Binary-search-2 problems solution

[Simranb10]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (problem1Solution.java)

Your solution is well-written and correctly implements the required algorithm. Here are a few points for improvement:

1. Condition Clarity: In firstSearch, the condition nums[mid] > nums[mid-1] is correct but might be slightly confusing. It is equivalent to checking if the previous element is not the same as the current one, which is the standard way to find the first occurrence. However, the reference solution uses nums[mid - 1] != target, which is more explicit. Similarly, in lastSearch, nums[mid] < nums[mid+1] is correct but could be written as nums[mid + 1] != target for clarity.

2. Edge Cases: Your solution handles edge cases such as an empty array (when n=0) correctly because the initial call to firstSearch with low=0 and high=-1 will not enter the while loop and return -1.

3. Efficiency: The solution is efficient. However, note that in lastSearch, you start the search from first to n-1, which is a good optimization because the last occurrence must be at or after the first occurrence.

4. Code Comments: The comments at the top are good, but consider adding comments within the methods to explain the conditions, especially for someone reading the code for the first time.

Overall, your solution is correct and efficient. Keep up the good work!

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (problem2Solution.java)

Let's think step by step before we give our evaluation.

First, let's understand the problem: We need to find the minimum element in a rotated sorted array with unique elements. The algorithm must run in O(log n) time.

The reference solution uses binary search. The key ideas:

1. If the current segment [low, high] is sorted (i.e., nums[low] <= nums[high]), then the minimum is nums[low].
2. Otherwise, compute mid. Check if mid is the minimum by comparing with its neighbors (if it exists). The condition: (mid==0 or nums[mid] < nums[mid-1]) and (mid==n-1 or nums[mid] < nums[mid+1]).
3. If the left part is sorted (nums[low] <= nums[mid]), then the minimum must be in the right part (because the array is rotated), so set low = mid+1.
4. Otherwise, the right part is sorted, so the minimum must be in the left part, so set high = mid-1.

Now, let's look at the student's solution:

The student also uses binary search. The code structure is similar.

Let's check step by step:

1. The student initializes low=0, high=n-1, and result=Integer.MAX_VALUE (which is a fallback, but might not be needed if the array is non-empty? The constraints say n>=1, so it's safe).

2. The while loop condition: low<=high.

3. First check: if nums[low] <= nums[high], then return nums[low]. This is correct because if the segment is sorted, the first element is the minimum.

4. Then compute mid = low + (high-low)/2.

5. Then the student has a print statement (which should be removed in production code, but for learning it's okay).

6. Then the student checks: if(mid>0 && nums[mid] < nums[mid-1] && nums[mid] < nums[mid+1]) return nums[mid]. This is almost correct, but there is a potential issue: what if mid is at the last element? Then mid+1 would be out of bounds. Similarly, the reference solution handles this by checking if mid==0 or mid==n-1. The student's condition requires mid>0 (so safe for mid-1) but does not check for mid+1. So if mid is the last element (high), then mid+1 is beyond the array. This will cause an ArrayIndexOutOfBoundsException.

For example, consider an array of length 1: [1]. Then low=0, high=0. The first condition (nums[low]<=nums[high]) is true, so it returns 1. So no problem.

But consider a case where the minimum is at the last element. For example, [2,1] (rotated once).

• Initially, low=0, high=1. nums[0]=2, nums[1]=1 -> not sorted (2<=1? false). So we compute mid = 0 + (1-0)/2 = 0.
• Then we check: mid>0? 0>0 -> false. So we skip the condition.
• Then check: if nums[low] <= nums[mid] -> nums[0]=2 <= nums[0]=2 -> true. So low = mid+1 = 1.
• Now low=1, high=1. Then we check: nums[low]=1 <= nums[high]=1 -> true, so return 1. So it works.

But what if the minimum is at the last element and the array is larger? Actually, the condition that checks for the mid being the minimum is only triggered when mid>0 and both neighbors are greater. So if the minimum is at the last index, we might not catch it with that condition. However, in the binary search, we eventually get to a segment that is sorted (like [1] in the above) and return.

But the problem is when mid is the last element and we try to access mid+1. For instance, consider an array [3,1,2] (rotated: the minimum is 1 at index1). But let's see if we ever get mid at the last element and try to check mid+1? Actually, in the condition, we have mid>0, which is true for mid=2 (if we have index2). But then we check nums[mid] < nums[mid+1] -> but mid+1 is beyond. So if mid is the last element (index n-1), then the condition will throw an exception.

How can mid be the last element? For example, if we have an array [2,3,1].

• low=0, high=2. nums[0]=2, nums[2]=1 -> not sorted. mid = (0+2)/2 = 1.
• Check: mid>0 -> true. Then check nums[1]=3 < nums[0]=2? -> false. So the condition fails.
• Then check: nums[low]=2 <= nums[mid]=3 -> true, so low = mid+1 = 2.
• Now low=2, high=2. Then we check: nums[2]=1 <= nums[2]=1 -> true, so return 1. So it works.

