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.
Input: X = 5
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
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:
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