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81. Search in Rotated Sorted Array II 👍

  • Time: $O(\log n) \to O(n)$
  • Space: $O(1)$
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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) / 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;
  }
};

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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) / 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;
  }
}

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class Solution:
  def search(self, nums: list[int], target: int) -> bool:
    l = 0
    r = len(nums) - 1

    while l <= r:
      m = (l + r) // 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