81. Search in Rotated Sorted Array II(Solution || Leetcode medium || Java)

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

Follow up: This problem is similar to Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

SOLUTION:

class Solution {
public boolean search(int[] nums, int target) {
int start = 0;//determining range for binary search
int end = nums.length — 1;

while(start <= end){//applying binary search
int mid = start + (end — start)/2;

if(nums[mid] == target){//If found target return ture
return true;
}

//This shows that our mid is in the rotated part of array
else if(nums[mid] > nums[start]){

//This condition will check if our target might lies between mid and start or not and if not lie then increase start
if(target >= nums[start] && target < nums[mid]){
end = mid — 1;
}else{
start = mid + 1;
}
}

//If our middle is less than start it means that our middle is in un-rotated part
else if(nums[mid] < nums[start]){

//This condition check if our target lies between mid and end or not and if not then decrease end and move toward start
if(target > nums[mid] && target <= nums[end]){
start = mid + 1;
}else{
end = mid — 1;
}
}
//if nothing happen increase start
else{
start++;
}

}
return false;//if not found target return false
}
}

TIME COMPLEITY :- O(log(n))

SPACE COMPLEXITY :- O(1)

Thank you for reading. If you have any queries then, please let me know in the comment section, I will surely be responsive toward it.