My professor was trying to get me to understand running time and worst case but I'm still confused. He say look at the for loop and see how many times it iterate and I guess for this code i have below it iterate n number of times. I just not too sure though. It for the Fibonacci sequence.
for (int i = 0; i < t; i++) {
j[i] = q;
int A = q;
q = u;
u = A + q;
}
for (int m = 0; m < b; m++) {
if (j[m] <= b) {
System.out.print(j[m]);
}
}
Yes. The complexity of the code is O(t) as the loop is running t times. Inside loop you are calculating the next fibonacci number and storing them into an array j[].
Next you are printing the content of the array, which is also simple iteration over the array.
I would suggest you to always use relevant and meaningful variable name like n instead of t and b while looping. Also you should name the array properly like int fibonacci[] instead of int j[]. This type of code is always self explanatory.
Related
I assume that if I can't convert my while loop into a for loop, then I don't fully understand the concepts here. Here's my working while loop:
(I am trying to implement a program, which calculates the sum 1+2+3+...+n where n is given as user input.)
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a number: ");
int number = Integer.valueOf(scanner.nextLine());
int sum = 0;
int i = 0;
while (i < number) {
i++;
sum += i;
}
System.out.println(sum);
I was looking at this user's answer here: https://stackoverflow.com/a/36023451/11522261
and tried to implement it, but it's not working.
.....
for (i = 0; i < number; i++){
sum += i;
}
System.out.println(sum);
It looks like the conditions of For Init expression statement and ForUpdate are set right, but I'm having trouble understanding the differences here.
Also, it looks like this exercise is trying to teach loops to solve iterative problems. But I suspect that this isn't helping me practice recursion. Perhaps there is a recursive solution to this problem which would be better. Just thinking out loud here. Thanks!
The difference is that in your while loop, you're incrementing i before adding it to sum, but in your for loop, it's after.
If the while is producing the right result, you can adjust your for by starting at 1 instead of 0 and continuing while <= number instead of < number.
For completeness, here's how your current for loop works:
i = 0
If i < number is not true, exit the loop; otherwise, continue to Step 3
sum += i (i is still 0)
i++
Goto Step 2
On the second pass, i < number is 1 < number so still true if number is greater than 1, so you go to Step 3, do sum += i while i is 1, then continue.
Eventually i < number isn't true anymore, and the loop exits.
For problems like this, the best approach is usually to use the debugger built into your IDE to step through the code statement by statement, looking at the values of variables as you go. That can reveal how things work really well. This article may be helpful: How to debug small programs
Since you are incrementing the value of i before adding it to sum in the while loop, the same thing needs to be done in case of for loop as well. Given below is the implementation using the for loop:
for (i = 0; i++ < number; ){
sum += i;
}
System.out.println(sum);
Use <= instead of <. This will solve Your problem, and make sure you understand why will the following code would work. I would recommand you to use a paper and pencil and start writing down the values of i and sum after every iteration.
for (i = 1; i <= number; i++) {
sum += i;
}
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I am working on a problem. Out of 17 test cases 10 works fine and gives the result in less than a second but in 7 cases it is taking 2 seconds which are beyond the time limit. Following is the code
import java.util.*;
import java.io.*;
class TestClass
{
static PrintWriter wr = new PrintWriter(System.out);
public static void func1(int arr[], int n)
{
int temp = arr[0];
for (int jj = 0; jj < n; jj++)
{
if (jj == (n - 1))
arr[jj] = temp;
else
arr[jj] = arr[jj + 1];
}
}
public static void func2(int arr[], int n, int rt)
{
int count = 0;
for (int a = 0; a < n; a++)
{
for (int b = a; b < n; b++)
{
if (arr[a] > arr[b])
count++;
}
}
if (rt == (n - 1))
wr.print(count);
else
wr.print(count + " ");
}
public static void main(String args[]) throws Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine().trim();
StringTokenizer st = new StringTokenizer(str);
int t = Integer.parseInt(st.nextToken());
for (int i = 0; i < t; i++) //for test cases
{
str = br.readLine().trim();
st = new StringTokenizer(str);
int n = Integer.parseInt(st.nextToken());
int arr[] = new int[n];
str = br.readLine().trim();
st = new StringTokenizer(str);
for (int j = 0; j < n; j++) //to take input of array for each test case
{
arr[j] = Integer.parseInt(st.nextToken());
}
for (int rt = 0; rt < n; rt++) //for number of times circular shifting of array is done
{
func1(arr, n); //circularly shifts the array by one position
func2(arr, n, rt); //prints the number of inversion counts
}
if (i != (t - 1))
wr.println();
}
wr.close();
br.close();
}
}
Can someone suggest how to optimize the code so that it takes less time in execution.
I know BufferReader and PrintWriter takes less time as compared to Scanner and System.out.print. I was using scanner and System.out.print earlier but changed it later in hope of getting less time but it didn't help. Also I did it earlier without the use of func1 and func2 and did all the operations in main only. The time in both the cases remains the same.
