Last updated on October 25th, 2024 at 10:26 pm

Here, We see Search in Rotated Sorted Array II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Search in Rotated Sorted Array II LeetCode Solution

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

1. Search in Rotated Sorted Array II Leetcode Solution C++

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int l = 0, r = nums.size() - 1;
        while (l <= r) {
            while (l < r && nums[l] == nums[l + 1]) {
                l++;
            }
            while (l < r && nums[r] == nums[r - 1]) {
                r--;
            }
            int m = l + (r - l) / 2;
            if (nums[m] == target) {
                return true;
            }
            if (nums[m] > target) {
                if (nums[l] > nums[m] || nums[l] <= target) {
                    r = m - 1;
                } else {
                    l = m + 1;
                }
            } else {
                if (nums[l] <= nums[m] || nums[l] > target) {
                    l = m + 1;
                } else {
                    r = m - 1;
                }
            }
        }
        return false;   
    }
};

2. Search in Rotated Sorted Array II Leetcode Solution Java

class Solution {
    public boolean search(int[] nums, int target) {
        int start = 0, end = nums.length - 1, mid = -1;
        while(start <= end) {
            mid = (start + end) / 2;
            if (nums[mid] == target) {
                return true;
            }
            if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
                if (target > nums[mid] && target <= nums[end]) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            } else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
                if (target < nums[mid] && target >= nums[start]) {
                    end = mid - 1;
                } else {
                    start = mid + 1;
                }
            } else {
                end--;
            }
        }
        return false;        
    }
}

3. Search in Rotated Sorted Array II Leetcode Solution JavaScript

var search = function(nums, target) {
    function findPivot(low, high) {
        if(high - low === 1 && nums[high] < nums[low]) return high;
        if(high - low <= 1) return 0;
        const mid = Math.floor((low + high) / 2);
        if(nums[mid] > nums[high]) return findPivot(mid, high);
        if(nums[mid] < nums[low]) return findPivot(low, mid);
        return Math.max(findPivot(low, mid), findPivot(mid, high));
    }
    function binarySearch(start, end) {
        while(start <= end) {
            const mid = Math.floor((start + end) / 2);    
            if(nums[mid] === target) return true;
            if(nums[mid] > target) end = mid - 1;
            else start = mid + 1;
        }
        return false;
    }
    const minIdx = findPivot(0, nums.length-1)
    return binarySearch(0, minIdx-1) || binarySearch(minIdx, nums.length-1)  
};

4. Search in Rotated Sorted Array II Leetcode Solution Python

class Solution(object):
    def search(self, nums, target):
        left,right = 0,len(nums)-1
        while left <=  right:
	        mid = (left+right) // 2
	        if nums[mid] == target:
		        return True
	        if nums[left] <= nums[mid]:
		        if nums[left] == nums[mid] and mid != left:
			        left+=1
			        continue
		        if nums[left] <= target< nums[mid]:
			        right = mid-1
		        else: left = mid+1
	        elif nums[left] > nums[mid]:
		        if nums[mid] < target <= nums[right]:
			        left = mid+1
		        else: right = mid-1
        return False