How do you solve Shifting Letters on LeetCode?

Shifting Letters (LeetCode #848) 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: Cumulative shifts allow for efficient calculations.

What is the brute force solution for Shifting Letters?

The brute force approach for Shifting Letters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves directly applying each shift to the corresponding letters of the string. For every index, we shift the first i + 1 letters of the string by the specified amount, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Shifting Letters?

Explain your thought process clearly while coding. Consider edge cases, such as maximum shift values. Always check if your solution can be optimized after the brute force.

What are common mistakes when solving Shifting Letters?

Not considering the wrap-around for characters. Forgetting to reverse the result after building it.

#848

Shifting Letters

Medium

ArrayStringPrefix SumPrefix SumArray

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 calculate the total shifts needed for each character in a single pass. This reduces the number of operations significantly and avoids nested loops.

⚙️

Algorithm

3 steps

1Step 1: Create a new array to store the cumulative shifts.
2Step 2: Traverse the shifts array from the end to the beginning, accumulating the shifts for each character.
3Step 3: For each character in the string, apply the total shift calculated and convert it back to a character.

solution.py11 lines

1# Full working Python code
2s = 'abc'
3shifts = [3, 5, 9]
4result = []
5total_shift = 0
6for i in range(len(shifts) - 1, -1, -1):
7    total_shift += shifts[i]
8    result.append(chr((ord(s[i]) - ord('a') + total_shift) % 26 + ord('a')))
9result.reverse()
10final_string = ''.join(result)
11print(final_string)

ℹ

Complexity note: The time complexity is O(n) because we traverse the shifts array once, and the space complexity is O(n) due to the result storage.

1Cumulative shifts allow for efficient calculations.
2Processing from the end of the shifts array simplifies the logic.

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