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

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]

Solution : 1

int firstPosition(vector<int>& nums, int target, int low, int high)
    {
        if(high >= low)
        {
            int mid = low + (high-low) / 2;
            if((mid == 0 || target > nums[mid-1]) && nums[mid] == target)
                return mid;
            else if(target > nums[mid])
                return firstPosition(nums, target, mid+1, high);
            else
                return firstPosition(nums, target, low, mid-1);
        }
        return -1;
    }

    int lastPosition(vector<int>& nums, int target, int low, int high, int size)
    {
        if(high >= low)
        {
            int mid = low + (high-low) / 2;
            if((mid == size - 1 || target < nums[mid+1]) && nums[mid] == target)
                {return mid;}
            else if(target < nums[mid])
                return lastPosition(nums, target, low, (mid-1), size);
            else
                return lastPosition(nums, target, (mid+1), high, size);
        }
        return -1;
    }
    
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result;
        int first = firstPosition(nums, target, 0, nums.size()-1);
        int last = lastPosition(nums, target, 0, nums.size()-1, nums.size());
        result.push_back(first);
        result.push_back(last);
        return result;
    }

Solution : 2

vector<int> searchRange(vector<int>& nums, int target) {
	vector<int> result;
	int first=-1, last=-1;

	if(nums.size() > 0)
	{
		for(int i=0; i<nums.size(); i++)
		{
			if(nums[i] == target)
			{
				if(first == -1)
					first = i;
				last = i;
			}
		}
	}
	result.push_back(first);
	result.push_back(last);
	return result;
}

Solution : 3

Iterator lower_bound (Iterator first, Iterator last, const val) 
Iterator upper_bound (Iterator first, Iterator last, const val) 

lower_bound returns an iterator pointing to the first element in the range [first,last) which has a value not less than ‘val’. 
upper_bound returns an iterator pointing to the first element in the range [first,last) which has a value greater than ‘val’. 

vector<int> searchRange3(vector<int>& nums, int target) {
	vector<int> result;
	int first=-1, last=-1;
	first = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
	last = upper_bound(nums.begin(), nums.end(), target) - nums.begin() -1;
	result.push_back(first);
	result.push_back(last);
	return result;
}

Complete Code :

/*
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]
*/

#include <bits/stdc++.h>
using namespace std;

vector<int> searchRange(vector<int>& nums, int target) {
	vector<int> result;
	int first=-1, last=-1;

	if(nums.size() > 0)
	{
		for(int i=0; i<nums.size(); i++)
		{
			if(nums[i] == target)
			{
				if(first == -1)
					first = i;
				last = i;
			}
		}
	}
	result.push_back(first);
	result.push_back(last);
	return result;
}

vector<int> searchRange3(vector<int>& nums, int target) {
	vector<int> result;
	int first=-1, last=-1;
	first = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
	last = upper_bound(nums.begin(), nums.end(), target) - nums.begin() -1;
	result.push_back(first);
	result.push_back(last);
	return result;
}

int firstPosition(vector<int>& nums, int target, int low, int high)
{
	//cout << "low: " << low << "high: " << high <<"mid: " << endl;
	if(high >= low)
	{
		int mid = low + (high-low) / 2;
		//cout << "low: " << low << " high: " << high <<" mid: " << mid << endl;
		if((mid == 0 || target > nums[mid-1]) && nums[mid] == target)
			return mid;
		else if(target > nums[mid])
			return firstPosition(nums, target, mid+1, high);
		else
			return firstPosition(nums, target, low, mid-1);
	}
	return -1;
}

int lastPosition(vector<int>& nums, int target, int low, int high, int size)
{
	//cout << "low: " << low << " high: " << high << endl;
	if(high >= low)
	{
		int mid = low + (high-low) / 2;
		//cout << "low: " << low << " high: " << high <<" mid: " << mid << endl;
		if((mid == size - 1 || target < nums[mid+1]) && nums[mid] == target)
			{return mid;}
		else if(target < nums[mid])
			return lastPosition(nums, target, low, (mid-1), size);
		else
			return lastPosition(nums, target, (mid+1), high, size);
		//cout << "mid: " << mid << endl;
	}
	return -1;
}

vector<int> searchRangeUsingBinarySearch(vector<int>& nums, int target) {
	vector<int> result;
	int first = firstPosition(nums, target, 0, nums.size()-1);
	int last = lastPosition(nums, target, 0, nums.size()-1, nums.size());
	result.push_back(first);
	result.push_back(last);
	return result;
}

int main()
{
	vector<int> v1 = {5,7,7,8,8,10};
	int target = 8;
	//vector<int> ret = searchRangeUsingBinarySearch(v1, target);
	//vector<int> ret = searchRange(v1, target);
	vector<int> ret = searchRange3(v1, target);
	for(int i=0; i<ret.size(); i++)
		cout << ret[i] << " ";
	cout << endl;
	return 0;
}