competitive-programmingDec 29, 20211 min read

[LeetCode] 153. Find Minimum in Rotated Sorted Array Explained

Updated: Dec 29, 2021

Binary search

Available in:enko

#Problem

153. Find Minimum in Rotated Sorted Array

#Approach

Use binary search to compare the $nums[l]$ , $nums[m]$ , $nums[r]$ values.

If $nums[l] \le nums[m]$ :
the minimum is to the right of index $m$ .

l <- m+1;

Otherwise:
the minimum is either $nums[m]$ itself or to the left of index $m$ .

r <- m;

#Code

/**
 * written: 2021. 12. 29. Wed. 15:51:07 [UTC+9]
 * jooncco의 mac에서.
 **/

typedef vector<int> vi;

class Solution {
public:
    int findMin(vi &nums) {
        int n= nums.size();
        // size가 1이거나 rotate가 일어나지 않은 경우
        if (n == 1 || nums[0] < nums[n-1]) return nums[0];

        int l= 0, r= n-1;
        while (l < r) {
            int m= l+(r-l)/2;
            // nums[l], nums[m], nums[r] 값을 비교.
            // nums[l] < nums[r]인 경우에 대한 예외처리를 포함
            if (nums[l] <= nums[m] && nums[l] > nums[r]) l= m+1;
            else r= m;
        }
        // l: 최소 index
        return nums[l];
    }
};

#Complexity

Time: $O(\log n)$
Space: $O(1)$