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

Problem Statement

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

Intuition

Approach 2:
Find using BS

Find the position of Left side, if left side exists then only right boundary exixts

Then find this target once for left , Take target +1 for right boundary
Then return ans

Links

https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/submissions/1052473508/

Video Links

Approach 1:

C++

class Solution {
public:
    int left_index(vector<int>& nums, int target){
        int low=0, high=nums.size()-1;
        while(low<=high){
            int mid = low+(high-low)/2;

            if(nums[mid] == target){
                if(mid == 0)
                    return mid;
                else if(nums[mid] != nums[mid-1])
                    return mid;
                else
                    high = mid-1;
            }
            else if(nums[mid] > target)
                high = mid-1;
            else
                low = mid+1;
        }

        return -1;
    }

    int right_index(vector<int>& nums, int target){
        int low=0, high=nums.size()-1;
        while(low<=high){
            int mid = low+(high-low)/2;

            if(nums[mid] == target){
                if(mid == nums.size()-1)
                    return mid;
                else if(nums[mid] != nums[mid+1])
                    return mid;
                else
                    low = mid+1;
            }
            else if(nums[mid] > target)
                high = mid-1;
            else
                low = mid+1;
        }

        return -1;
    }

    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ans;
        ans.push_back(left_index(nums,target));
        ans.push_back(right_index(nums,target));

        return ans;
    }
};

Approach 2:

C++

class Solution {
private:
    int lower_bound(vector<int>& nums, int low, int high, int target){
        while(low <= high){
            int mid = (low + high) >> 1;
            if(nums[mid] < target){
                low = mid + 1;
            }
            else{
                high = mid - 1;
            }
        }
        return low;
    }
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int low = 0, high = nums.size()-1;
        int startingPosition = lower_bound(nums, low, high, target);
        int endingPosition = lower_bound(nums, low, high, target + 1) - 1;
        if(startingPosition < nums.size() && nums[startingPosition] == target){
            return {startingPosition, endingPosition};
        }
        return {-1, -1};
    }
};

Approach 3:

C++

Approach 4:

C++