[GeeksForGeeks] Minimum length unsorted subarray, sorting which makes the array sorted

Given an array, find the minimum length unsorted subarray. After soring this unsorted subarray,

the whole array is sorted.

 

Example, if the input array is [10, 12, 20, 30, 25, 40, 32, 31, 45, 50, 60], your program should be 

able to find that the minimum length unsorted subarray lies between the indices 3 and 7.

 

Solution 1. O(n * logn) runtime, O(n) space

1. make a copy of the given array.

2. sort the copy.

3. find the first and last indices where the given array and its sorted copy don't match .

 

Solution 2. O(n) runtime, O(1) space

The core idea of this solution is to first find the start index i of the unsorted subarray where arr[i] < arr[i- 1], 

and use arr[i - 1] as the current max value. 

Then keep scaning the rest of the array, as long as arr[i] < arr[i - 1] or arr[i] is smaller than the max value, 

we know that arr[i] is not in its sorted place. Update end index to i.

If arr[i] is bigger than max, update the current max to arr[i].

 

public int[] getMinLenUnsortedSubarray(int[] arr) {
    int[] r = {-1, -1};
    if(arr == null || arr.length <= 1) {
        return r;
    }
    int i = 1;
    for(; i < arr.length; i++){
        if(arr[i] < arr[i - 1]) {
            break;
        }
    }
    if(i == arr.length - 1){
        return r;            
    }
    r[0] = i - 1;
    r[1] = i;
    int max = arr[i - 1];
    for(i++; i < arr.length; i++){
        if(arr[i] < arr[i - 1] || arr[i] < max){
            r[1] = i;
        }
        else if(arr[i] >= max){
            max = arr[i];
        }
    }
    return r;
}