Pseudo code:
1 Selection sort (A)
{
2 For i 1 to A.lenght-1
{
3 imin i
4 for j i+1 to A.length
{
5 If (A[j] < A[imin])
Imin j
}
6 Temp A[i]
7 A[i] A[imin]
8 A[imin] Temp
}
}

Question No. 03: The step by stem analysis of the algorithm designed in question 2 is as follow,

The time taken by each statement (step) is given as follows,

Selection sort (A) Cost Times
{
For i 1 to A.length-1 c1 n
{
imin i-------------------- c2 n-1
for j i+1 to A.length
{ c3 n+n-1+…1=n(n+1)/2
If (A[j] < A[imin])
imin j
}
Temp A[i] c4 n-1
A[i] A[imin] c5 n-1
A[imin] Temp c6 n-1
}
}

Total time T(n) can be calculated as follows,

T(n)= c1n+c2(n-1)+c3(n(n+1)/2)+c4(n-1)+c5(n-1)+c6(n-1)
c1n+c2n-c2+c3 (n2+n/2) +c4n-c4+c5n-c5+c6n-c6 collect the like terms
c3n2/2+ (c1+c2+c3/2+c4+c5+c6) n + (-c2-c4-c5-c6)

This is equivalent to a polynomial an2+bn+c
Therefore T (n) = Big O (n2) by taking the highest order term of the equation.
Note: a, b, c in polynomial represents constants
// Ignoring constant terms

T(n) = O (n2)

Pseudo code:
1 Selection sort (A)
{
2 For i 1 to A.lenght-1
{
3 imin i
4 for j i+1 to A.length
{
5 If (A[j] < A[imin])
Imin j
}
6 Temp A[i]
7 A[i] A[imin]
8 A[imin] Temp
}
}

Question No. 03: The step by stem analysis of the algorithm designed in question 2 is as follow,

The time taken by each statement (step) is given as follows,

{
For i 1 to A.length-1 |c1 | n
{
imin i--------------------> c2 n-1
for j i+1 to A.length
{ c3 n+n-1+…1=n(n+1)/2
If (A[j] < A[imin])
imin j
}
Temp A[i] c4 n-1
A[i] A[imin] c5 n-1
A[imin] Temp c6 n-1
}
}

Total time T(n) can be calculated as follows,

T(n)= c1n+c2(n-1)+c3(n(n+1)/2)+c4(n-1)+c5(n-1)+c6(n-1)
c1n+c2n-c2+c3 (n2+n/2) +c4n-c4+c5n-c5+c6n-c6 collect the like terms
c3n2/2+ (c1+c2+c3/2+c4+c5+c6) n + (-c2-c4-c5-c6)

This is equivalent to a polynomial an2+bn+c
Therefore T (n) = Big O (n2) by taking the highest order term of the equation.
Note: a, b, c in polynomial represents constants

// Ignoring constant terms

T(n) = O (n2)

You are required to calculate (Step by Step) the worst case time complexity T(n) of the algorithm designed in Question No. 02.

ANS:
C1 =  (n-1)
C2 = (n-1)+(n-2)+…+1
   =  
C3 = (n-1)
T(n) = (n-1) . C1 +  . C2 +  (n-1)C3
T(n) = an2 + bn + c
Here a,b and c are constants in term of C1, C2 and C3
T(n) = O(n2)

You are required to design (write) the algorithm (Only Pseudo code) determined in Question 1.

Selectionsort (A , n)
{
For (i=0 to n-1)                                             
{
min  =  i      C1                         
For (  j = i+1 to n-1)                                         
{                                                             
C2
}
C3
}
}

Pseudo code:
1 Selection sort (A)
{
2 For i 1 to A.lenght-1
{
3 imin i
4 for j i+1 to A.length
{
5 If (A[j] < A[imin])
Imin j
}
6 Temp A[i]
7 A[i] A[imin]
8 A[imin] Temp
}
}

Question No. 03: The step by stem analysis of the algorithm designed in question 2 is as follow,

The time taken by each statement (step) is given as follows,

Selection sort (A) Cost Times
{
For i 1 to A.length-1 c1 n
{
imin i-------------------- c2 n-1
for j i+1 to A.length
{ c3 n+n-1+…1=n(n+1)/2
If (A[j] < A[imin])
imin j
}
Temp A[i] c4 n-1
A[i] A[imin] c5 n-1
A[imin] Temp c6 n-1
}
}

Total time T(n) can be calculated as follows,

T(n)= c1n+c2(n-1)+c3(n(n+1)/2)+c4(n-1)+c5(n-1)+c6(n-1)
c1n+c2n-c2+c3 (n2+n/2) +c4n-c4+c5n-c5+c6n-c6 collect the like terms
c3n2/2+ (c1+c2+c3/2+c4+c5+c6) n + (-c2-c4-c5-c6)

This is equivalent to a polynomial an2+bn+c
Therefore T (n) = Big O (n2) by taking the highest order term of the equation.
Note: a, b, c in polynomial represents constants
// Ignoring constant terms

T(n) = O (n2)