Java - Check if array is sorted descendant

Question

I need to check if array was sorted strictly descendant. I wrote following code

public boolean isSortedDescendant(int [] array){
    if ((array.length == 0) || (array.length == 1)) {
        return true;
    } else {
        for(int i = 0; i < array.length - 1; i++){
            if (array[i] > array[i + 1]) {
                return true;
            }
        }
        return false;
    }
}

But it not working correctly. for

   int[] array2 = {3, 2, 2};

at least. I spend a lot of time for different approaches, but without any luck.

Your code can return true on the first i - how do you know the entire array is correct by only checking the first pair? — Bernhard Barker
– Bernhard Barker, Commented Feb 11, 2018 at 13:02
(The synonym of descending is descendent (which is used where descendant may be - word usage is neither reflexive nor transitive).) — greybeard
– greybeard, Commented Feb 11, 2018 at 13:17

Eran · Accepted Answer · 2018-02-11 13:23:34Z

7

You should only return true after checking all the pair of elements:

public boolean isSortedDescendant(int [] array){
    if ((array.length == 0) || (array.length == 1)) {
        return true;
    } else {
        for(int i = 0; i < array.length - 1; i++){
            if (array[i] <= array[i + 1]) {
                return false;
            }
        }
        return true;
    }
}

edited Feb 11, 2018 at 13:23

answered Feb 11, 2018 at 13:01

Eran

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3 Comments

greybeard Over a year ago

It should be enough to follow for (int i = 1 ; i < array.length; i++) if (array[i-1] < array[i]) return false; with return true; (no for explicit special cases or using else where if always returns).

Dmitry Sokolov Over a year ago

Thanks. but need little fix if (array[i] <= array[i + 1]) {

greybeard Over a year ago

(Oops - just noted strictly in the question: "we" need to include equality as a reason to return false.)

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