CODEWITHSUNDEEP
javaalgorithm
Mayur Kulkarni
Mayur Kulkarni

Reputation: 1316

Effectively print a long sequence

I've to print a number which is a huge sequence of 5's and 3's (upto 100,000 ints). Instead of storing it in array, I just kept their count in noOfThrees and noOfFives.

For simplicity call this number x. .

Since I have to print the largest number in the sequence, x will be initially 5's and then followed with 3's (I have working logic to print if there's no 5's or no 3's)

To print the number, I am using a for loop like this : 

for(int i=0; i<noOfFives; i++) 
    System.out.print(5);
for(int i=0; i<noOfThrees; i++)
    System.out.print(3);

But if x is a 100,000 long int number, it takes about 4-5sec to print it on console which is not desirable.

My take:

But the problem here is if it's odd, it will again end up printing in steps of 1. How do I effectively print this sequence?

Upvotes: 0

Views: 257

Answers (3)

Andy Turner
Andy Turner

Reputation: 140318

If you think that the number of print calls is the issue, you could reduce it to just 1: put the right number of 3s and 5s into a char array, then print that:

char[] cs = new char[noOfFives + noOfThrees];
Arrays.fill(cs, 0, noOfFives, '5');
Arrays.fill(cs, noOfFives, cs.length, '3');
System.out.print(cs);

Upvotes: 4

He Yuntao
He Yuntao

Reputation: 96

I think you can use loop unrolling, it can reduce execute time of loop.
for example:

    for(int i=0; i<noOfFives; i+=5) {
        System.out.print(5);
        System.out.print(5);
        System.out.print(5);
        System.out.print(5);
        System.out.print(5);
    }

For more details look at https://en.wikipedia.org/wiki/Loop_unrolling

Upvotes: 0

Elemental
Elemental

Reputation: 7476

Your question seems a bit strange - I wonder whether the logical way to make your code faster is to look elsewhere at how these values are calculated.

However I think it would significantly increase your performance to first build the string in memory and then print it.

I also think you should look at Why is printing "B" dramatically slower than printing "#"? which might shed some interesting light on your issue.

Upvotes: 1

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