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[sundeep4sandy]

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[super30admin]

Find the First and Last Position of an Element in given Sorted Array (34FindFirstAndLastPositionOfElementInSortedArray.js)

Your solution is well-written and efficient. Here are some strengths and areas for improvement:

Strengths:

• You correctly implemented two binary search functions to find the first and last occurrences.
• The code is clean and uses appropriate variable names.
• You handled the empty array case at the beginning.

Areas for improvement:

• Consider optimizing by checking if the first occurrence is found before proceeding to find the last occurrence. Currently, even if the target is not found, both binary searches run. You can modify the code to return [-1, -1] immediately if findFirst() returns -1.
• The function name findSecond might be misleading; consider renaming it to findLast for clarity.
• While using bit shifting for integer division is efficient, it might be less readable. You could use Math.floor and Math.ceil for clarity, but this is a minor point.

Here's a slightly optimized version of your code:

var searchRange = function (nums, target) {
  if (nums.length === 0) return [-1, -1];

  function findFirst() {
    let lo = 0,
      hi = nums.length - 1;
    while (lo < hi) {
      const mid = lo + ((hi - lo) >> 1);
      if (target <= nums[mid]) hi = mid;
      else lo = mid + 1;
    }
    return nums[lo] === target ? lo : -1;
  }

  function findLast() {
    let lo = 0,
      hi = nums.length - 1;
    while (lo < hi) {
      const mid = lo + ((hi - lo + 1) >> 1);
      if (target >= nums[mid]) lo = mid;
      else hi = mid - 1;
    }
    return nums[lo] === target ? lo : -1;
  }

  const first = findFirst();
  if (first === -1) return [-1, -1];
  const last = findLast();
  return [first, last];
};

This way, if the target is not found, we avoid the second binary search.

Overall, your solution is correct and efficient. With these minor improvements, it would be even better.

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (153FindMinimumInRotatedSortedArray.js)

Your solution is excellent! It correctly solves the problem with optimal time and space complexity. The code is clean and easy to understand. Here are a few points to consider:

1. Clarity: The use of (left + right) >>> 1 is a good practice to avoid integer overflow in languages like Java, but in JavaScript, since array indices are limited (up to 5000 in this problem), it is not strictly necessary. However, it is a good habit for larger arrays. You might also use Math.floor((left + right) / 2) for clarity, but the current approach is fine.

2. Edge Cases: Your solution handles edge cases such as when the array has one element (left == right) or when the array is fully sorted. The loop condition while (left < right) ensures that when left and right converge, we have found the minimum.

3. Comparison: Your solution is actually more efficient than the reference solution in terms of code simplicity, as it avoids extra checks for the minimum element at the mid point. The reference solution checks if the mid is the minimum by comparing with neighbors, which is redundant because the binary search logic naturally converges to the minimum.

4. Best Practices: Your code follows best practices for binary search. The variable names are descriptive, and the logic is straightforward.

One minor suggestion: You might add a comment explaining why you compare nums[mid] with nums[right] instead of nums[left]. This could help others understand the intuition: by comparing with the right element, we can determine which half is sorted and where the minimum must lie.

Overall, great job! Your solution is correct and efficient.

VERDICT: PASS

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Find the Peak Element (162FindPeakElement.js)

The student's solution is well-implemented and meets all problem requirements. It is efficient and concise. One minor point for improvement is code readability: using >>> for midpoint calculation might be unfamiliar to some readers. Consider using Math.floor((lo + hi) / 2) for clarity, although the current method is correct and efficient. Additionally, adding comments to explain the logic could be helpful for maintainability. For instance, explaining why comparing with the right neighbor is sufficient and how the algorithm ensures finding a peak.

Overall, this is a strong solution.

VERDICT: PASS