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