how to find length till specific value in an array in java

Question

In a number of array i need length till max value without using sort and

{6,8,2,7,10,3,1}

is it possible to return max-first minimum is that array(10-2=8)?

It's not clear what you want. In your example `(10-2=8)` where does the 2 comes from ? is 10 the max value of the array ? is 8 from `array[1]` or it's completely different ? do you need the index until the max value (here it should be 5)? — Asoub, Sep 29 '16 at 15:28

score 1 · Answer 1 · edited May 23 '17 at 12:07

1

A simple solution for it is the function Array.short, you can do that:

/*Code not tested*/
Arrays.sort(array);
System.out.println(array[0]); //min value
System.out.println(array[array.length-1]); //max value

You can see more about this function here: Java: Sort an array

Edit

edited May 23 '17 at 12:07

Community

answered Sep 27 '16 at 22:10

Dante

@Rishi Thanks is true I'm through the cell phone, sorry. – Dante Sep 27 '16 at 22:21
@DanteFañaBadia without using sort i need 10-2 not 10-1 – Digs Sep 28 '16 at 00:29
@Digs well you can use System.out.println(array[1]); , but the code will be very specific and useless and if you have 2 "1" doesn't work. – Dante Sep 28 '16 at 00:34
@DanteFañaBadia ik but it should pass all test cases – Digs Sep 28 '16 at 00:36
@Digs use sort algorithm here a example http://www.java2novice.com/java-sorting-algorithms/bubble-sort/ – Dante Sep 28 '16 at 00:38

score 0 · Answer 2 · answered Sep 27 '16 at 22:27

IntSummaryStatistics stats = IntStream.of(intArray).summaryStatistics();
int maxDiff = stats.getMax() - stats.getMin();

This approach has O(n) time complexity, which is better than O(n log n) time complexity of sorting the array.

There are faster O(n) algorithms that just calculate the max diff, but this approach is probably good enough for most purposes.

2 Answers2