How do you solve Restore The Array on LeetCode?

Restore The Array (LeetCode #1416) 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 range of valid integers helps in efficiently checking splits.

What is the time complexity of Restore The Array?

The optimal solution for Restore The Array runs in O(n) time and uses O(n) space. This complexity is efficient because we only traverse the string once and use a fixed amount of space for the dp array.

What is the brute force solution for Restore The Array?

The brute force approach for Restore The Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible ways to split the string into valid integers and counting those that fall within the specified range. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Restore The Array?

Restore The Array uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, String Manipulation.

What are the interview tips for Restore The Array?

Always clarify constraints and edge cases before diving into coding. Think about how to optimize your solution early on to avoid brute force pitfalls. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Restore The Array?

Not considering leading zeros in splits, which can lead to invalid numbers. Failing to handle large numbers correctly, especially when converting from string to integer.

#1416

Restore The Array

Hard

StringDynamic ProgrammingDynamic ProgrammingString Manipulation

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)

The optimal approach uses dynamic programming to build a solution incrementally. We keep track of the number of ways to split the string at each index, reducing redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array where dp[i] represents the number of ways to split the substring s[i:].
2Step 2: Iterate through the string from the end to the beginning.
3Step 3: For each position, check possible splits of length 1 to 10 (since k can be at most 10 digits long). If the number is valid, add the ways from dp to the current position.

solution.py17 lines

1# Full working Python code
2def count_valid_splits(s, k):
3    mod = 10**9 + 7
4    n = len(s)
5    dp = [0] * (n + 1)
6    dp[n] = 1  # Base case: one way to split an empty string
7
8    for i in range(n - 1, -1, -1):
9        for length in range(1, 11):  # Check lengths from 1 to 10
10            if i + length <= n:
11                num = int(s[i:i + length])
12                if 1 <= num <= k:
13                    dp[i] = (dp[i] + dp[i + length]) % mod
14    return dp[0]
15
16# Example usage
17print(count_valid_splits('1317', 2000))

ℹ

Complexity note: This complexity is efficient because we only traverse the string once and use a fixed amount of space for the dp array.

1Understanding the range of valid integers helps in efficiently checking splits.
2Dynamic programming can significantly reduce the number of calculations by storing intermediate results.

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