Time_Complexity.txt

The actual input on which the irations depends is total number of credit hours for each subject for each teacher for each class so,
n= Total Number of credit hours in whole input.
so the complexity of first few lines can maximum n. Therefore we are starting calculatng inside while loop which is runing n times.
Time_Table_Generator(Class_Details[][][], Available_Slots[45])
I am assuming maximum credit hours of a subject are 5
 
	
	Final_Time_Table[][Available_Slots.length]
	Teachers[][45];
	for class in dim1 of classDetails
		Lecture_NO = 1;
		int Total_Credit_Hours =0;
		for(i=0; i<Class_Details[class].length_of_dim2; i++)

			Total_Credit_Hours +=Class_Details[class][i][4]
		
		while(Total_Credit_Hours=>1)									
			For each teacher in dim2 of classDetails[class]						...............n+1
				Teacher_No=cd[class][teacher][2];						...............n
				if(availble_slots[lectureNo]==0])						.............n
				
					lecture_No++;								.............n
				
				else if(Teachers[Teacher_No][lecture_No]==NULL && Class_Details[class][teacher][4]!=0 && Class_Details[class][teacher][5]=0)      ............n
				
					Final_Time_Table[class][lecture_No]==Class_Details[class][teacher][1]+Class_Details[class][teacher][3];			  ............n
					Teachers[Teacher_No][ln]=class+Class_Details[class][teacher][3];
					Class_Details[class][teacher][4]--;
					Total_Credit_Hours--;
					lecture_No++;														   ............n
				
				else if(Class_Details[class][teacher][5]=1 && Class_Details[class][teacher][4]!=0 )						 .............n
				
					Allow=1;
					for(i=lecture_No; i<(lecture_No+Class_Details[class][teacher][4]); i++)                                                  ............ n*(5+1)
					
						if(teacher[Teacher_No][i]==!NULL or Available_Slots[i]!==1)                                                       ............n*(5) 
						
							Allow=0;
							break;
						
					
					if(Allow)														..............n
					
								for(i=lecture_No; i<(lecture_No+Class_details[class][teacher][4]); i++)                             ...........n*(5+1)
								
									Final_Time_Table[class][i]==Class_details[class][teacher][1]+Class_Details[class][teacher][3];    ........n(5)
									Teachers[Teacher_No][i]=class+Class_Details[class][teacher][3];
									Class_details[class][teacher][4]--;
									Total_Credit_Hours--;
									lecture_No++;
								
IF we add all of them they will be kn .... k is constant so the complexity is O(n)
We know that n depend on number of classes, number of teachers for each class and number of credit hours, if we say them p, q r respectively then
Complexity is O(p*q*r)
					
							
Minimum_Room_Allocator(Final_Time_Table[n][Available_Slots.length])

	for(Room_No=1; Room_No<=Final_Time_Table.dim1_length; Room_No++)
	
		for(lecture_No=1; lecture_No<=Final_Time_Table.dim2_length; lecture_No++)
		
			for(i=Room_No; i<= Final_Time_Table_dim1_length; i++)
			
				if(Available_Slot[lecture_No]=1 && (Final_Time_TableT[i][Lecture_No] not contain "R. No . " ) && 
				Final_Time_Table[i][lecture_No]!=null && Final_Time_Table[i][lecture_No-1]!=Final_Time_Table[i][lecture_No])
					for(int j=lecture_No; j< Final_Time_Table.dim2_length; J++)	
					
						Final_Time_Table[i][j]+= "R. No . "+Room_No;
						if(Final_Time_Table[i][j]!=Final_Time_Table[i][j+1])
						
							break;
						
					break;
In this fuction also 3rd loop is executing n( hours) times and next loop can maximum be the number of classes and third one can run maximum five times.
We can assume that the number of classes is logn So the complexity is O(nlogn)
We know that hours is a function of classes and and total no. of weakly lectures represented by p and q respectively.
So Complexity O(p*q*log(P*q))