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Find array with k number of merge sort calls in Python
Suppose we have two numbers a and b, we have to find an array containing values in range [1, a] and requires exactly b number of calls of recursive merge sort function.
So, if the input is like a = 10, b = 15, then the output will be [3,1,4,6,2,8,5,9,10,7]
To solve this, we will follow these steps −
- Define a function solve() . This will take left, right, array,b
- if b < 1 or left + 1 is same as right, then
- return
- b := b - 2
- mid := (left + right) / 2
- temp := array[mid - 1]
- array[mid-1] := array[mid]
- array[mid] := temp
- solve(left, mid, array, b)
- solve(mid, right, array, b)
- From the main method do the following −
- if b mod 2 is same as 0, then
- display "None"
- return
- array := an array of size n + 1, and fill with 0
- array[0] := 1
- for i in range 1 to a, do
- array[i] := i + 1
- b := b - 1
- solve(0, a, array, b)
- return array, a
Example
Let us see the following implementation to get better understanding −
def solve(left,right,array,b):
if (b < 1 or left + 1 == right):
return
b -= 2
mid = (left + right) // 2
temp = array[mid - 1]
array[mid-1] = array[mid]
array[mid] = temp
solve(left, mid, array, b)
solve(mid, right, array, b)
def find_arr(a,b):
if (b % 2 == 0):
print("None")
return
array = [0 for i in range(a + 2)]
array[0] = 1
for i in range(1, a):
array[i] = i + 1
b -=1
solve(0, a, array, b)
return array, a
a = 10
b = 15
array, size = find_arr(a, b)
print(array[:size])Input
10,15
Output
[3, 1, 4, 6, 2, 8, 5, 9, 10, 7]
Updated on: 28-Aug-2020
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