581. Shortest Unsorted Continuous Subarray

581. Shortest Unsorted Continuous Subarray (Easy)

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 <=.

Solution 1. Insertion Sort

Naive insertion sort. For nums[i], if its insertion position pos[i] is not the same as i, numbers between pos[i] and i are unsorted. Keep record of the leftmost pos[i] (as lb) and rightmost i (as rb) that pos[i] != i. The answer is rb - lb + 1.

// OJ: https://leetcode.com/problems/shortest-unsorted-continuous-subarray
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
  int findUnsortedSubarray(vector<int>& nums) {
    int lb = INT_MAX, rb = INT_MIN;
    for (int i = 1; i < nums.size(); ++i) {
      int val = nums[i], j = i;
      for (; j - 1 >= 0 && nums[j - 1] > val; --j) nums[j] = nums[j - 1];
      nums[j] = val;
      if (j != i) {
        lb = min(lb, j);
        rb = i;
      }
    }
    return lb == INT_MAX ? 0 : rb - lb + 1;
  }
};