Java Solution, Naive and Optimized Method

  • 2

    The naive way to find minimum value is to treat the array as unsorted array and traverse the whole array. It results in O(n) complexity.

    public class Solution {
        public int findMin(int[] nums) {
            int min = nums[0];
            for(int i = 1; i < nums.length; i++){
                if(min > nums[i]) min = nums[i];
            return min;

    But since the array is partially sorted, we can do binary search to find at what index the rotation started.
    To do this, we divide array "nums" into 2 sub-array bounded by left <= mid < right.
    The rotation is located in either [left, mid] or [mid+1, right] in which the array is not sorted. And then we process the sub-array that is not sorted to find rotation index. The time complexity is O(lg n)

    public int findMin(int[] nums) {
            int left = 0;
            int right = nums.length - 1;
            int mid = 0;
            while(left < right) {
                //the subarray [left, right] is sorted perfectly
                if (nums[left] < nums[right]) return nums[left];
                mid = (left + right)>>1;
                if(nums[mid] >= nums[left]) {
                    //subarray [left, mid] is sorted
                    //but we have checked above that nums[left] is not the smallest element
                    //so continue with subarray [mid + 1, right]
                    left = mid + 1;
                } else {
                    //subarray [mid + 1, right] is sorted
                    //so the rotation is in [left, mid]
                    right = mid;
            return nums[left];

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