Find the number of sub arrays in the permutation of first N natural numbers such that their median is M in Python

Suppose we have an array A containing the permutation of first N natural numbers and another number M is also given, where M ≤ N, we have to find the number of sub-arrays such that the median of the sequence is M. As we know the median of a sequence is defined as the value of the element which is in the middle of the sequence after sorting it according to ascending order. For even length sequence, the left of two middle elements is used.

So, if the input is like A = [3, 5, 6, 4, 2] and M = 5, then the output will be 4 as the required subarrays are [3, 5, 6], [5], [5, 6] and [5, 6, 4].

To solve this, we will follow these steps −

• n := size of arr

• my_map := a new map

• my_map[0] := 1

• has := False, add := 0, result := 0

• for i in range 0 to n, do

  • if arr[i] < m, then

    • add := add - 1

  • otherwise when arr[i] > m, then

    • add := add + 1

  • if arr[i] is same as m, then

    • has := True

  • if has is true, then

    • if add present in my_map, then

      • result := result + my_map[add]

    • if add-1 present in my_map , then

      • result := result + my_map[add - 1]

  • otherwise,

    • my_map[add] := (value of my_map[add], if present, otherwise 0) + 1

• return result

Example

Let us see the following implementation to get better understanding −

 Live Demo

def solve(arr, m):
   n = len(arr)
   my_map = {}
   my_map[0] = 1
   has = False
   add = 0
   result = 0
   for i in range(n):
      if (arr[i] < m):
         add -= 1
      elif (arr[i] > m):
         add += 1
      if (arr[i] == m):
         has = True
      if (has):
         if(add in my_map):
            result += my_map[add]
         if add-1 in my_map:
            result += my_map[add - 1]
      else:
         my_map[add] = my_map.get(add, 0) + 1
   return result
arr = [3, 5, 6, 4, 2]
m = 5
print(solve(arr, m))

Input

[3, 5, 6, 4, 2] , 5

Output

3