81. Search in Rotated Sorted Array II

Description

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?

Solutions

Solution 1: Binary Search

We define the left boundary $l=0$ and the right boundary $r=n-1$ for the binary search, where $n$ is the length of the array.

During each binary search process, we get the current midpoint $mid=(l+r)/2$.

If $nums[mid] \gt nums[r]$, it means that $[l,mid]$ is ordered. At this time, if $nums[l] \le target \le nums[mid]$, it means that $target$ is in $[l,mid]$, otherwise $target$ is in $[mid+1,r]$.
If $nums[mid] \lt nums[r]$, it means that $[mid+1,r]$ is ordered. At this time, if $nums[mid] \lt target \le nums[r]$, it means that $target$ is in $[mid+1,r]$, otherwise $target$ is in $[l,mid]$.
If $nums[mid] = nums[r]$, it means that the elements $nums[mid]$ and $nums[r]$ are equal. At this time, we cannot determine which interval $target$ is in, so we can only decrease $r$ by $1$.

After the binary search ends, if $nums[l] = target$, it means that the target value $target$ exists in the array, otherwise it means it does not exist.

The time complexity is approximately $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array.

Edge case, if there are repeated values, there are two situations to consider: [3 1 1] and [1 1 3 1]. In these situations, when the middle value is equal to the rightmost value, the target value 3 could be either on the left or on the right.

class Solution {
    public boolean search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] > nums[r]) {
                if (nums[l] <= target && target <= nums[mid]) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            } else if (nums[mid] < nums[r]) {
                if (nums[mid] < target && target <= nums[r]) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            } else {
                --r;
            }
        }
        return nums[l] == target;
    }
}

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

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        if not nums:
            return False

        i = 0  # left pointer
        j = len(nums) - 1  # right pointer

        while i <= j:
            mid = (i + j) // 2

            if nums[mid] == target:
                return True

            if nums[i] <= nums[mid]:  # left half ordered, right half not ordered
                if nums[i] <= target <= nums[mid]:
                    j = mid
                else:
                    i += 1
            else:  # right half ordered, left half not ordered
                if nums[mid] <= target <= nums[j]:
                    i = mid
                else:
                    j -= 1

        return False

############

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        l, r = 0, len(nums) - 1
        while l <= r:
            mid = (l + r) >> 1
            if nums[mid] == target:
                return True
            if nums[mid] < nums[r] or nums[mid] < nums[l]:
                if target > nums[mid] and target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1
            elif nums[mid] > nums[l] or nums[mid] > nums[r]:
                if target < nums[mid] and target >= nums[l]:
                    r = mid - 1
                else:
                    l = mid + 1
            else:
                r -= 1
        return False

func search(nums []int, target int) bool {
	l, r := 0, len(nums)-1
	for l < r {
		mid := (l + r) >> 1
		if nums[mid] > nums[r] {
			if nums[l] <= target && target <= nums[mid] {
				r = mid
			} else {
				l = mid + 1
			}
		} else if nums[mid] < nums[r] {
			if nums[mid] < target && target <= nums[r] {
				l = mid + 1
			} else {
				r = mid
			}
		} else {
			r--
		}
	}
	return nums[l] == target
}

function search(nums: number[], target: number): boolean {
    let [l, r] = [0, nums.length - 1];
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] > nums[r]) {
            if (nums[l] <= target && target <= nums[mid]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        } else if (nums[mid] < nums[r]) {
            if (nums[mid] < target && target <= nums[r]) {
                l = mid + 1;
            } else {
                r = mid;
            }
        } else {
            --r;
        }
    }
    return nums[l] === target;
}

81 - Search in Rotated Sorted Array II

81. Search in Rotated Sorted Array II

Description

Solutions

All Problems

All Solutions

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81. Search in Rotated Sorted Array II

Description

Solutions

All Problems

All Solutions