Search in Rotated Sorted Array II LeetCode Solution | Easy Approach

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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?

Search in Rotated Sorted Array II Solutions

Time: O(\log n) \to O(n)O(logn)→O(n)
Space:O(1)

C++

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

    while (l <= r) {
      const int m = l + (r - l) / 2;
      if (nums[m] == target)
        return true;
      if (nums[l] == nums[m] && nums[m] == nums[r]) {
        ++l;
        --r;
      } else if (nums[l] <= nums[m]) {  // nums[l..m] are sorted
        if (nums[l] <= target && target < nums[m])
          r = m - 1;
        else
          l = m + 1;
      } else {  // nums[m..n - 1] are sorted
        if (nums[m] < target && target <= nums[r])
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return false;
  }
};

Java

 
 class Solution {
  public boolean search(int[] nums, int target) {
    int l = 0;
    int r = nums.length - 1;

    while (l <= r) {
      final int m = l + (r - l) / 2;
      if (nums[m] == target)
        return true;
      if (nums[l] == nums[m] && nums[m] == nums[r]) {
        ++l;
        --r;
      } else if (nums[l] <= nums[m]) { // nums[l..m] are sorted
        if (nums[l] <= target && target < nums[m])
          r = m - 1;
        else
          l = m + 1;
      } else { // nums[m..n - 1] are sorted
        if (nums[m] < target && target <= nums[r])
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return false;
  }
}

Python

  Will be updated Soonclass Solution:
  def search(self, nums: List[int], target: int) -> bool:
    l = 0
    r = len(nums) - 1

    while l <= r:
      m = l + (r - l) // 2
      if nums[m] == target:
        return True
      if nums[l] == nums[m] == nums[r]:
        l += 1
        r -= 1
      elif nums[l] <= nums[m]:  # nums[l..m] are sorted
        if nums[l] <= target < nums[m]:
          r = m - 1
        else:
          l = m + 1
      else:  # nums[m..n - 1] are sorted
        if nums[m] < target <= nums[r]:
          l = m + 1
        else:
          r = m - 1

    return False

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