How do you solve Shifting Letters II on LeetCode?

Shifting Letters II (LeetCode #2381) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a difference array allows us to efficiently track cumulative shifts.

What is the brute force solution for Shifting Letters II?

The brute force approach for Shifting Letters II is Brute Force, with O(n²) time complexity and O(1) space complexity. We will directly apply each shift operation to the string for the specified range. This is straightforward but inefficient because we may end up shifting the same characters multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Shifting Letters II?

Shifting Letters II uses the following concepts: Array, String, Prefix Sum. The recommended patterns to study are: Prefix Sum, Difference Array.

What are the interview tips for Shifting Letters II?

Explain your thought process clearly and ask clarifying questions. Discuss the trade-offs of different approaches before diving into coding. Test your solution with edge cases to ensure robustness.

What are common mistakes when solving Shifting Letters II?

Not considering wrap-around when shifting characters. Failing to properly initialize and use the shift array.

#2381

Shifting Letters II

Medium

ArrayStringPrefix SumPrefix SumDifference Array

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of applying shifts directly, we can track the cumulative effect of shifts using a difference array. This allows us to compute the final shifts in a single pass, making the solution much more efficient.

⚙️

Algorithm

4 steps

1Step 1: Create a shift array initialized to zero, with the same length as the string.
2Step 2: For each shift operation, update the start index with +1 and the index after the end with -1 to mark the beginning and end of shifts.
3Step 3: Compute the prefix sum on the shift array to determine the total shifts for each index.
4Step 4: Apply the computed shifts to the original string, adjusting for the wrap-around using modulo 26.

solution.py14 lines

1def shiftingLetters(s, shifts):
2    n = len(s)
3    shift = [0] * (n + 1)
4    for start, end, direction in shifts:
5        shift[start] += 1 if direction == 1 else -1
6        if end + 1 < n:
7            shift[end + 1] -= 1 if direction == 1 else -1
8    for i in range(1, n):
9        shift[i] += shift[i - 1]
10    result = []
11    for i in range(n):
12        new_char = chr((ord(s[i]) - ord('a') + shift[i]) % 26 + ord('a'))
13        result.append(new_char)
14    return ''.join(result)

ℹ

Complexity note: This complexity arises because we make a single pass to mark shifts and another pass to compute the final characters, leading to O(n) time overall.

1Using a difference array allows us to efficiently track cumulative shifts.
2Prefix sums can simplify the application of multiple range updates.

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