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:
- Sum of square-sums of first n natural numbers
- Find m-th summation of first n natural numbers.
- Find the average of first N natural numbers
- Find if given number is sum of first n natural numbers
- Program to find sum of first n natural numbers
- Find sum of N-th group of Natural Numbers
- Find the good permutation of first N natural numbers
- Find permutation of first N natural numbers that satisfies the given condition
- Find the number of sub arrays in the permutation of first N natural numbers such that their median is M
- Find ways an Integer can be expressed as sum of n-th power of unique natural numbers
- Count square and non-square numbers before n
- Find maximum product of digits among numbers less than or equal to N
- Program to find the maximum difference between the index of any two different numbers
- Sum of sum of first n natural numbers
- LCM of First n Natural Numbers
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