81. Search in Rotated Sorted Array II

Problem Statement

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

Intuition

Approach:

            else if(nums[low]==nums[mid] and nums[mid]==nums[high])
                low++; high--;

Only this addition to
1 1 1 1 1 0  1 1 1 1 1 
To takle this edge case

Links

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

Video Links

https://www.youtube.com/watch?v=w2G2W8l__pc&ab_channel=takeUforward

Approach 1:

C++

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int low=0, high=nums.size()-1;

        while(low<=high){
            int mid=low+(high-low)/2;
            if(nums[mid]==target)   return true;           // 1 0 1 1 1
            else if(nums[low]==nums[mid] and nums[mid]==nums[high])
                low++; high--;
            
            else if(nums[low]<=nums[mid]){
                if(nums[low]<=target and target<=nums[mid])
                    high=mid-1;
                else
                    low=mid+1;
            }
            else{
                if(nums[mid]<=target and target<=nums[high])
                    low=mid+1;
                else
                    high=mid-1;
            }
        }

        return false;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++