Time complexity for inner for loop

Question

I have a question about time complexity of the following program below. I know why the first for loop is n + 2 but for the second for loop why is it (n+1)/2?

// sample for loop // this is not a fully coded program just for loop time complexity analysis

for (int i=n; i>=0; i--) //n + 2
    for (int j=i; j<n; j++) // (n+2)(n+1)/2
        cout << i << “,” << j <<endl; //(n+1)(n)/2

Answer 1

The inner loop runs ni times; ie it runs 0 times when i==n , then 1 time when i == n-1 , etc, up to n times when i == 0 ; for a total of n*(n+1)/2 times. (not (n+1)*(n+2)/2 as stated in your comments).

Since the inner loop is initiated n+1 times (not n+2 , as you state), it's fair to say that the average number of times it runs is n/2 .

Answer 2

First your first loop has time complexity of O(n) and nested loop normal has a time complexity of O(n^2), but the inner loop is executed i times, for each value of i. The outer loop is executed n times, thus you see a pattern of execution like this: 1 + 2 + 3 + 4 + ... + n times

Specific Info:

1. O(1): Time complexity of a function (or set of statements) is considered as O(1) if it doesn't contain loop, recursion and call to any other non-constant time function.

2. O(n): Time Complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.

3. O(n^c): Time complexity of nested loops is equal to the number of times the innermost statement is executed.

Source - > Link

Answer 3

Iterating i from 1 to N , and j from 1 to the current value of i inclusive, is like exploring one half of a rectangle of dimensions N by N+1 , split diagonally. To appreciate this, draw out the table for yourself on paper: i on the rows, j on the columns, and a tick in each cell that the nested loop visits.

Answer 4

The time complexity of the for loop is O(N^2).

I just answered a similar question on codereview.stackexchange.com .

Here's a simple program that illustrates the time complexity of the loop.

#include <iostream>
using namespace std;

int main()
{
   const int N = 10;

   int numTotalOps = 0;
   for (int i = 0; i < N; ++i)
   {
      size_t numOps = 0;
      for (int j=i; j < N; j++)
      {
         cout << i << "," << j << endl;
         ++numTotalOps;
         ++numOps;
      }
      cout << "\nNumber of operations for i = " << i << ": " << numOps << endl;
   }

   cout << "\nNumber of total operations: " << numTotalOps << endl;
   return 0;
}

Output:

0,0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9

Number of operations for i = 0: 10
1,1
1,2
1,3
1,4
1,5
1,6
1,7
1,8
1,9

Number of operations for i = 1: 9
2,2
2,3
2,4
2,5
2,6
2,7
2,8
2,9

Number of operations for i = 2: 8
3,3
3,4
3,5
3,6
3,7
3,8
3,9

Number of operations for i = 3: 7
4,4
4,5
4,6
4,7
4,8
4,9

Number of operations for i = 4: 6
5,5
5,6
5,7
5,8
5,9

Number of operations for i = 5: 5
6,6
6,7
6,8
6,9

Number of operations for i = 6: 4
7,7
7,8
7,9

Number of operations for i = 7: 3
8,8
8,9

Number of operations for i = 8: 2
9,9

Number of operations for i = 9: 1

Number of total operations: 55