O(1):
// Here c is a constant
   for (int i = 1; i <= c; i++) {
        // some O(1) expressions
   }

O(n):
// Here c is a positive integer constant
   for (int i = 1; i <= n; i += c) {
        // some O(1) expressions
   }

O(n^2):
for (int i = 1; i <=n; i += c) {
       for (int j = 1; j <=n; j += c) {
          // some O(1) expressions
       }
   }

O(logn):
for (int i = 1; i <=n; i *= c) {
       // some O(1) expressions
   }

Explanation:
1, c, c^2, c^3, … c^k

c^k <= n
k <= logn


O(sqrt(n)):
for (int i = 1; i*i <=n; i++) {
       // some O(1) expressions
   }

Explanation:
1, 4, 9, 16, .... , k^2

k^2 <= n
k <= sqrt(n)

O(n+m):
for (int i = 1; i <=m; i += c) {
        // some O(1) expressions
   }
   for (int i = 1; i <=n; i += c) {
        // some O(1) expressions
   }
if n==m : O(2n) equivalent to O(n)

//For Exercise:

O(loglog(n)):
// Here c is a constant greater than 1
   for (int i = 2; i <=n; i = pow(i, c)) {
       // some O(1) expressions
   }

Explanation:
 In this case, i takes values 2, 2^c, (2^c)^c = 2^c2, (2^c2)^c = 2^c3, …, 2^C^k

2^C^k <= n
C^k <= logn
k <= loglogn

O(sqrt(n)):
c = 0;
for(int i = 1; c<=n; i++)
{
    c = c+i;
}

Explanation:
i   c
--  --
1   1
2   1+2
3   1+2+3
4   1+2+3+4
.   .
.   .
k   1+2+3+4+....+k

1+2+3+4+....+k <= n
k(k+1)/2 <= n
k^2 <= n
k = sqrt(n)

O(log(n)):
c = 0
for(int i = 1; i<=n; i=i*2)
{
    c = c+1;
}
for(int j = 1; j<=c; j=j*2)
{
    // O(1) operation
}

Explanation:
complexity of 1st for loop : c = O(logn)
complexity of 2nd for loop : k = O(logC)
O(logN + logC)
O(logN + loglogN)

HOME WORK:
while(m!=n)
{
    if(m>n) m = m-n;
    else n = n-m;
}
Ans: O(n)

///Prefix Sum:
    5 1 2 4  1  0  2  3  5  7  1  2
    5 6 8 12 13 13 15 18 23 30 31 33
sum[0] = 0
sum[i] = sum[i-1] + a[i];

ans = sum[R] - sum[L-1];