Well, you've probably already seen the O(n) solution here. If you've ever wondered how it is done, here is a detailed explanation. It will begin with the naive O(nlogn)
solution by sorting, then transition to the O(n)
twopass solution and eventually the O(n)
onepass solution.
I  Naive solution by sorting
The original problem is asking for the shortest subarray that needs to be sorted in order to make the whole input array sorted. Assume nums
is the input array with length n
, nums_sorted
is the sorted version of nums
, nums[i, j]
(both indices inclusive) is such a subarray of the shortest length. Since we only need to sort nums[i, j]
in order to transform nums
into nums_sorted
, all other elements in nums
are left intact, which means the subarray nums[0, i  1]
will be exactly the same as subarray nums_sorted[0, i  1]
, and the subarray nums[j + 1, n  1]
exactly the same as nums_sorted[j + 1, n  1]
. We can also conclude nums[i] != nums_sorted[i]
and nums[j] != nums_sorted[j]
. Otherwise the subarray nums[i, j]
will not be the shortest (for example, if nums[i] == nums_sorted[i]
, sorting nums[i + 1, j]
will also render nums
sorted).
Therefore, the shortest subarray can be constructed as follows: its leftmost index can be obtained by finding the first index at which the two elements of nums
and nums_sorted
are different while its rightmost index can be obtained by finding the last such index.
The straightforward way for finding the two boundary indices would be comparing elements from nums
and nums_sorted
one by one. But we need to construct nums_sorted
from nums
first. This is the following O(nlogn)
time and O(n)
space solution:
public int findUnsortedSubarray(int[] nums) {
int[] nums_sorted = Arrays.copyOf(nums, nums.length);
Arrays.sort(nums_sorted);
int i = 0, j = nums.length  1;
while (i < nums.length && nums[i] == nums_sorted[i]) i++;
while (i < j && nums[j] == nums_sorted[j]) j;
return j  i + 1;
}
II  O(n) time twopass solution
It turns out that the two boundary indices i
and j
can be found in linear time, if we take advantage of the following three properties:

nums[0, i  1]
andnums[j + 1, n  1]
are both sorted. 
nums[i] != nums_sorted[i]
andnums[j] != nums_sorted[j]
. 
nums[i  1] <= min
andmax <= nums[j + 1]
, wheremin
andmax
are the minimum and maximum values of subarraynums[i, j]
.
The first and third properties guarantee that the subarray nums[0, i  1]
will be exactly the same as subarray nums_sorted[0, i  1]
, and the subarray nums[j + 1, n  1]
exactly the same as nums_sorted[j + 1, n  1]
, while the second property ensures that i
will be the first index at which the two elements of nums
and nums_sorted
are different and j
be the last such index.
Since we aim at the shortest subarrays, from the first property alone, we need to find the two longest sorted subarrays starting at index 0
and ending at index n  1
, respectively. Assume the two subarrays are nums[0, l]
and nums[r, n  1]
. If there is overlapping between these two subarrays, i.e.l >= r
, then the whole array is sorted so 0
will be returned. Otherwise, the input array is not sorted. However, we cannot say sorting nums[l, r]
will leave the whole array sorted, because at this moment the third property may not be satisfied.
To guarantee the third property, assume min
and max
are the minimum and maximum values of subarray nums[l, r]
, then we need to decrease l
as long as nums[l] > min
, and increase r
as long as nums[r] < max
. After this is done, it can be shown that the second property will be met automatically, and nums[l + 1, r  1]
will be the shortest subarray we are looking for (that is, i = l + 1
and j = r  1
).
