Find First and Last Position of Element in Sorted Array LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Find First and Last Position of Element in Sorted Array in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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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