Find Minimum in Rotated Sorted Array: the idea to remember.

The core insight

Compare the middle value to the rightmost value and keep the half containing the minimum.

Why it works

When nums[mid] > nums[right], the pivot is strictly to the right. Otherwise mid can itself be the minimum.

The mistake to avoid

Use right = mid, not mid minus one, when mid might be the minimum. Distinct values are assumed.

Commented Java solution

Use java.util.* imports and a surrounding solution class. Tree snippets assume the standard TreeNode definition; linked lists use ListNode and Clone Graph uses a Node with val and neighbors; Copy Random List uses Node with val, next and random. Read the input assumptions above.

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

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