But what if the minimum is at the last index and we have a larger array? Actually, the condition that checks for the minimum at mid is only applied when mid is not at the boundaries? Actually, the student's condition requires mid>0 and also checks mid+1. So if mid is the last index (n-1), then mid>0 is true (if n>1), but then we access mid+1 which is out of bounds. So the code will crash when mid is the last index and we try to check nums[mid+1].

When does mid become the last index? For example, when low = high-1 and we compute mid = low + (high-low)/2 = low. So if we have low = n-2 and high = n-1, then mid = n-2. Then we might set low = mid+1 = n-1, so then we have low=high=n-1 and return. So mid is never n-1 in the binary search? Actually, in the while loop, we have low<=high. The mid is computed as the floor of (low+high)/2. So if low and high are both n-1, then mid = n-1. But in that case, the first condition (nums[low]<=nums[high]) is true because it's the same element, so we return. So we never get to the condition check for mid when mid is the last index and the segment is not sorted? Actually, if the segment has one element, we return at the first condition.

But what if the segment has two elements? For example, [5,4] (rotated: the minimum is 4 at index1).

• low=0, high=1. Check: nums[0]=5 <= nums[1]=4? -> false.
• mid = (0+1)/2 = 0.
• Then check: mid>0? false -> skip the condition.
• Then check: nums[low]=5 <= nums[mid]=5 -> true, so set low = mid+1 = 1.
• Now low=1, high=1. Then we check: nums[1]=4<=nums[1]=4 -> true, return 4. So it works.

So when does mid become the last index? Actually, it seems that mid is never the last index when the segment has more than one element and is not sorted? Because if the segment has more than one element, then high>low, so mid is always strictly less than high? For example, if low=2, high=3, then mid=2. So mid is never equal to high when high>low. The only time mid equals high is when low==high, but then we return at the first condition.

Therefore, the student's condition "if(mid>0 && nums[mid] < nums[mid-1] && nums[mid] < nums[mid+1])" is actually safe because when mid is computed, we always have low<=mid<high (unless low==high, but then we return early). Why? Because mid = low + (high-low)/2. Since (high-low)/2 is integer division (floor), we have mid >= low and mid < high when low<high. So mid is always less than high when low<high. Therefore, mid+1 is at most high, which is within the array. So the condition is safe.

Wait, let me verify: if low=0, high=1: mid=0 -> then mid+1=1 which is within.
If low=0, high=2: mid=1 -> then mid+1=2 which is within.
If low=1, high=2: mid=1 -> mid+1=2 which is within.
So indeed, when we have at least two elements (so low<high), mid is always strictly less than high. Therefore, mid+1 is always <= high, which is within the array. So the student's condition is safe.

But what about the left neighbor? The condition requires mid>0, so we avoid mid-1 when mid=0. So it's safe.

So the condition is correct.

Now, the rest of the code: if the left part is sorted (nums[low] <= nums[mid]), then we set low=mid+1. Otherwise, we set high=mid-1. This is

VERDICT: NEEDS_IMPROVEMENT

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Find the Peak Element (problem3Solution.java)

Your solution is on the right track by using binary search to achieve logarithmic time complexity. However, there is a critical error in your code: when mid is the last index of the array, accessing nums[mid+1] will cause an ArrayIndexOutOfBoundsException. You should check if mid is at the end before comparing with the next element, just like you do for the left neighbor. For example, in the condition (mid == nums.length-1 || nums[mid] > nums[mid+1]), the right-hand side is safe because if mid is the last element, the condition short-circuits and does not evaluate nums[mid+1].

Also, consider the case when the array has only one element. Your code should handle that without errors.

Additionally, for compatibility with online judges, it's important to name your class Solution (unless otherwise specified) and ensure the method signature matches exactly. Your class is named problem3Solution, which might not be accepted.

Here is a corrected version of your code:

class Solution {
    public int findPeakElement(int[] nums) {
        int low = 0;
        int high = nums.length - 1;
        
        while (low <= high) {
            int mid = low + (high - low) / 2;
            boolean greaterThanLeft = (mid == 0) || (nums[mid] > nums[mid - 1]);
            boolean greaterThanRight = (mid == nums.length - 1) || (nums[mid] > nums[mid + 1]);
            
            if (greaterThanLeft && greaterThanRight) {
                return mid;
            } else if (mid < nums.length - 1 && nums[mid + 1] > nums[mid]) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return -1;
    }
}

This version explicitly checks the boundaries and avoids array index errors. Keep up the good work on using binary search, but always remember to test edge cases like empty arrays, single-element arrays, and the first and last elements.

VERDICT: NEEDS_IMPROVEMENT