Finding the longest subarrays and the maximum and minimum values of the middle subarray takes onepass. Ensuring the third property requires a second pass. Therefore we have this twopass solution:
public int findUnsortedSubarray(int[] nums) {
int l = 0, r = nums.length  1, max = Integer.MIN_VALUE, min = Integer.MAX_VALUE;
while (l < r && nums[l] <= nums[l + 1]) l++;
if (l >= r) return 0;
while (nums[r] >= nums[r  1]) r;
for (int k = l; k <= r; k++) {
max = Math.max(max, nums[k]);
min = Math.min(min, nums[k]);
}
while (l >= 0 && min < nums[l]) l;
while (r < nums.length && nums[r] < max) r++;
return (r  l  1);
}
III  O(n) time onepass solution
To understand this onepass solution, we need to introduce some equivalent mathematical models for describing a sorted array (assuming in ascending order). Suppose the given array is nums
with length n
, these models are as follows:

nums[k] <= nums[k + 1]
for all0 <= k < n  1
. 
nums[k] == max[k]
for all0 <= k <= n  1
, wheremax[k]
is the maximum value of subarraynums[0, k]
. 
nums[k] == min[k]
for all0 <= k <= n  1
, wheremin[k]
is the minimum value of subarraynums[k, n  1]
.
The first model is the most common one (and probably the most familiar one) while the last two are less common. It's easy to show that the second model is equivalent to the first by noting that for any index k < n  1
, we have max[k] <= max[k + 1]
, then nums[k] = max[k] <= max[k + 1] = nums[k + 1]
. Similar results hold for the third model: nums[k] = min[k] <= min[k + 1] = nums[k + 1]
.
With these models in place, we can show that if indices i
and j
satisfy the following conditions, then nums[i, j]
will be the shortest subarray we are looking for:
i
is the smallest index such thatnums[i] != min[i]
;j
is the largest index such thatnums[j] != max[j]
.
The proof proceeds by showing that the two conditions above are equivalent to the three properties in part II
.
Firstly we will show that the first property in part II
is held true. From condition 1
, we have nums[k] == min[k]
for all 0 <= k < i
. Then nums[k] = min[k] <= min[k + 1] = nums[k + 1]
for all k < i  1
. By definition, nums[0, i  1]
is sorted. Similarly from condition 2
, nums[k] == max[k]
for all j < k <= n  1
. Then nums[k] = max[k] <= max[k + 1] = nums[k + 1]
for all j < k < n  1
. By definition, nums[j + 1, n  1]
is sorted.
Then we will show the third property is satisfied. Let min_m
and max_m
be the minimum and maximum values of subarray nums[i, j]
, respectively, then we have min_m >= min[i] >= min[i  1] = nums[i  1]
and max_m <= max[j] <= max[j + 1] = nums[j + 1]
.
Lastly we will show that the second property is also valid. Note that if the first and third properties are both true, then we know the subarray nums[0, i  1]
will be exactly the same as subarray nums_sorted[0, i  1]
, and the subarray nums[j + 1, n  1]
exactly the same as nums_sorted[j + 1, n  1]
. In this case just suppose we have nums[i] == nums_sorted[i]
and nums[j] == nums_sorted[j]
, let's see what will happen. Since the subarrays nums[i, n  1]
and nums_sorted[i, n  1]
contain exactly the same elements (though the order may be different), then the minimum element of the former will be the same as the latter. Since nums_sorted[i, n  1]
is sorted in ascending order, we will have min[i] = nums_sorted[i] = nums[i]
, which contradicts the assumption that nums[i] != min[i]
. Similarly we can show that nums[j] == nums_sorted[j]
implies nums[j] == max[j]
, which contradicts the assumption that nums[j] != max[j]
.
Finding the smallest index i
such that nums[i] != min[i]
and the largest index j
such that nums[j] != max[j]
can be done in onepass, as shown below. Note that we don't really need arrays to hold values for min[r]
and max[l]
, by taking advantage of the recurrence relation min[r] = Math.min(min[r + 1], nums[r])
and max[l] = Math.max(max[l  1], nums[l])
. Also we initialized the indices i
and j
such that correct results will be returned even if the input array is already sorted (which requires initially j  i + 1 = 0
).
public int findUnsortedSubarray(int[] nums) {
int i = 0, j = 1, max = Integer.MIN_VALUE, min = Integer.MAX_VALUE;
for (int l = 0, r = nums.length  1; r >= 0; l++, r) {
max = Math.max(max, nums[l]);
if (nums[l] != max) j = l;
min = Math.min(min, nums[r]);
if (nums[r] != min) i = r;
}
return (j  i + 1);
}