153. Find Minimum in Rotated Sorted Array

Description

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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

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

Output: 1

Example 2

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

Output: 0

Solutions

Solution 1

/**
 * Time complexity : O(N)
 * Space complexity : O(1)
 */

class Solution {
    public int findMin(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int n = nums.length;
        int i = 1;
        
        while(i < n && nums[i-1] < nums[i]) {
            i++;
        }
        
        if(i >= n) return nums[0];
        
        return nums[i];
    }
}

Solution 2

/**
 * Time complexity : O(logN)
 * Space complexity : O(1)
 */
 
class Solution {
    public int findMin(int[] nums) {
        if(nums == null || nums.length == 0) return -1;
        
        if(nums.length == 1) return nums[0];
        
        int left = 0, right = nums.length-1;
        
        if(nums[left] < nums[right]) return nums[0];
        
        while(left <= right) {
            int mid = left+(right-left)/2;
            
            if(nums[mid] > nums[mid+1]) return nums[mid+1];
            
            if(nums[mid-1] > nums[mid]) return nums[mid];
            
            if(nums[mid] > nums[0]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return -1;
    }
}

Solution 3

/**
 * Time complexity : O(logN)
 * Space complexity : O(1)
 */

class Solution {
    public int findMin(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        
        int left = 0, right = nums.length-1;
        
        while(left <= right) {
            
            if(nums[left] <= nums[right]) {
                return nums[left];
            }
            
            int mid = (left+right)/2;
            if(nums[mid] >= nums[left]) {
                left = mid+1;
            } else {
                right = mid;
            }
        }
        
        return -1;
    }
}

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