Find the sum of medians of all odd length subarrays - GeeksforGeeks

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: 18Explanation : 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++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to find sum of medians// of all odd-length subarraysint 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 Codeint main(){    vector<int> arr = { 4, 2, 5, 1 };    cout << solve(arr);} // This code is contributed by Samim Hossain Mondal.
// Java program for the above approachimport 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
# Python program for the above approach # Function to find sum of medians# of all odd-length subarraysdef 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 Codeif __name__ == "__main__":    arr = [4, 2, 5, 1]    print(solve(arr))    
// C# program for the above approachusing 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
<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 Codevar arr = [ 4, 2, 5, 1 ];document.write(solve(arr)); // This code is contributed by shikhasingrajput</script>
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 
Use max and min heaps to calculate the running median.Traverse each and every element in the array.While creating a new subarray, add an element into the heaps and return median if the size is odd else return 0.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.The difference between both the heaps should not be greater than one, and one extra element is always placed in max_heap.
Use max and min heaps to calculate the running median.
Traverse each and every element in the array.
While creating a new subarray, add an element into the heaps and return median if the size is odd else return 0.
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.
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 approachfrom 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 subarraysdef 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 Codeif __name__ == "__main__":    arr = [4, 2, 5, 1]    print(solve(arr))
18
Time Complexity: O(N2 * Log(N))  Auxiliary Space: O(N)
29AjayKumar
_saurabh_jaiswal
samim2000
sumitgumber28
shikhasingrajput
avtarkumar719
nnr223442
median-finding
subarray
Arrays
Heap
Mathematical
Sorting
Arrays
Mathematical
Sorting
Heap
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