Search in Rotated Sorted Array II
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# LeetCode #81: Search in Rotated Sorted Array II (C/C++).

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source: https://leetcode.com/problems/search-in-rotated-sorted-array
C/C++ Solution to LeetCode problem 81. Search in Rotated Sorted Array II.

## Problem

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, nums, ..., 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.

## Examples

### 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
• -104 <= nums[i] <= 104
• nums is guaranteed to be rotated at some pivot.
• -104 <= target <= 104

## Solution

Solution is same as in problem #33, the only difference is that numbers can be repeated so:

• While lower index < higher index:
• While the value at the lower index is equal to the next one, move the lower index up.
• While the value at the higher index is equal to the previous one, move the higher index down.
• Perform binary search as in problem #33
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 class Solution { public: int search(vector<int>& nums, int target) { int l = 0; int h = nums.size() - 1; int m; while (l <= h) { while (l < h && nums[l] == nums[l+1]) l ++; while (h > l && nums[h] == nums[h-1]) h --; m = (l + h) / 2; if (nums[m] == target) return true; if (nums[l] <= nums[m]) { if (target <= nums[m] && target >= nums[l]) h = m - 1; else l = m + 1; } else { if (target >= nums[m] && target <= nums[h]) l = m + 1; else h = m - 1; } } return false; } };