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 −

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

Output