Shortest Unsorted Continuous Subarray LT581

Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.

You need to find the shortest such subarray and output its length.

Example 1:

Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

Note:

1. Then length of the input array is in range [1, 10,000].
2. The input array may contain duplicates, so ascending order here means <=.

Idea 1. Similar to Max Chunks To Make Sorted II LT768, find the chunks and merge chunks by updating the end index, start index is the staring point of first chunk, the lengh = end index - start index.

The key point is: The array is considered sorted if the left elements are smaller or equal than the right elements, that is, the maximum of the left subarray is smaller or equal than the minimum of the right subarray.

The unsorted chunks can be found by:

a. store the minValue from right to left

b. stack in ascending order.

Time complexity: O(n)

Space complexity: O(n)

1.a

 class Solution {
     public int findUnsortedSubarray(int[] nums) {
         int sz = nums.length;
         int[] rightMin = new int[sz];
         rightMin[sz-1] = nums[sz-1];

         for(int i = sz-2; i >= 0; --i) {
             rightMin[i] = Math.min(rightMin[i+1], nums[i]);
         }

         int startIndex = -1;
         int endIndex = -1;
         int currMax = Integer.MIN_VALUE;
         for(int i = 0; i + 1 < sz; ++i) {
             currMax = Math.max(currMax, nums[i]);

             if(currMax > rightMin[i+1]) {
                 if(startIndex == -1) {
                     startIndex = i;  // first unsorted chunk
                 }
                 else {
                     endIndex = -1;  // new unsorted chunk, mark endIndex = -1 to merge
                 }
             }
             else if(startIndex != -1 && endIndex == -1){
                 endIndex = i;   // if curMax <= rightMin[i+1], means nums(i+1...end) is sorted
             }
         }

         if(startIndex == -1) {
             return 0;
         }

         if(endIndex == -1) {
             endIndex = sz - 1;
         }

         return endIndex - startIndex + 1;
     }
 }

1.b stack

 class Solution {
     public int findUnsortedSubarray(int[] nums) {
        Deque<Integer> ascendingStack = new LinkedList<>();

         int startIndex = nums.length;
         int endIndex = -1;
         int currMaxIndex = 0;
         for(int i = 0; i < nums.length; ++i) {
             if(currMaxIndex == nums.length || nums[currMaxIndex] < nums[i] ) {
                 currMaxIndex = i;
             }
             while(!ascendingStack.isEmpty() && nums[ascendingStack.peek()] > nums[i]) {
                 startIndex = Math.min(startIndex, ascendingStack.pop());  // unsorted chunk
                 endIndex = -1;
             }
             ascendingStack.push(currMaxIndex);
             if(startIndex != nums.length && endIndex == -1) {
                 endIndex = i;  // the end of unsorted chunk: the current index i
             }
         }

         if(startIndex == nums.length) {
             return 0;   // the array is sorted
         }

         return endIndex - startIndex + 1;
     }
 }

Idea 2. The boundary of unsorted subarray is determined by the leftmost and rightmost elements not in the correct position(the sorted position). Borrow the idea from selection sort, for any pair of integeras (0 < i < j < nums.length), if num[i] > nums[j],  this pair will get swapped to get the right position for the sorted array, since we only require the bounday, there is no need to sort it, just note down the index of elements which mark the boundary of the unsorted subarray. Hence, out of all pairs, the leftmost index i not at it's correct position is the left boundary, the rightmost index j is the right boundary.

Time complexity: O(n2)

Space complexity: O(1)

 class Solution {
     public int findUnsortedSubarray(int[] nums) {
        int startIndex = nums.length;
        int endIndex = -1;

        for(int i = 0; i < nums.length; ++i) {
            for(int j = i; j < nums.length; ++j) {
                if(nums[i] > nums[j]) {
                    startIndex = Math.min(startIndex, i);
                    endIndex = Math.max(endIndex, j);
                }
            }
        }

        if(startIndex == nums.length) {
            return 0;
        }

        return endIndex - startIndex + 1;
     }
 }

Idea 3. Comparing with the correct position in the sorted array, mark down the leftmost index and rightmost index different from the sorted position.

