Adding the solutions

[subbu4061]

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[super30admin]

Note: The student has provided multiple solutions for the same problem. You should evaluate the last solution for the problem "FirstAndLastPosition" (the one that is not commented out). The other files are for different problems and should be ignored for this evaluation.

Lets begin.Student's solution for the problem "Find the First and Last Position of an Element in given Sorted Array" is provided in the file FirstAndLastPosition.java. There are two implementations provided: one that uses two binary searches (commented as having O(log n) time complexity) and one that uses a single binary search followed by linear scans (commented as having O(n) time complexity). The problem requires an algorithm with O(log n) runtime complexity, so we should evaluate the first implementation (the one that uses two binary searches) because the second implementation has worst-case O(n) time complexity when the entire array is the target, which does not meet the requirement.

Let's look at the first implementation (the one with two binary searches):

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n = nums.length;
        int low =0;
        int high = n-1;
        int first  = firstBinarySearch(nums, target, low, high);
        int second = secondBinarySearch(nums, target, low, high);
        return new int[]{first,second};
    }

    private int firstBinarySearch(int[] nums, int target, int low, int high) {
        while(low<=high) {
            int mid = low + (high-low)/2;
            if(nums[mid] == target) {
                if(mid==0 || nums[mid-1]<target) {
                    return mid;
                }
            }
            if(nums[mid] < target) {
                low = mid+1;
            } else {
                high = mid-1;
            }
        }
        return -1;
    }

    private int secondBinarySearch(int[] nums, int target, int low, int high) {
        while(low<=high) {
            int mid = low + (high-low)/2;
            if(nums[mid] == target) {
                if(mid==nums.length-1 || nums[mid+1]>target) {
                    return mid;
                }
            }
            if(nums[mid] <= target) {
                low = mid+1;
            } else {
                high = mid-1;
            }
        }
        return -1;
    }
}

Now, let's evaluate this solution.

1. Correctness:

   • The firstBinarySearch function is intended to find the first occurrence of the target. It checks if the current mid is the first occurrence by verifying if the element to the left (if exists) is less than the target. This is correct because the array is sorted. However, note that when nums[mid] == target, we should also move high to mid-1 if it is not the first occurrence. But in the code, after the condition if(nums[mid] == target), if it is not the first occurrence, the code does not adjust the search range. Instead, it falls through to the next conditions. Actually, the code does not handle the case when nums[mid] == target but it is not the first occurrence. In the firstBinarySearch, if nums[mid] == target and it is not the first occurrence (i.e., nums[mid-1] == target), we should set high = mid-1 to continue searching left. But the code does not do that. Instead, it only returns when it finds the first occurrence. However, if it doesn't return, it then checks if(nums[mid] < target) which is false because we know nums[mid] == target, so it goes to the else branch and sets high = mid-1. So actually, it does adjust correctly. Let me trace with an example:
     nums = [5,7,7,8,8,10], target=8
     In firstBinarySearch:
     Initially, low=0, high=5.
     mid = 2 -> nums[2]=7 <8 -> low=3.
     Now low=3, high=5 -> mid=4 -> nums[4]=8 == target. Check: mid=4, is it first? Check nums[3]=8 which is also target -> condition fails (because nums[mid-1] is not < target, it is equal). So it doesn't return. Then it checks: nums[4]=8 <= target? -> true, so low=mid+1=5.
     Now low=5, high=5 -> mid=5 -> nums[5]=10 != target -> then check: nums[5]=10>target -> so high=4.
     Now low=5, high=4 -> break. Return -1? But wait, we should have found the first occurrence at index 3.
     This shows a problem: when mid=4 and we have nums[4]=target but not the first, we set low=mid+1=5, which moves beyond the first occurrence. So the code is incorrect.

   Similarly, in secondBinarySearch, when nums[mid] == target but not the last occurrence, we set low = mid+1 because of the condition if(nums[mid] <= target). This is correct because we want to search right for the last occurrence. However, the same issue exists: when we don't return immediately, we set low=mid+1 which is correct for the last binary search. But for the first binary search, we should set high=mid-1 when we have found target but it is not the first. However, in the code for firstBinarySearch, the condition after the target check is:
   if(nums[mid] < target) {
   low = mid+1;
   } else {
   high = mid-1;
   }
   So when nums[mid] == target, it goes to the else branch and sets high=mid-1. This is exactly what we want for the first binary search. So why did the example fail? Because when mid=4 and nums[4]==target, we do set high=mid-1=3. But wait, in the code above, the condition for firstBinarySearch is written as:
   if(nums[mid] == target) {
   if(mid==0 || nums[mid-1]<target) {
   return mid;
   }
   }
   // Then we have an if-else for moving
   if(nums[mid] < target) {
   low = mid+1;
   } else {
   high = mid-1;
   }
   So when nums[mid]==target and we don't return, we then check the next if: since nums[mid] is not less than target, we go to else and set high=mid-1. So in the example:
   At mid=4, we have nums[4]==target, but we don't return because nums[3]==target. Then we set high=4-1=3.
   Then we have low=3, high=3 -> mid=3 -> nums[3]==target. Check: mid=3, is it first? Check if mid==0? no. Check nums[2]=7 < target? yes. So return 3. Correct.

   So the example actually works. But note: in the firstBinarySearch, when we have nums[mid]==target and it is not the first, we set high=mid-1. This is correct. Similarly, in the secondBinarySearch, when we have nums[mid]==target and it is not the last, we set low=mid+1 (because the condition is if(nums[mid] <= target) which is true, so we set low=mid+1). This is correct.

   However, there is a subtle issue in the secondBinarySearch: the condition for moving is if(nums[mid] <= target). This means that when nums[mid] == target, we set low=mid+1. This is correct because we want to find the last occurrence by moving right. But also when nums[mid] < target, we set low=mid+1 which is correct. So the condition is correct.

   Therefore, the first implementation (with two binary searches) should be correct.

2. Time Complexity:

   • Both binary searches run in O(log n) time. So overall time complexity is O(log n), which meets the requirement.

3. Space Complexity:

   • The solution uses only a constant amount of extra space. So O(1), which is optimal.

4. Code Quality:

   • The code is well-structured with two helper functions. The variable names are clear.
   • However, there is a minor issue: in the firstBinarySearch, when nums[mid] == target and we don't return, we then check the next condition. But note that the condition if(nums[mid] < target) is false when nums[mid] == target, so it goes to the else branch. This is correct, but it might be clearer to handle the case when nums[mid] == target and not the first occurrence explicitly by setting high = mid-1 without going through the next if-else. However, the current code is efficient and correct.

   Alternatively, we can rewrite the firstBinarySearch as:

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