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Is there any difference in time space complexity for these two?
for (int i=0; i<= 100; i++) {
System.out.println("hi");
}
for (int i=100; i>= 0; i--) {
System.out.println("hi");
}
And what if the loop is larger and complex? Like when i <= 1000000
No, there is no difference in space or time complexity between the two.
I'd be curious to hear why you thought there might be.
P.S. Of course, if the actual code is different, then the answer might be different too.
For these simple loops? No. You're just working with a constant string, some of System.out.println()'s overhead, and an iterated primitive.
However, for more complex nested loops, there could be differences, especially if loops are nested with changing inner loop lengths.
No both has same time space complexity.
As such no difference in time complexity i.e O(N).
Deciding on which approach to follow is your design specific. For example if I have to find largest prime factor of a given number I would probably use 2nd approach. So basically it is design(of your intended project) specific.
For this particular example is not. For larger arrays it might be slower on some systems.
Usually systems are optimized for accessing memory sequentially. As long as memory access pattern is predictable both loops have same efficiency but on older machines or different architectures things may be different.
For space complexity is not (i miss-read the question)
As long as you don't break out of the loop with some logic (depending on index variable) there should be no difference in space/complexity.
If you need to iterate on every single value of your index, going in reverse order is exactly like going in forward.
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I am actually new to java libraries. I dont know what this code is doing.
I have been given a task to optimize this. Any help will be useful.
I mean to ask whether str.equals(local) checks for entire string aur for particular index value in the for loop.
public void fnc(String str, int[] ar1)
{
String local = "findnumber";
for(int i=0; i<ar1.length; i++)
{
if(str.equals(local) && ar1[i] * 2 > 10)
{
Integer ip = new Integer(ar1[i]);
ip = ip * 2;
System.out.print(ip.toString());
}
}
}
The big O complexity of the algorithm is O(N) where N is the array size. You cannot improve on that ... for the arguments provided and producing the same output.
There are some things that can be done to improve efficiency though.
Hint: look for a computation that is performed on each loop iteration that could be performed once.
Hint: look for some unnecessary object creation1
Hint: unnecessary use of a reference type.
There are one or two other questionable micro-optimizations, but see if you can spot them without any hints. (I say "questionable" because I suspect that the JIT compiler would do the same optimization itself.)
One final note: the actual speed of the code will be dominated by the print statement, and its ability of the OS to write stuff to (for example) the console. And probably by JVM startup / warmup effects ... unless the fnc method is called many times.
1 - Notwithstanding anything else, new Integer(...) is the wrong way to convert an int to an Integer.
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Code 1
String ccMailAddresses="abc#co.in,xyz#co.in";
String ccMailAddressArray[]=ccMailAddresses.split(",");
for (int index = 0; index < ccMailAddressArray.length; index++) {
System.out.println(ccMailAddressArray[index]);
}
Code 2
String ccMailAddresses="abc#co.in,xy";
for (String ccMailAddress : ccMailAddresses.split(",")){
System.out.println(ccMailAddress );
}
Which is best practice ? Is any performance hit in any one of them ?
Obviously, the second piece of code is shorter and working on a higher level of abstraction - as it is using a for-each loop; instead of that counting-for loop.
In that sense: code2 requires less efforts for the reader to grasp what is going on. So, from that perspective, it is definitely "better" than code1.
Performance wise, there shouldn't be any different between the two.
The code does nearly the same thing. The if-statement in Code1 is useless. In my opinion this is alway true.
The difference is betweeen those two line:
for (int index = 0; index < ccMailAddressArray.length; index++)
Here your code will be executed until your index is bigger than the lenght of your array.
for (String ccMailAddress : ccMailAddresses.split(","))
The code in this loop is executed for every element in your array.
But keep in mind the result is the same!
hope that helps!
No difference between both code:
code1 is traditional for loop , and code2 is for each loop introduced for java collection to iterate.
Code 1
String ccMailAddresses="abc#co.in,xyz#co.in";
String ccMailAddressArray[]=ccMailAddresses.split(",");
if(ccMailAddressArray!=null){
for (int index = 0; index < ccMailAddressArray.length; index++) {
System.out.println(ccMailAddressArray[index]);
}
}
In this example you make an unnecessary check against ccMailAddressArray because split() will never return null. The for loop here provides you an index. This can be useful sometimes e.g. if you have a color array that has a color stored for each element. Then you could use color[index] to select the color for each element and draw it somewhere. Please now, that this is a bit more complex than just iterating over all items. So you use this version whenever you need the index.
