Reference: LeetCode
Difficulty: Hard

Post: 4 Solutions Including Divide-and-Conquer, Binary Search (+ Follow-Up)

Problem

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

Find the minimum element.

You may assume duplicate exists in the array.

Example: 

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Input: [3,4,5,1,2] 
Output: 1

Input: [2,2,2,0,1]
Output: 0

Input: [3,3,3,3]
Output: 3

Input: [3,6,3]
Output: 3

Input: [3,1,3]
Output: 1

Follow-up Questions:

  • Would allow duplicates affect the run-time complexity? How and why?

Analysis

Sorting & Brute-Force as in 153.

Divide & Conquer

Reference: Huahua

In the mid3 case, the result could be calculated in $O(1)$.

Note: Duplicates affect the run-time complexity. Consider the following case:

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public:
  int findMin(vector<int>& nums)
  {
    return findMin(nums, 0, nums.size() - 1);
  }

private:
  int findMin(const vector<int>& nums, int lo, int hi)
  {
    // only one element
    if (lo == hi) 
      return nums[lo];

    if (isSorted(nums, lo, hi)) 
      return nums[lo];
    
    int mid = lo + (hi - lo) / 2;
    int left = findMin(nums, lo, mid);
    int right = findMin(nums, mid + 1, hi);
    return min(left, right);
  }

  bool isSorted(const vector<int>& nums, int lo, int hi)
  {
    return nums[lo] < nums[hi];  // must be < (<= will fail in the case [3, 1, 3])
  }

Time: $O(\log{N})$

  • False: $T(N) = 2T(N/2) = O(N)$ ❌
    • At each level, at least one side could be done in $O(1)$.
  • True: $T(N) = O(1) + T(N/2) = O(\log{N})$

Space: $O(\log{N})$

The basic idea is that if nums[mid] >= nums[lo] we assure that the inflection point is on the right part.

However, unlike the solution in 153, we need 3 cases:

Recursion:

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public:
  int findMin(vector<int>& nums)
  {
    return findMin(nums, 0, nums.size() - 1);
  }

private:
  int findMin(const vector<int>& nums, int lo, int hi)
  {
    // only one element
    if (lo == hi) 
      return nums[lo];
    
    // the subarray is sorted
    if (nums[lo] < nums[hi])
      return nums[lo];
    
    int mid = lo + (hi - lo) / 2;

    // changes here --> 3 cases
    if (nums[mid] > nums[lo])
    {
      return findMin(nums, mid + 1, hi);
    }
    else if (nums[mid] < nums[lo])
    {
      return findMin(nums, lo, mid);
    }
    else  // nums[mid] == nums[lo]
    {
      return min(findMin(nums, mid + 1, hi), findMin(nums, lo, mid));
    }
  }

Iteration:

Invalid!

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public:
  int findMin(vector<int>& nums)
  {
    int lo = 0;
    int hi = nums.size() - 1;
    // stops when there is only one element
    // in other words, it makes sure that there are at least 2 elements
    while (lo < hi)
    {
      if (nums[lo] < nums[hi]) return nums[lo];
      
      int mid = lo + (hi - lo) / 2;
      if (nums[mid] > nums[lo])
      {
        lo = mid + 1;
      }
      else if (nums[mid] > nums[lo])
      {
        hi = mid;
      }
      else
      {
        // ???
      }
    }
    return nums[lo];
  }

Time: $O(\log{N})$

  • In the worst case scenario it would become $O(N)$.

Space: $O(\log{N})$