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Find the sum of medians of all odd length subarrays

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  • Last Updated : 14 Jan, 2022

Given an array arr[] of size N, the task is to find the sum of medians of all sub-array of odd-length.

Examples:

Input: arr[] = {4, 2, 5, 1}
Output: 18
Explanation : Sub-Arrays of odd length and their medians are :

  • [4]  -> Median is 4
  • [4, 2, 5]  -> Median is 4
  • [2]  -> Median is 2
  • [2, 5, 1]  -> Median is 2
  • [5]  -> Median is 5
  • [1]  -> Median is 1

Their sum = 4 + 4+ 2 + 2 + 5 +1 = 18

Input: arr[] = {1, 2}
Output: 3

 

Pre-requisites: Median of Stream of Running Integers using STL

Naive Approach: Generate each and every sub-array. If the length of the sub-array is odd, then sort the sub-array and return the middle element.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of medians
// of all odd-length subarrays
int solve(vector<int> arr)
{
    int ans = 0;
    int n = arr.size();
 
    // Loop to calculate the sum
    for(int i = 0; i < n; i++)
    {
        vector<int> new_arr;
        for(int j = i; j < n; j++)
        {
            new_arr.push_back(arr[j]);
             
            // Odd length subarray
            if ((new_arr.size() % 2) == 1)
            {
                sort(new_arr.begin(), new_arr.end());
                int mid = new_arr.size() / 2;
                ans += new_arr[mid];
            }
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 4, 2, 5, 1 };
    cout << solve(arr);
}
 
// This code is contributed by Samim Hossain Mondal.


Java




// Java program for the above approach
import java.util.*;
 
class GFG {
    // Function to find sum of medians
    // of all odd-length subarrays
    static int solve(int[] arr) {
        int ans = 0;
        int n = arr.length;
 
        // Loop to calculate the sum
        for (int i = 0; i < n; i++) {
            List<Integer> new_arr = new LinkedList<Integer>();
            for (int j = i; j < n; j++) {
                new_arr.add(arr[j]);
 
                // Odd length subarray
                if ((new_arr.size() % 2) == 1) {
                    Collections.sort(new_arr);
                    int mid = new_arr.size() / 2;
                    ans += new_arr.get(mid);
                }
            }
        }
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args) {
        int[] arr = { 4, 2, 5, 1 };
        System.out.println(solve(arr));
    }
}
// This code is contributed by 29AjayKumar


Python3




# Python program for the above approach
 
# Function to find sum of medians
# of all odd-length subarrays
def solve(arr):
        ans = 0
        n = len(arr)
         
        # Loop to calculate the sum
        for i in range(n):
            new_arr = []
            for j in range(i, n, 1):
                new_arr.append(arr[j])
 
                # Odd length subarray
                if (len(new_arr)) % 2 == 1:
                    new_arr.sort()
                    mid = len(new_arr)//2
                    ans += new_arr[mid]
        return (ans)
 
# Driver Code
if __name__ == "__main__":
    arr = [4, 2, 5, 1]
    print(solve(arr))
    


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to find sum of medians
  // of all odd-length subarrays
  static int solve(int[] arr) {
    int ans = 0;
    int n = arr.Length;
 
    // Loop to calculate the sum
    for (int i = 0; i < n; i++) {
      List<int> new_arr = new List<int>();
      for (int j = i; j < n; j++) {
        new_arr.Add(arr[j]);
 
        // Odd length subarray
        if ((new_arr.Count % 2) == 1) {
          new_arr.Sort();
          int mid = new_arr.Count / 2;
          ans += new_arr[mid];
        }
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void Main() {
    int[] arr = { 4, 2, 5, 1 };
    Console.Write(solve(arr));
  }
}
 
// This code is contributed by Saurabh Jaiswal


Javascript




<script>
 
// javascript program for the above approach
    // Function to find sum of medians
    // of all odd-length subarrays
    function solve(arr) {
        var ans = 0;
        var n = arr.length;
 
        // Loop to calculate the sum
        for (var i = 0; i < n; i++) {
            var new_arr= new Array();
            for (var j = i; j < n; j++) {
                new_arr.push(arr[j]);
 
                // Odd length subarray
                if ((new_arr.length % 2) == 1) {
                    new_arr.sort();
                    var mid = Math.floor(new_arr.length / 2);
                     
                    // document.write(mid);
                    ans += new_arr[mid];
                }
            }
        }
        return ans;
    }
 
    // Driver Code
var arr = [ 4, 2, 5, 1 ];
document.write(solve(arr));
 
// This code is contributed by shikhasingrajput
</script>


Output

18

Time Complexity: O(N3 * Log(N))  
Auxiliary Space: O(N)

Note: Instead of sorting array each time, which costs (N*logN), insertion sort can be applied. But still, overall Time Complexity will be O(N3).

Efficient Approach: The median of the sorted array is the value separating the higher half from the lower half in the array. For finding out the median, we only need the middle element, rather than the entire sorted array. The approach of Median of Stream of Running Integers can be applied over here. Follow the steps mentioned below 

  1. Use max and min heaps to calculate the running median.
  2. Traverse each and every element in the array.
  3. While creating a new subarray, add an element into the heaps and return median if the size is odd else return 0.
  4. Max_heap is used to store lower half elements such that the maximum element is at the top and min_heap is used to store higher half elements such that the minimum element is at the top.
  5. The difference between both the heaps should not be greater than one, and one extra element is always placed in max_heap.

Note: Here max_heap is implemented using min_heap, by just negating the values so that the maximum negative element can be popped.

Below is the implementation of the above approach:  

Python3




# Python program for the above approach
from heapq import heappush as push, heappop as pop
 
# Find the sum of medians of odd-length
# subarrays
 
 
class find_median():
 
    # Constructor to declare two heaps
    def __init__(self):
 
        # Store lower half elements such that
        # maximum element is at top
        self.max_heap = []
 
        # Store higher half elements such that
        # minimum element is at top
        self.min_heap = []
 
    def add(self, val):
 
        # len(max_heap) == 0 or curr_element
        # smaller than max_heap top
        if (len(self.max_heap) == 0 or
            self.max_heap[0] > val):
            push(self.max_heap, -val)
 
        else:
            push(self.min_heap, val)
 
        # If size of max_heap + 1 greater
        # than min_heap
        if (len(self.max_heap)+1 >
            len(self.min_heap)):
            val = pop(self.max_heap)
            push(self.min_heap, -val)
 
        # If size of min_heap
        # greater than max_heap
        if (len(self.min_heap) >
            len(self.max_heap)):
            val = pop(self.min_heap)
            push(self.max_heap, -val)
 
        # Finally if sum of sizes is odd,
        # return median
        if (len(self.min_heap) +
            len(self.max_heap)) % 2 == 1:
            return (-self.max_heap[0])
 
        # Else return 0
        else:
            return 0
 
# Function to calculate the sum
# of all odd length subarrays
def solve(arr):
    ans = 0
     
    # Size of the array
    n = len(arr)
    for i in range(n):
         
        # Create an object
        # of class find_median
        obj = find_median()
        for j in range(i, n, 1):
             
            # Add value to the heaps
            # using object
            val = obj.add(arr[j])
            ans += val
     
    return (ans)
 
# Driver Code
if __name__ == "__main__":
    arr = [4, 2, 5, 1]
    print(solve(arr))


Output

18

Time Complexity: O(N2 * Log(N))  
Auxiliary Space: O(N)


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