How do you solve Number of Ways to Separate Numbers on LeetCode?

Number of Ways to Separate Numbers (LeetCode #1977) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Valid numbers cannot have leading zeros unless they are '0'.

What is the brute force solution for Number of Ways to Separate Numbers?

The brute force approach for Number of Ways to Separate Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible ways to insert commas in the string and checking if the resulting numbers are valid. This is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Number of Ways to Separate Numbers?

Number of Ways to Separate Numbers uses the following concepts: String, Dynamic Programming, Suffix Array. The recommended patterns to study are: Dynamic Programming, Backtracking.

What are the interview tips for Number of Ways to Separate Numbers?

Always clarify the constraints and edge cases with the interviewer. Think about how to build solutions incrementally, especially with DP. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving Number of Ways to Separate Numbers?

Forgetting to check for leading zeros in numbers. Not considering the non-decreasing property of the sequence.

#1977

Number of Ways to Separate Numbers

Hard

StringDynamic ProgrammingSuffix ArrayDynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n)

The optimal approach uses dynamic programming to build up the number of valid ways to split the string. We keep track of valid splits using a DP array, leveraging previously computed results to avoid redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the number of ways to split the substring num[0:i].
2Step 2: Iterate through each position in the string, checking all possible previous split points.
3Step 3: For each split, check if the resulting number is valid (non-decreasing and no leading zeros). Update the DP array accordingly.

solution.py19 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def countWays(num):
5    n = len(num)
6    dp = [0] * (n + 1)
7    dp[0] = 1
8
9    for i in range(1, n + 1):
10        for j in range(i):
11            if isValid(num[j:i]):
12                dp[i] = (dp[i] + dp[j]) % MOD
13    return dp[n]
14
15
16def isValid(sub):
17    if len(sub) > 1 and sub[0] == '0':
18        return False
19    return True

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops iterating through the string. The space complexity is O(n) for the DP array used to store results.

1Valid numbers cannot have leading zeros unless they are '0'.
2The sequence of numbers must be non-decreasing.

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