Find maximum N such that the sum of square of first N natural numbers is not more than X in Python

Suppose we have a given integer X, we have to find the maximum value N so that the sum of first N natural numbers should not exceed the value X.

So, if the input is like X = 7, then the output will be 2 as 2 is the maximum possible value of N, for N = 3, the sum of the series will exceed X = 7 So, 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14.

To solve this, we will follow these steps −

• Define a function sum_of_squares() . This will take N

• res :=(N *(N + 1) *(2 * N + 1)) / 6

• return res

• From the main method, do the following −

• low := 1

• high := 100000

• N := 0

• while low −= high, do

  • mid :=(high + low) / 2

  • if sum_of_squares(mid) −= X, then

    • N := mid

    • low := mid + 1

  • otherwise,

    • high := mid - 1

• return N

Example

Let us see the following implementation to get better understanding −

 Live Demo

def sum_of_squares(N):
   res = (N * (N + 1) * (2 * N + 1)) // 6
   return res
def get_max(X):
   low, high = 1, 100000
   N = 0
   while low <= high:
      mid = (high + low) // 2
      if sum_of_squares(mid) <= X:
         N = mid
         low = mid + 1
      else:
         high = mid - 1
   return N
X = 7
print(get_max(X))

Input

7

Output

2

Arnab Chakraborty

Published on 25-Aug-2020 10:56:04

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