Day 17 Leet code series, today we will be picking the problem Find First and Last Position of Element in Sorted Array (https://leetcode.com/problems/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]

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int start = -1;
        int end = -1;
        int count = 0;
        vector<int> res;
        for(int i=0;i<nums.size();i++){
            if(nums[i] == target){
                count++;
                end = i;
            }
        }
        if(count){
            start = end - count + 1;
        }
        res.push_back(start);
        res.push_back(end); 
        return res;
    }
};

Explaination:

1. Just take start and end pointer.
2. If the nums[i] matches the target keep the count and end will the the end pos of the occurance.
3. Now to find start just subtract count from end and add 1 because indices are from 0.
4. Return the resulted array.