Reference: LeetCode
Difficulty: Medium

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 no duplicate exists in the array.

Example: 

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

Input: [4,5,6,7,0,1,2]
Output: 0

Analysis

Sorting

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public:
  int findMin(vector<int>& nums)
  {
    sort(nums.begin(), nums.end());
    return nums[0];
  }

Time: $O(N\log{N})$
Space: $O(1)$

Brute-Force

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public:
  int findMin(vector<int>& nums)
  {
    int min = nums[0];
    for (auto& val : nums)
    {
      if (val < min)
      {
        min = val;
      }
    }
    return min;
  }

Time: $O(N)$
Space: $O(1)$

Divide & Conquer

Reference: Huahua

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

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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];
  }

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.

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;

    if (nums[mid] >= nums[lo])
    {
      return findMin(nums, mid + 1, hi);
    }
    else
    {
      return findMin(nums, lo, mid);
    }
  }

Iteration:

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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
      {
        hi = mid;
      }
    }
    return nums[lo];
  }

Time: $O(\log{N})$
Space: $O(\log{N})$ or $O(1)$