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Search in Rotated Sorted Array II⚓︎

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Description⚓︎

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0 Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3 Output: false

Constraints:

  • 1 <= nums.length <= 5000
  • -10^4 <= nums[i] <= 10^4
  • nums is guaranteed to be rotated at some pivot.
  • -10^4 <= target <= 10^4

Solution⚓︎

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class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int R = nums.size() - 1;
        while (R && nums[R] == nums[0]) R--;
        if (R < 0) return nums[0] == target;

        int l = -1, r = R + 1;
        while (l + 1 < r) {
            int mid = l + r >> 1;
            if (nums[mid] >= nums[0]) l = mid;
            else r = mid;
        }

        if (target >= nums[0]) r = l + 1, l = -1;
        else r = nums.size();

        while (l + 1 < r) {
            int mid = l + r >> 1;
            if (nums[mid] >= target) r = mid;
            else l = mid;
        }
        if (r < nums.size() && nums[r] == target) return true;
        return false;
    }
};

Note: The worst time complexity of this problem is \(O(n)\) when all the numbers in nums are equal.

The elements of this problem are not unique. This means that we cannot directly divide the array into two segments based on the relationship with nums[0], i.e., we cannot find the rotation point by 'bisection'.

Because the nature of 'bisection' is bipartition, not monotonicity. As long as one segment satisfies a certain property and the other segment does not satisfy a certain property, we can use 'bisection'.