Table of Contents
What is best time complexity for find the sum of n natural numbers?
The running time of summing, one after the other, the first n consecutive numbers is indeed O(n). But the complexity of the result, that is the size of “sum from 1 to n” = n(n – 1) / 2 is O(n ^ 2).
What is the sum of all numbers from 1 to 100?
5,050
The sum of the numbers 1-100 would be equal to the number of pairs (50) multiplied by the sum of each pair (101), or 50 x 101 = 5,050.
What is the time complexity of sum of N digits?
Explain why time complexity for summing digits in a number of length N is O(logN) I understand this code, and that the code will take the ones place digit, add that digit to sum, and remove that digit. It keeps doing this until n is equal to 0, at which point it will return sum.
How to find the sum of all numbers from 1 to N?
Given a number n, find the sum of digits in all numbers from 1 to n. A naive solution is to go through every number x from 1 to n and compute the sum in x by traversing all digits of x. Below is the implementation of this idea.
What is the runtime complexity of sum of first n consecutive numbers?
You are confusing complexity of runtimeand the size (complexity) of the result. The running timeof summing, one after the other, the first nconsecutive numbers is indeed O(n).1 But the complexity of the result, that is the size of “sum from 1 to n” = n(n– 1) / 2 is O(n^ 2).
How to find the answer of sum of series of n natural?
answer of sum of series of n natural can be found using two ways. first way is by adding all the numbers in loop. in this case algorithm is linear and code will be like this int sum = 0; for (int i = 1; i <= n; i++) { sum += n; } return sum;
What is the sum of sum (299) from 100 to 199?
Note that sum of sum (299) is sum (99) + sum of digits from 100 to 199 + sum of digits from 200 to 299. Sum of 100 to 199 is sum (99) + 1*100 and sum of 299 is sum (99) + 2*100. In general, this sum can be computed as w*msd + (msd* (msd-1)/2)*10 d b) Sum of digits in msd * 10 d to n.