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
Table of Contents
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
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;
}
};Code language: PHP (php)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;
}
}Code language: PHP (php)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)
};Code language: JavaScript (javascript)Search in Rotated Sorted Array II Leetcode Solution Python
class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: bool
"""
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