think this way
1)sum = 0;
for( i = 0; i < n; i++)
sum++
this executes n times look like
n-1
---
\
/ 1 = n-1 So T(n) = O(n)
---
i=1
this executes n times
2)sum = 0;
for( i = 0; i < n; i++)
for( j = 0; j < i*i; j++)
sum++
n-1 i^2-1 n-1
--- --- ---
\ \ \
/ / 1 = / i^2-1 = So T(n) = O(n^3)
--- --- ---
i=0 j=0 i=0
let consider loop imitial condition start from 1
3) sum = 0;
for( i = 1; i <= n; i++)
for( j = 1; j <= i*i; j++)
for( k = 1; k <= j; k++)
sum++;
n i^2 j n i^2 n
--- --- --- --- --- ---
\ \ \ \ \ \
/ / / 1 = / / j = / i2(i2+1)/2
--- --- --- --- --- ---
i=1 j=1 k=1 i=1 j=1 i=1
then
n
---
\
/ (i4+i2)/2 =
---
i=1
Sigma n^4 = n*(n+1)*(2*n+1)*(3*n2+3*n-1)/30 =O(n5)
Sigma (n^2) = n(n+1)(2n+1)/6 =O(n3)
so overall O(n5)