Code 2
String ccMailAddresses="abc#co.in,xy";
for (String ccMailAddress : ccMailAddresses.split(",")){
System.out.println(ccMailAddress );
}
In this example you do essentially the same. The for-each loop you're using here does the same for you. The only difference is that you don't have the index here. It's just a bit shorter and simpler which makes it also easier to read.
So: Use for-each (Code 2) whenever you can because it is a bit simpler to understand what you are dong. If you use the index based for loop (Code 1) I assume you want to do something with index which also implies that some more complicated things may be involved.
Thanks for an answer stackoverflowgigs.
My question is solved now!. As one of my senior gave me the review comment about code 2 is performance hit by repeatedly calling ccMailAddresses.split(",") in for each loop.
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enter image description here
This is the code screenshot link.
I am edit many time but every time the output is not completely sorted.
Your problem is very simply: you are increasing k far too often within your code!
Meaning: you are already looping with k; so you got:
for (int k=0; k < a.length; k++) { // dont use hardcoded "9" here btw!
and then you have
k++ again in the loop body. You simply dont have to do that!
Meaning: your k is growing faster than it should. But as your loop stops when k reaches 9; you are not processing all elements in your array!
Plus: insertion sort doesn't work with iterating your array once!
You have to keep iterating until all elements are in their place! You really want to study/think more about this algorithm. You, not us that is!
And as said, dont use hard-coded limits for arrays. You already say once that your array should contain 10 elements. From there on, you should be using a.length only! Besides: only use one-letter names for loop counters and such things. "a" is a pretty bad name for an array; why dont you call it "numbers" for example.
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How can I go through several possibilities without exhausting loops in a program that sum of the proper divisors(divisors including 1 but not itself) of the number is greater than the number, but no subset of those divisors sums to the number itself.
For Example:
12=(factors sum)1+2+3+4+6
Atleast Any Subset Sum should be equal to the number i.e 12 in this case.
Question link
:http://www.practice.geeksforgeeks.org/problem-page.php?pid=301
Approach:
Step1 (Doubt)
By looking at the problem I found that its optimal solution will be Dynamic Programming.(Since I don't know that).
Apart from that I thought of the solution like this
For Example 12=1,2,3,4,6(divisors)
So,Solving it through like this with 1 starting having all permutations: 1+2+3+4+6 or 1+3+4+6 or 1+4+6 or 1+6(checking at each step whether it is <=12.
Similarly,I checked starting with 2: 2+3+4+6 or 2+4+6(I got and moved out of the loop)
The solution I am thinking here is very long.I have to put seperate loop for each number and also I am saving the divisors in a string.
Can anyone give me "Hint" how to start the problem without the Dynamic Programming(Dp) approach as I am learning Dp these days.
I assume you have a method
int[] getDivisors(int number)
the hard task here is to evaluate the possible sums.
An easy but long way would be, to iterate over all of those like that.
Think of such a sum as a binary code, 1 if its in sum, 0 if its not.
Now you can write a for-loop which, from 0...0 to 1...1 (in binary) with #1s = divisors.length;
Maybe a better way would be recursion methods. Like:
boolean hasSameSum(int number, Stack<Integer> divisors, int sum){
if(sum==number)
return false;
if(divisors.isEmpty())
return true;
int div = divisor.pop();
return hasSameSum(number, divisors, sum)&&hasSameSum(number, divisors, sum+div);
}
But honestly, that's not a programming problem, but more a mathematical approach problem :-)
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The Boolean array is initialized to true. I need to know how to turn every third value to false and also going through the array over and over. It is essentially duck duck goose without the randomness.
This sounds like a beginners programming exercise, so I'm going to just give you a hint or two:
Go back to your textbook, lecture notes, tutorial and reread the stuff on for loops. Focus on the old style ones.
Think about how to write a for loop that steps through integer index values in the pattern that you require.
Re "... going through the array over and over" - not sure what you mean, but maybe the hint for this is to think about using nested loops; i.e. a loop inside another loop.
But the most important advice is to try and work this out for yourself.
Well I'm really not sure what you mean by going through the array over and over but the following code will turn every third value to false.
for (int i = 0; i < myVar.length; i++) {
if (i % 3 == 0) {
myVar[i] = false;
}
}
Edit: Oops someone beat me to it while I was typing lol.