When all elements in the given array are the same and the pivot is either chosen as the first or last element what case of time complexity would occur code example

Example: analysis of quick sort

T(n) = 2*T(n/2) + n                        // T(n/2) = 2*T(n/4) + (n/2)    

        = 2*[ 2*T(n/4) + n/2 ] + n
	= 22*T(n/4) + n + n
	= 22*T(n/4) + 2n                       // T(n/4) = 2*T(n/8) + (n/4)

	= 22*[ 2*T(n/8) + (n/4) ] + 2n
	= 23*T(n/8) + 22*(n/4) + 2n
	= 23*T(n/8) + n + 2n
	= 23*T(n/8) + 3n

	= 24*T(n/16) + 4n
	and so on....

	= 2k*T(n/(2k)) + k*n      // Keep going until: n/(2k) = 1  <==> n = 2k    

	= 2k*T(1) + k*n
	= 2k*1 + k*n
	= 2k + k*n               // n = 2k
	= n + k*n
	= n + (lg(n))*n
        = n*( lg(n) + 1 )
       ~= n*lg(n))