O(n) algorithm. No code

  • 0

    From left to right, find the potential start s where nums[s+1]<nums[s]
    From right to left, find the potential end e where nums[e]<nums[e-1]
    Find the min value and max value of the elements between start and end, compare them with the numbers from 0 to s and e to length -1. Update the start and end.
    Return end - start + 1

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