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Link for the Problem – Find First and Last Position of Element in Sorted Array– LeetCode Problem
Find First and Last Position of Element in Sorted Array– LeetCode Problem
Problem:
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
Find First and Last Position of Element in Sorted Array– LeetCode Solutions
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { const int l = lower_bound(begin(nums), end(nums), target) - begin(nums); if (l == nums.size() || nums[l] != target) return {-1, -1}; const int r = upper_bound(begin(nums), end(nums), target) - begin(nums) - 1; return {l, r}; } };
class Solution { public int[] searchRange(int[] nums, int target) { final int l = firstGreaterEqual(nums, target); if (l == nums.length || nums[l] != target) return new int[] {-1, -1}; final int r = firstGreaterEqual(nums, target + 1) - 1; return new int[] {l, r}; } // find the first index l s.t A[l] >= target // return A.length if can't find private int firstGreaterEqual(int[] A, int target) { int l = 0; int r = A.length; while (l < r) { final int m = l + (r - l) / 2; if (A[m] >= target) r = m; else l = m + 1; } return l; } }
class Solution: def searchRange(self, nums: List[int], target: int) -> List[int]: l = bisect_left(nums, target) if l == len(nums) or nums[l] != target: return -1, -1 r = bisect_right(nums, target) - 1 return l, r
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