Question

What would be the running time for the code below

int sum = 1;
for (int i = 1; i <= n; i++)
for (int j = i; j <= n; j++)
for (int k = i; k <= j; k++)
sum++;

Answer 1

The running time for the given code is approximately O(n^3) due to its three nested loops.

Step-by-step explanation:

The running time for the given code can be determined by analyzing the nested for loops. The outer loop runs n times, the second loop runs n - i + 1 times, and the innermost loop runs j - i + 1 times. Summing up all the iterations, the total number of iterations will be:

1. (n - 1) + (n - 2) + ... + 1 = n * (n - 1) / 2
2. (n - 2) + (n - 3) + ... + 1 = (n - 1) * (n - 2) / 2
3. ...
4. 1

As a result, the running time for the code is approximately O(n^3) because it has three nested loops.