How to check if array is already sorted

Question

so how to make such logic

int[] arr = {2, 5, 3};

if (/* arr is sorted */)
    ....
else 
    ...

Its bad that method Array.sort is void

Answer 1

You don't need to sort your array to check if it's sorted. Loop over each consecutive pair of elements and check if the first is less than the second; if you find a pair for which this isn't true, the array is not sorted.

boolean sorted = true;

for (int i = 0; i < arr.length - 1; i++) {
    if (arr[i] > arr[i+1]) {
        sorted = false;
        break;
    }
}

Answer 2

Well you can check it in O(n) worst case linear time. A non-sorted array (assuming you mean sorting in ascending order) will have a trip point. That is at some point arr[i] > arr[i+1]

All you need to do is

boolean is_array_sorted(int arr[]) {
  for(int i=0; i < arr.len-1; i++) {
    if(arr[i] > arr[i+1]) {
       return false;
    }
  }
  return true;
}

Just change the > to < if your array sort is supposed to be descending order

Answer 3

public static boolean isSorted(int[] arr) {
    for (int i = 0; i < arr.length - 1; i++) {
        if (a[i + 1] < a[i]) {
            return false;
        };
    }
    return true;
}

Answer 4

function isArraySorted(arr){
    for(let i=0;i<arr.length;i++){
        if(arr[i+1] && (arr[i+1] > arr[i])){
            continue;
        } else if(arr[i+1] && (arr[i+1] < arr[i])){
            return false;
        }
    }
    return true;
}

this worked for me.

Answer 5

public static <T>
boolean isArraySorted(T[] elements, Comparator<? super T> cmp) {
  int n = elements.length;
  for (int i = 1; i < n; ++i) {
    if (cmp.compare(elements[i-1], elements[i]) > 0) { return false; }
  }
  return true;
}

Answer 6

A shorter version:

[0,1,2,3,4].reduce((a,v) => (a!==false) && (a <= v) ? v : false, -Infinity)
[4,3,1,2,0].reduce((a,v) => (a!==false) && (a >= v) ? v : false, +Infinity)

Be careful, as in some cases it isn't effective because it will loop through entire array without breaking prematurely.

Array.prototype.reduce()