I am getting the currect output in all the cases so code is correct, I just need help in optimizing it.
The website you are using acquires questions from past programming competitions. I recognize this as a familiar problem
Like most optimization questions, the preferred steps are:
Do less.
Do the same in fewer instructions.
Don't use functions.
Use faster instructions.
In your case, you have an array, and you wish to rotate it a number of times, and then to process it from the rotated position.
Rotating an array is an incredibly expensive operation, because you typically need to copy every element in the array into a new location. What is worse for you is that you are doing it the simplest way, you are rotating the array one step for every step needing rotation.
So, if you have a 100 element array that needs rotated 45 steps, you would then have (3 copies per element swap) 100 * 45 * 3 copies to perform your rotation.
In the above example, a better approach would be to figure out a routine that rotates an array 45 elements at a time. There are a number of ways to do this. The easiest is to double the RAM requirements and just have two arrays
b[x] = a[(mod(x+45), a.length)]
An even faster "do less" would be to never rotate the array, but to perform the calculation in reverse. This is conceptually the function of the desired index in the rotated array to the actual index in the pre-rotated array. This avoids all copying, and the index numbers (by virtue of being heavily manipulated in the math processing unit) will already be stored in the CPU registers, which is the fastest RAM a computer has.
Note that once you have the starting index in the original array, you can then calculate the next index without going through the calculation again.
I might have read this problem a bit wrong; because, it is not written to highlight the problem being solved. However, the core principles above apply, and it will be up to you to apply them to the exact specifics of your programming challenge.
An example of a faster rotate that does less
public static void func1(int arr[], int shift) {
int offset = shift % arr.length;
int [] rotated = new int[arr.length];
// (arr.length - 1) is the last index, walk up till we need to copy from the head of arr
for (int index = 0; index < (arr.length - 1) - offset; index++) {
rotated[index] = arr[index+offset];
}
// copy the end of the array back into the beginning
for ( int index = (arr.length - 1) - offset; index < arr.length; index++ ) {
rotated[index] = (offset - ((arr.length - 1) - index) - 1);
}
System.arraycopy(rotated, 0, arr, 0, arr.length);
}
This copies the array into its rotated position in one pass, instead of doing a pass per index to be rotated.
The first rule of optimisation (having decided it is necessary) is to use a profiler. This counts how many times methods are invoked, and measures the accumulated time within each method, and gives you a report.
It doesn't matter if a method is slow if you only run it a few times. If you run it hundreds of thousands of times, you need to either make it faster, or run it fewer times.
If you're using a mainstream IDE, you already have a profiler. Read its documentation and use it.
The other first rule of optimisation is, if there's already literature about the problem you're trying to solve, read it. Most of us might have invented bubble-sort independently. Fewer of us would have come up with QuickSort, but it's a better solution.
It looks as if you're counting inversions in the array. Your implementation is about as efficient as you can get, given that naive approach.
for(int i=0; i< array.length; i++) {
int n1 = array[i];
for(int j=i+1; j< array.length; j++) {
n2 = array[j];
if(n1 > n2) {
count++;
}
}
}
For an array of length l this will take ( l - 1) + ( l - 2 ) ... 1 -- that's a triangular number, and grows proportionally to the square of l.
So for l=1000 you're doing ~500,000 comparisons. Then since you're repeating the count for all 1000 rotations of the array, that would be 500,000,000 comparisons, which is definitely the sort of number where things start taking a noticeable amount of time.
Googling for inversion count reveals a more sophisticated approach, which is to perform a merge sort, counting inversions as they are encountered.
Otherwise, we need to look for opportunities for huge numbers of loop iterations. A loop inside a loop makes for big numbers. A loop inside a loop inside another loop makes for even bigger numbers.
You have:
for (int i = 0; i < t; i++) {
// stuff removed
for (int rt = 0; rt < n; rt++) {
// snip
func2(arr, n, rt); //prints the number of inversion counts
}
// snip
}
public static void func2(int arr[], int n, int rt) {
// snip
for (int a = 0; a < n; a++) {
for (int b = a; b < n; b++) {
// stuff
}
}
// snip
}
That's four levels of looping. Look at the input values for your slow tests, and work out what n * n * n * t is -- that an indicator of how many times it'll do the work in the inner block.
We don't know what your algorithm is supposed to achieve. But think about whether you're doing the same thing twice in any of these loops.
It looks as if func1() is supposed to rotate an array. Have a look at System.arrayCopy() for moving whole chunks of array at a time. Most CPUs will do this in a single operation.
