionary<int, int> lhm)
  {
    int count = 1;
    foreach (KeyValuePair<int, int> it in lhm)
    {
      if (count == 1)
      {
        return it;
      }
      count++;
    }
 
    return default(KeyValuePair<int, int>);
  }
 
  static KeyValuePair<int, int> getLast(SortedDictionary<int, int> lhm)
  {
    int count = 1;
 
    foreach (KeyValuePair<int, int> it in lhm)
    {
      if (count == lhm.Count)
      {
        return it;
      }
      count++;
    }
 
    return default(KeyValuePair<int, int>);
  }
 
  public static void Main(string[] args)
  {
    // Given array arr[]
    int[] arr = { 5, 4, 3, 2, 1, 6, 3, 5, 4, 2, 1 };
 
    // Subarray size
    int k = 3;
    int n = arr.Length;
 
 
    // Function Call
    maxSubarray(arr, n, k);
  }
}
 
// This code is contributed by akashish__

                    

Javascript

// Function to find the minimum and
// maximum element for each subarray
// of size K
function maxSubarray(arr, n, k) {
  // To store the frequency of element
  // for every subarray
  let map = new Map();
 
  // To count the subarray array size
  // while traversing array
  let l = 0;
 
  // Traverse the array
  for (let i = 0; i < n; i++) {
    // Increment till we store the
    // frequency of first K element
    l++;
 
    // Update the count for current
    // element
    if (map.has(arr[i])) {
      map.set(arr[i], map.get(arr[i]) + 1);
    } else {
      map.set(arr[i], 1);
    }
 
    // If subarray size is K, then
    // find the minimum and maximum
    // for each subarray
    if (l === k) {
      // Iterator points to end
      // of the Map
      let itMax = map.keys().next().value;
      for (let key of map.keys()) {
        itMax = Math.max(itMax, key);
      }
 
      // Iterator points to start of
      // the Map
      let itMin = map.keys().next().value;
      for (let key of map.keys()) {
        itMin = Math.min(itMin, key);
      }
 
      // Print the minimum and maximum
      // element of current sub-array
      console.log(itMin + " " + itMax);
 
      // Decrement the frequency of
      // arr[i - K + 1]
      map.set(arr[i - k + 1], map.get(arr[i - k + 1]) - 1);
 
      // if arr[i - K + 1] is zero
      // remove from the map
      if (map.get(arr[i - k + 1]) === 0) {
        map.delete(arr[i - k + 1]);
      }
 
      l--;
    }
  }
  return 0;
}
 
// Given array arr[]
let arr = [5, 4, 3, 2, 1, 6, 3, 5, 4, 2, 1];
 
// Subarray size
let k = 3;
 
let n = arr.length;
 
// Function Call
maxSubarray(arr, n, k);
 
// This code is contributed by akashish__

                    
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.



Last Updated : 12 Jan, 2023
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