How do you solve Find Smallest Letter Greater Than Target on LeetCode?

Find Smallest Letter Greater Than Target (LeetCode #744) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Binary search is a powerful technique for searching in sorted arrays, significantly reducing time complexity.

What is the brute force solution for Find Smallest Letter Greater Than Target?

The brute force approach for Find Smallest Letter Greater Than Target is Brute Force, with O(n) time complexity and O(1) space complexity. We can simply iterate through the list of letters and check each one to see if it is greater than the target. This approach is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Find Smallest Letter Greater Than Target?

Find Smallest Letter Greater Than Target uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Array.

What are the interview tips for Find Smallest Letter Greater Than Target?

Always clarify the constraints and properties of the input data before jumping into coding. Think about edge cases, such as when the target is greater than the largest letter. Practice explaining your thought process clearly as you code, as communication is key in interviews.

What are common mistakes when solving Find Smallest Letter Greater Than Target?

Not recognizing that the array is sorted and using a linear search instead of binary search. Failing to handle the wrap-around case where the target is greater than all elements.

#744

Find Smallest Letter Greater Than Target

Easy

ArrayBinary SearchBinary SearchArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log n)
Space	O(1)	O(1)

💡

Intuition

Time O(log n)Space O(1)

Since the letters are sorted, we can use binary search to find the smallest letter greater than the target efficiently. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Initialize left pointer to 0 and right pointer to the length of letters.
2Step 2: While left pointer is less than right pointer, calculate mid index.
3Step 3: If the letter at mid index is less than or equal to the target, move the left pointer to mid + 1.
4Step 4: If the letter at mid index is greater than the target, move the right pointer to mid.
5Step 5: After the loop, return the letter at the left pointer, which will be the smallest letter greater than target.

solution.py9 lines

1def nextGreatestLetter(letters, target):
2    left, right = 0, len(letters)
3    while left < right:
4        mid = (left + right) // 2
5        if letters[mid] <= target:
6            left = mid + 1
7        else:
8            right = mid
9    return letters[left % len(letters)]

ℹ

Complexity note: The time complexity is O(log n) due to binary search, which is much faster than linear search. The space complexity remains O(1) as we are using a constant amount of space.

1Binary search is a powerful technique for searching in sorted arrays, significantly reducing time complexity.
2Understanding the properties of sorted arrays can lead to more efficient algorithms.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.