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153. Find Minimum in Rotated Sorted Array

Problem Statement

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

  • [4,5,6,7,0,1,2] if it was rotated 4 times.
  • [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

Solution
Initial Approach

// if left is smaller and right is smaller -> right is the answer
// if left is smaller and right is greater -> call on right
// if left is smaller and current is the last element -> first element is the answer
- concern. if the array is rotated more than n+1 times, this would fail

Approach 2
  • low,high pointers
    • if midValue > highValue -> update lowIndex to midIndex+1
    • if midValue <= highValue -> update highValue to mindIndex;
    • continue till low < high
    • return value for nums\[low]
class Solution {
    public:
        int findMin(vector<int>& nums) {
            int low = 0;    
            int high = nums.size()-1;
            int highValue;
            int midIndex, midValue;
            while(low < high){
                midIndex = (low + high)/2;
                midValue = nums[midIndex];
                highValue = nums[high];

                if(midValue > highValue)
                    low = midIndex+1;
                else
                    high = midIndex;
            }
            return nums[low];
        }
};

class Solution {
public:
    int findMin(vector<int>& nums) {
        int low = 0;
        int high = nums.size()-1;
        int ans = INT_MAX;

        while(low <= high){
            int mid = low + (high-low)/2;
            if(nums[mid] < ans)
                ans = nums[mid];
            if(nums[low] <= nums[mid]){ // left side of the array is sorted
                // since left is sorted, we update min
                ans = min(ans,nums[low]);
                low = mid+1;
                // now we have smallest value from sorted array in ans (left->mid)
            }else{ // answer not in left since left > mid
                ans = min(ans, nums[mid]);
                high = mid-1;
            }
        }

        return ans;
    }
};