Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value

It's quite similar to binary search. The modification is to not to stop when you get the target find. Search for the upper and lower elements to check if they are the same or not.

Here goes the full algorithm,

1. Initialize the result range to be [-1,-1]
2. If array size is 0
   return result;
3. Initialize left = 0, right = n-1;

4. while(left<=right)
       mid=(left+right)/2;
       if(nums[mid]==target) //once found search for range
           set i=mid;
           set j=mid;
           while(i>=left && nums[i]==target)
               Decrease i;
           while(j<=right && nums[j]==target)
               Increase j;
           Result is [i+1,j-1]            		
       else if(nums[mid]<target)
           left=mid+1;
       else
           right=mid-1;
       end if
   End while
   

5. Result stores the starting and ending point of range. If target is not found then the range will be the initial one, [-1,-1]

Let's solve with the first test case,

left = 0
right = 5
mid = 2
a[mid]<target, left = mid+1 = 3

a[mid] == target
so traverse left and right from this
range found to be [3,4]

For the second test case,
It simple comes out of the loop

C++ Implementation:

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

vector<int> searchRange(vector<int>& nums, int target)
{
    int n = nums.size();
    
    vector<int> p(2, -1);
    
    if (n == 0)
        return p;
    
    int left = 0, right = n - 1;
    int mid;
    
    while (left <= right) {
        mid = (left + right) / 2;
        if (nums[mid] == target) { //once found search for range
            int i = mid;
            int j = mid;
            while (i >= left && nums[i] == target)
                i--;
            while (j <= right && nums[j] == target)
                j++;
            p[0] = i + 1;
            p[1] = j - 1;
            return p;
        }
        else if (nums[mid] < target)
            left = mid + 1;
        else
            right = mid - 1;
    }
    return p;
}

int main()
{
    int t;
 
    cout << "Enter number of testcases\n";
    cin >> t;
 
    while (t--) {
        int n;
        
        cout << "Enter length of array\n";
        cin >> n;
        
        cout << "Enter the sorted vector\n";
        vector<int> arr(n);
        for (int i = 0; i < n; i++)
            cin >> arr[i];
        
        int k;
        
        cout << "Enter target no\n";
        cin >> k;
        
        vector<int> result = searchRange(arr, n);
        cout << "range is: [" << result[0] << "," << result[1] << "]\n";
    }
    
    return 0;
}

Output:

Enter number of testcases
2
Enter length of array
6
Enter the sorted vector
5 6 7 8 8 10
Enter target no
8
range is: [3,4]
Enter length of array
6
Enter the sorted vector
5 7 7 8 8 10
Enter target no
6
range is: [-1,-1]