How do you solve Shuffle String on LeetCode?

Shuffle String (LeetCode #1528) 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: Understanding the relationship between indices and characters is crucial.

What is the brute force solution for Shuffle String?

The brute force approach for Shuffle String is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves creating a new string and placing each character from the original string at its new position as specified by the indices array. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Shuffle String?

Shuffle String uses the following concepts: Array, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Shuffle String?

Explain your thought process clearly while coding. Test your solution with edge cases, like minimum and maximum input sizes. Discuss the time and space complexity of your solution.

What are common mistakes when solving Shuffle String?

Not initializing the result array correctly. Confusing the indices while assigning characters.

#1528

Shuffle String

Easy

ArrayStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a similar approach as the brute force but is more efficient in terms of clarity and execution. We directly assign characters to their new positions in a single pass.

⚙️

Algorithm

4 steps

1Step 1: Create an empty array of the same length as s.
2Step 2: Loop through each index and character in s simultaneously.
3Step 3: Assign each character to its new position in the result array using the indices array.
4Step 4: Convert the result array back to a string and return it.

solution.py5 lines

1def shuffleString(s, indices):
2    result = [''] * len(s)
3    for i in range(len(s)):
4        result[indices[i]] = s[i]
5    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string and indices. The space complexity is O(n) due to the auxiliary array used to store the result.

1Understanding the relationship between indices and characters is crucial.
2Using an auxiliary array helps manage character placement effectively.

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