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34. Find First and Last Position of Element in Sorted Array

Description

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

 

Constraints:

  • 0 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • nums is a non-decreasing array.
  • -109 <= target <= 109

Solutions

We can perform two binary searches to find the left boundary and the right boundary.

The time complexity is $O(\log n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

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class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        l = bisect_left(nums, target)
        r = bisect_left(nums, target + 1)
        return [-1, -1] if l == r else [l, r - 1]

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class Solution {
    public int[] searchRange(int[] nums, int target) {
        int l = search(nums, target);
        int r = search(nums, target + 1);
        return l == r ? new int[] {-1, -1} : new int[] {l, r - 1};
    }

    private int search(int[] nums, int x) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

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class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int l = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
        int r = lower_bound(nums.begin(), nums.end(), target + 1) - nums.begin();
        if (l == r) {
            return {-1, -1};
        }
        return {l, r - 1};
    }
};

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func searchRange(nums []int, target int) []int {
    l := sort.SearchInts(nums, target)
    r := sort.SearchInts(nums, target+1)
    if l == r {
        return []int{-1, -1}
    }
    return []int{l, r - 1}
}

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

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impl Solution {
    pub fn search_range(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let n = nums.len();
        let search = |x| {