Algorithm of:
for ( j = 0; j < n; j++ )
{
for ( k = j; k < n; k++ )
{
<some operation>
}
}
will result in a number of iterations of <some operation> given by the expression:
<number of iterations> = n + (n-1) + (n-2) + (n-3) + ........ + (n - n)
Reduce the above series expression to an algebraic expression.
After determining the algebraic expression express the performance in Big O Notation.
No one is going to do the homework.
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