Is it possible to have two loops in a function?
public static void reduce(Rational fraction){
int divisorNum = 0;
int n = 2;
while(n < fraction.num){
if(fraction.num % n == 0){
divisorNum = n;
System.out.println("n: " + divisorNum);
n++;
}
}
int divisorDenom = 1;
int m = 2;
while(m<fraction.denom){
if(fraction.denom % m == 0){
divisorDenom = m;
System.out.println("m: " + divisorDenom);
m++;
}
}
}
I'm trying to get the greatest common denominator. I know this is the very long way about doing this problem but I just wanted to try having two loops. When I call this function, only the first loop gets printed and not the second. I originally had an if statement, but seeing that the second loop doesn't execute I figured that I fix this part first.
Here's my other part of the code:
public static void main(String[] args){
Rational fraction = new Rational();
fraction.num = 36;
fraction.denom = 20;
reduce(fraction);
}
Absolutely. There are no limitations
Watch your conditional test = is not quite ==
Based on your edit I suspect fraction.denom is initialized at 1 or 0
Hence you will never get in the second loop
You can have any number of loops in your function :-
1.You can have nested loops;
2.Two loops side by side.
SO,your piece of code is fine enough considering the value of n, until the conditions for loop execution are met :-
public static void ....
while(n<x){
do this
add to counter
}
while(m<x){
do this
add to counter
}
if(y==z){ // NOTE :- Here you have committed mistake, compare using ==, not by =(it will be always true else and your condition will always be met else)
print this
}
Yup. You even can have 3 if you try hard enough.
EDIT: The didactic version:
Loops, as the name suggest, are constructs that allow you to repeat blocks of code several times (post conditional loops -> until certain condition is met keep running, pre conditional loops -> if certain condition is met, keep running). This is often called "iteration". So in a typical for-loop:
for ( int i = 0; i < 10; ++i )
print(array[i]);
You can say you're "iterating" over the array 10 times.
This has nothing to do with functions. You can have several loops inside a function, or functions being called inside loops. As long as you define your "blocks" of code (with begining and ending braces) you do what you think its best.
Yes, there are no limitations when it comes to looping. You could do 1000 while loops if you wanted.
An example here could be doing something like making a square out of *...
Here's an example
for(int i = 0; i < 4; i++)
{
for(int j = 0; j < 4; j++)
{
System.out.print("*");
}
System.out.println();
}
It would look like:
****
****
****
****
Getting value of k right every time, but values in the array m[l] in the last for-loop are all same.
int l,j,k=0;
int m[]=new int[10];
for(l=0;l<len;l++){
for(j=0;j<10;j++){
char a= sub.charAt(j);
k=k*100+(int)a;
}
System.out.println("k=" +k);
m[l]=k;
System.out.println(m[l]);
}
for(l=0;l<len;l++)
System.out.println("outside all loop m[l] = " +m[l]);
This line accomplished nothing the first 9 times it runs in the inner loop:
k=k*100+(int)a;
Its value is just set by the last run of the inner loop, which is when m[l] gets set also. Fixing that should help with whatever you're trying to do.
Problem statement was a bit fuzzy but looks like changing k to long would do the trick.
long k = 0;
int l, j = 0;
long m[]=new long[10];
Arrayindex out of bounds error is showing up and I have particularly no idea why it is hapening. I am trying create 2N tuple objects and trying to put them in a array of size 2N
Tuple[] minuteunit = new Tuple[2*N];
if(!intervals.isEmpty())
{
for(i = 0; i < ((2*N)-1); i = 1+2)
{
minuteunit[i] = new Tuple(intervals.get(i).getBeginMinuteUnit(),"s");
minuteunit[i+1] = new Tuple(intervals.get(i).getEndMinuteUnit(),"e");
}
it is most likely the face you are using i in intervals.get(i), because i is incrementing +2. I would imagine you have N values in intervals and therefore when i >= (N/2) you get an overflow.
Try this:
for(i = 0; i < N; i++)
{
minuteunit[2*i] = new Tuple(intervals.get(i).getBeginMinuteUnit(),"s");
minuteunit[2*i+1] = new Tuple(intervals.get(i).getEndMinuteUnit(),"e");
}
Also, assuming intervals should contain N entries, you could update your intervals.isEmpty() check to:
if(intevals.size() == N)
{
...
for(i = 0; i < ((2*N)-1); i = 1+2)
At each iteration i equals 3, so it never finishes. I think you want to add 2 at each iteration so use:
for(i = 0; i < ((2*N)-1); i = i+2)
You need to make your loop's termination condition i<2*N-2, otherwise the iteration of adding 2 will push i over the end of your array when i is 2*N-2 (which it wi eventually be).
btw, I assume that your iteration of i=1+2 is a typo, and you have actually coded i=i+2
I'm going to take a shot in the dark here and guess that maybe intervals.get(i) is what's actually causing your problem here and not the array itself. Ignoring what appears to be a typo in the increment, the condition for your loop seems to be fine, since if i < 2*N - 1, then i + 1 < 2*N and so there's no out of bound index for the array. That leaves the intervals.get(i) method.