Time complexity: O(nlogn)

Space complexity: O(n)

 class Solution {
     public int findUnsortedSubarray(int[] nums) {
         int[] numsSorted = Arrays.copyOf(nums, nums.length);

         Arrays.sort(numsSorted);

         int startIndex = nums.length;
         int endIndex = -1;
         for(int i = 0; i < nums.length; ++i) {
             if(nums[i] != numsSorted[i]) {
                 startIndex = Math.min(startIndex, i);
                 endIndex = Math.max(endIndex, i);
             }
         }

         if(startIndex == nums.length) {
             return 0;
         }

         return endIndex - startIndex + 1;
     }
 }

Idea 4. The correct position of the minimum element in the unsorted subarray determins the left boundary, the correct position of the maximum element in the unsorted subarray determins the right boundary. Two steps: 1. Find the unsorted subarray 2. find the minimum and maximum

a. stack in ascending order for min, in descending order for maxmum

While traversing over the nums array starting from the begining, pushing elements over the stack if in asecending order, otherwise in a falling slope(unsorted subarray), an element nums[j] smaller than the element on the top of the stack, poping elements in the stack until the elemnts on the top is equal or smaller than nums[j], the last poped element is the correct position for nums[j]. For all the nums[j] not in correct position, the minmum of the correct positions determins the left boundary.

Time complexity: O(n)

Space complexity: O(n)

 class Solution {
     public int findUnsortedSubarray(int[] nums) {
         Deque<Integer> indexStack = new LinkedList<>();

         int startIndex = nums.length;
         int endIndex = -1;
         for(int i = 0; i < nums.length; ++i) {
             while(!indexStack.isEmpty() && nums[indexStack.peek()] > nums[i]) {
                 startIndex = Math.min(startIndex, indexStack.pop());
             }
             indexStack.push(i);
         }

         for(int i = nums.length-1; i >= 0; --i) {
             while(!indexStack.isEmpty() && nums[i] > nums[indexStack.peek()]) {
                 endIndex = Math.max(endIndex, indexStack.pop());
             }
             indexStack.push(i);
         }

         if(startIndex == nums.length) {
             return 0;
         }

         return endIndex - startIndex + 1;
     }
 }

b. without extra space, rising slope starting from the begining, falling slope staring from the end

Time complexity: O(n)

Space complexity: O(1)

C1 class Solution {
     public int findUnsortedSubarray(int[] nums) {
         int startIndex = 0;

         while(startIndex+1 < nums.length && nums[startIndex] <= nums[startIndex+1]) {
             ++startIndex;
         }

         if(startIndex == nums.length-1) {
             return 0;
         }

         int minItem = nums[startIndex+1];
         for(int i = startIndex+1; i < nums.length; ++i) {
             minItem = Math.min(minItem, nums[i]);
         }

         // nums[0..startIndex] is sorted,
         // the correct position is the index of the first element bigger than minItem
         // from 0 to startIndex (left to right)
         for(int i = 0; i <= startIndex; ++i) {
             if(nums[i] > minItem) {
                 startIndex = i;
                 break;
             }
         }

         int endIndex = nums.length-1;
         while(endIndex-1 >= 0 && nums[endIndex-1] <= nums[endIndex]) {
             --endIndex;
         }

         int maxItem = nums[endIndex-1];
         for(int i = endIndex-1; i >= 0; --i) {
             maxItem = Math.max(maxItem, nums[i]);
         }

         // nums[endIndex, nums.length-1] is sorted
         // the correct position of the index of the first element smaller than maxItem
         // from nums.length-1 to endIndex (right to left)
         for(int i = nums.length-1; i>= endIndex; --i) {
             if(nums[i] < maxItem) {
                 endIndex = i;
                 break;
             }
         }

         return endIndex - startIndex + 1;
     }
 }

Find the first unsorted subarray and the minimum element can be combine in one loop to make code more concise.

 class Solution {
     public int findUnsortedSubarray(int[] nums) {
         int minItem = Integer.MAX_VALUE;

         boolean isSorted = true;
         for(int i = 1; i < nums.length; ++i){
             if(nums[i-1] > nums[i]) {
                 isSorted = false;
             }

             if(!isSorted) {
                 minItem = Math.min(minItem, nums[i]);
             }
         }

         if(isSorted) {
             return 0;
         }

         int startIndex = 0;
         for(; startIndex < nums.length; ++startIndex) {
             if(nums[startIndex] > minItem) {
                 break;
             }
         }

         int maxItem = Integer.MIN_VALUE;
         isSorted = true;
         for(int i = nums.length-2; i >= 0; --i) {
             if(nums[i] > nums[i+1]) {
                 isSorted = false;
             }

             if(!isSorted) {
                 maxItem = Math.max(maxItem, nums[i]);
             }
         }

         int endIndex = nums.length-1;
         for(;endIndex >= 0; --endIndex) {
             if(nums[endIndex] < maxItem) {
                 break;
             }
         }

         return endIndex - startIndex + 1;
     }
 }