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# 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

7

Output

2

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