Python Program to Get K initial powers of N

Given size K and value N, the task is to write a Python Program to compute a list of powers of N till K.

Input : N = 4, K = 6
Output : [1, 4, 16, 64, 256, 1024]
Explanation : 4^i is output till i = K. 

Input : N = 3, K = 6
Output : [1, 3, 9, 27, 81, 243]
Explanation : 3^i is output till i = K. 

Method #1 : Using list comprehension + ** operator

In this, the values are incremented till K using list comprehension and ** is used to get required power of numbers. 

Python3

# Python3 code to demonstrate working of
# Get K initial powers of N
# Using list comprehension + ** operator

# initializing N
N = 4

# printing original list
print("The original N is : " + str(N))

# initializing K
K = 6

# list comprehension provides shorthand for problem
res = [N ** idx for idx in range(0, K)]

# printing result
print("Square values of N till K : " + str(res))

Output

The original N is : 4
Square values of N till K : [1, 4, 16, 64, 256, 1024]

Time Complexity: O(n)
Auxiliary Space: O(1)

Method #2 : Using pow() + list comprehension

In this, we perform the task of computing power using pow(), the rest of all the functions are performed using list comprehension.

Python3

# Python3 code to demonstrate working of
# Get K initial powers of N
# Using pow() + list comprehension
from math import pow

# initializing N
N = 4

# printing original list
print("The original N is : " + str(N))

# initializing K
K = 6

# list comprehension provides shorthand for problem
# squaring using pow()
res = [int(pow(N, idx)) for idx in range(0, K)]

# printing result
print("Square values of N till K : " + str(res))

Output

The original N is : 4
Square values of N till K : [1, 4, 16, 64, 256, 1024]

Time complexity: O(n), where n is the length of the test_list. The pow() + list comprehension takes O(n) time
Auxiliary Space: O(1), no extra space of is required

Method #3 : Using Numpy

Note: Install numpy module using command “pip install numpy”

Python3

# Python3 code to demonstrate working of
# Get K initial powers of N
# Using NumPy
import numpy as np

# initializing N
N = 4

# printing original list
print("The original N is : " + str(N))

# initializing K
K = 6

# using np.power() to get squares
res = np.power(N, np.arange(K))

# printing result
print("Square values of N till K : " + str(res))

Output:

The original N is : 4
Square values of N till K : [1, 4, 16, 64, 256, 1024]

Time Complexity: O(K)
Auxiliary Space: O(K)

Method #4 : Using operator.pow()

1. Initiated a for loop from i=1 to K
2. Raised N to power of i using operator.pow() and appended to output list
3. Displayed output list

Python3

# Python3 code to demonstrate working of
# Get K initial powers of N
import operator
# initializing N
N = 4

# printing original list
print("The original N is : " + str(N))

# initializing K
K = 6
res = []
for i in range(1,K+1):
    res.append(operator.pow(N,i))
# printing result
print("Square values of N till K : " + str(res))

Output

The original N is : 4
Square values of N till K : [4, 16, 64, 256, 1024, 4096]

Time Complexity: O(K)
Auxiliary Space: O(K)

Method #5: Using a simple loop

We can solve this problem using a simple loop that iterates from 1 to K and computes the power of N for each iteration. 

Python3

# Python3 code to demonstrate working of
# Get K initial powers of N
import operator

# initializing N
N = 4

# printing original list
print("The original N is : " + str(N))

# initializing K
K = 6

res = []
for i in range(1, K+1):
    res.append(N**i)

# printing result
print("Square values of N till K : " + str(res))

Output

The original N is : 4
Square values of N till K : [4, 16, 64, 256, 1024, 4096]

Time complexity: O(K) 
Auxiliary space: O(1) because it only requires a single variable to store the result at each iteration.

Method #6: Using recursion

step-by-step approach

• Define a function named “powers_of_n” that takes two arguments, N and K.
• Inside the function, check if the value of K is equal to 1. If it is, return a list containing the value of N.
• If the value of K is not equal to 1, call the “powers_of_n” function recursively with the arguments N and K-1.
• Store the result of the recursive call in a variable named “powers”.
• Append the value of N raised to the power of K to the “powers” list.
• Return the “powers” list.
• Define two variables, N and K, with values of 4 and 6 respectively.
• Call the “powers_of_n” function with the arguments N and K, and store the result in a variable named “res”.
• Print the “res” list using the “print” function and a formatted string.

Python3

def powers_of_n(N, K):
    if K == 1:
        return [N]
    else:
        powers = powers_of_n(N, K-1)
        powers.append(N**K)
        return powers

N = 4
K = 6
res = powers_of_n(N, K)
print("Square values of N till K : " + str(res))

Output

Square values of N till K : [4, 16, 64, 256, 1024, 4096]

Time complexity: O(K)
Auxiliary space: O(K) (for the recursive call stack)

Method #7: Using heapq:

Algorithm:

1. Define a function called powers_of_n that takes two arguments N and K.
2. Check if K is equal to 1.
   a. If yes, return a list with a single element which is the value of N.
3. If K is not equal to 1, recursively call the powers_of_n function with N and K-1 as arguments.
4. Append the value of N raised to the power of K to the list obtained in step 3.
5. Return the list obtained in step 4.
6. Define the values of N and K and call the powers_of_n function with these values.
7. Print the list obtained in step 5.

Python3

import heapq

def powers_of_n(N, K):
    res = []
    heap = []
    heapq.heappush(heap, N)
    while len(res) < K:
        x = heapq.heappop(heap)
        res.append(x)
        heapq.heappush(heap, x * N)
    return res

N = 4
K = 6
res = powers_of_n(N, K)
# printing original list
print("The original N is : " + str(N))
print("Square values of N till K : " + str(res))
#This code is contributed by Vinay Pinjala

Output

The original N is : 4
Square values of N till K : [4, 16, 64, 256, 1024, 4096]

Time complexity: O(K)

The function recursively calls itself K times, each time reducing the value of K by 1. Therefore, the time complexity of the function is O(K).
Space complexity: O(K)

The function uses a list to store the values of N raised to the power of K. The size of the list is K. Therefore, the space complexity of the function is O(K).