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

Given an integer X, the task is to find the maximum value N such that the sum of first N natural numbers is not more than X.

Examples:

Input: X = 5
Output: 2
2 is the maximum possible value of N because for N = 3, the sum of the series will exceed X
i.e. 12 + 22 + 32 = 1 + 4 + 9 = 14

Input: X = 25
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Solution: A simple solution is to run a loop from 1 till the maximum N such that S(N) ≤ X, where S(N) is the sum of square of first N natural numbers. Sum of square of first N natural numbers is given by the formula S(N) = N * (N + 1) * (2 * N + 1) / 6. The time complexity of this approach is O(N).

Efficient Approach: An efficient solution is to use Binary Search to find the value of N. The time complexity of this approach is O(log N).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long ` ` `  `// Function to return the sum of the squares ` `// of first N natural numbers ` `ll squareSum(ll N) ` `{ ` `    ``ll sum = (ll)(N * (N + 1) * (2 * N + 1)) / 6; ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to return the maximum N such that ` `// the sum of the squares of first N ` `// natural numbers is not more than X ` `ll findMaxN(ll X) ` `{ ` `    ``ll low = 1, high = 100000; ` `    ``int` `N = 0; ` ` `  `    ``while` `(low <= high) { ` `        ``ll mid = (high + low) / 2; ` ` `  `        ``if` `(squareSum(mid) <= X) { ` `            ``N = mid; ` `            ``low = mid + 1; ` `        ``} ` ` `  `        ``else` `            ``high = mid - 1; ` `    ``} ` ` `  `    ``return` `N; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``ll X = 25; ` `    ``cout << findMaxN(X); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the sum of the squares ` `// of first N natural numbers ` `static` `long` `squareSum(``long` `N) ` `{ ` `    ``long` `sum = (``long``)(N * (N + ``1``) * (``2` `* N + ``1``)) / ``6``; ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to return the maximum N such that ` `// the sum of the squares of first N ` `// natural numbers is not more than X ` `static` `long` `findMaxN(``long` `X) ` `{ ` `    ``long` `low = ``1``, high = ``100000``; ` `    ``int` `N = ``0``; ` ` `  `    ``while` `(low <= high)  ` `    ``{ ` `        ``long` `mid = (high + low) / ``2``; ` ` `  `        ``if` `(squareSum(mid) <= X)  ` `        ``{ ` `            ``N = (``int``) mid; ` `            ``low = mid + ``1``; ` `        ``} ` ` `  `        ``else` `            ``high = mid - ``1``; ` `    ``} ` ` `  `    ``return` `N; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``long` `X = ``25``; ` `    ``System.out.println(findMaxN(X)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the sum of the squares ` `// of first N natural numbers ` `static` `long` `squareSum(``long` `N) ` `{ ` `    ``long` `sum = (``long``)(N * (N + 1) * (2 * N + 1)) / 6; ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to return the maximum N such that ` `// the sum of the squares of first N ` `// natural numbers is not more than X ` `static` `long` `findMaxN(``long` `X) ` `{ ` `    ``long` `low = 1, high = 100000; ` `    ``int` `N = 0; ` ` `  `    ``while` `(low <= high)  ` `    ``{ ` `        ``long` `mid = (high + low) / 2; ` ` `  `        ``if` `(squareSum(mid) <= X)  ` `        ``{ ` `            ``N = (``int``) mid; ` `            ``low = mid + 1; ` `        ``} ` ` `  `        ``else` `            ``high = mid - 1; ` `    ``} ` ` `  `    ``return` `N; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `         `  `    ``long` `X = 25; ` `    ``Console.WriteLine(findMaxN(X)); ` `} ` `} ` ` `  `// This code contributed by ajit `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the sum of the  ` `# squares of first N natural numbers  ` `def` `squareSum(N):  ` ` `  `    ``Sum` `=` `(N ``*` `(N ``+` `1``) ``*` `(``2` `*` `N ``+` `1``)) ``/``/` `6` `    ``return` `Sum` ` `  `# Function to return the maximum N such  ` `# that the sum of the squares of first N  ` `# natural numbers is not more than X  ` `def` `findMaxN(X): ` ` `  `    ``low, high, N ``=` `1``, ``100000``, ``0` ` `  `    ``while` `low <``=` `high: ` `        ``mid ``=` `(high ``+` `low) ``/``/` `2` ` `  `        ``if` `squareSum(mid) <``=` `X:  ` `            ``N ``=` `mid  ` `            ``low ``=` `mid ``+` `1` `         `  `        ``else``: ` `            ``high ``=` `mid ``-` `1` `     `  `    ``return` `N  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``X ``=` `25` `    ``print``(findMaxN(X))  ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## PHP

 ` `

Output:

```3
```

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Improved By : rituraj_jain, Rajput-Ji, jit_t