1849. Splitting a String Into Descending Consecutive Values

Problem Description

This problem asks us to check whether a given string s, comprised exclusively of digits, can be split into two or more non-empty substrings such that:

1. The numerical values of these substrings, when converted from strings to numbers, form a sequence in descending order.
2. The difference between the numerical values of each pair of adjacent substrings is exactly 1.

For example, if s = "543210", it is possible to split it into ["54", "3", "2", "10"], where each number is 1 less than the previous, and so the condition is met.

The key constraints are that the substrings must be non-empty and must appear in the same order they do in the original string. There cannot be any rearrangement of the substrings.

Intuition

To solve this problem, we can use a depth-first search (DFS) algorithm. The DFS approach will try to partition the string s at different points and recursively check if the resulting substrates satisfy the conditions.

The solution uses three key ideas:

1. Recursion: Recursively split the string and check if the current split creates a number that is exactly 1 less than the preceding number.
2. Backtracking: If at any point the condition is not met, the function backtracks, trying a different split.
3. Early Stopping: If the given string has been entirely traversed and more than one substring meeting the conditions has been found, we can stop the search and return true.

To implement the solution:

• A recursive function dfs is defined, which takes parameters i (the current starting index in s), x (the last numerical value of the substring), and k (the count of valid splits found so far).
• Initially, dfs is called with starting index 0, x set as -1 (as there is no previous number), and k as 0 (no splits found yet).
• Within dfs, we iterate over the string starting from index i, attempting to form a new substring from s[i] to s[j].
• We check if our substring satisfies the conditions; if it does, we continue the search by recursively calling dfs with the next starting index (j+1), the new number y, and increment k.
• If at any point, a full pass through the string is made (i equals the length of s) and more than one valid substring (k > 1) has been found, the function returns true, indicating the splitting is possible. Otherwise, the function eventually returns false.

Using DFS, this solution effectively searches for all possible substrings that might fit the criteria, and upon finding the first valid series of splits, it returns true. If no valid series is found by the time the entire string has been traversed, it returns false.

Learn more about Backtracking patterns.

Solution Approach

The implementation of the solution revolves around a classic DFS algorithm. Here's a step-by-step explanation of the approach using the given Python code:

1. Initialization: A helper function, dfs, is defined inside the main class Solution. This recursive function is at the heart of the DFS implementation. It aims to try all possible splits of the string s beginning from index i.

2. Base Case: If dfs reaches the end of the string (i == len(s)), it checks whether at least two substrings with the required properties have been found (k > 1). If so, the function returns True; otherwise, False.

3. Recursive Search: Inside the dfs function, starting from index i, the code attempts to build a substring piece by piece by iterating from i to each j within the loop. The goal is to form a numerical value y of the current substring being considered (s[i:j+1]).

4. Conditions Check: After each additional character is included in y, we check if y is exactly one less than the previous number x. If x is -1 (indicating this is the first number), any y is acceptable (since there is no prior number to compare with).

   • If this condition is satisfied, the function immediately proceeds with a recursive call, dfs(j + 1, y, k + 1), where j + 1 is the new starting index for the next substring, y is now the last number, and k + 1 indicates one more valid split has been found.
   • If the returned value from this recursive call is True, indicating that proceeding from index j + 1 has led to a valid sequence, the current dfs also returns True.

5. Continuation and Backtracking: If the condition fails or the recursive call doesn't result in a valid solution, the function continues to the next iteration of the loop, effectively trying a longer substring (more characters included from s) or backtracking if all possibilities starting from index i have been exhausted.

6. Return Result: The dfs function will return False if none of the iterations and recursive calls result in a valid split sequence. On the other hand, the function will exit with a True at the first instance of finding a valid sequence of substrings.

7. Main Function Call: To start the process, the splitString method in the Solution class calls dfs(0, -1, 0) indicating it starts looking for splits from the beginning of the string (i = 0), with no prior number (x = -1), and no splits found yet (k = 0).

This algorithm effectively explores all potential ways to split the string through DFS while avoiding a full search when the first valid solution is found. It also uses recursion and backtracking systematically to traverse the decision tree until it either finds a valid sequence of splits or exhausts all possibilities and determines it's not possible.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach with the string s = "321". This string should be possible to split into a sequence of descending numbers that are each 1 less than the previous, specifically into ["3", "2", "1"].

Initial Call:

• The splitString method calls dfs(0, -1, 0) to start the recursive search.

Action in dfs - First Iteration from index 0:

• We try to form substrings starting from the first character.
• The loop runs from i = 0 to the end of the string.
• We start by picking s[0:1] which is "3", and convert it to number y = 3.
• Since the last number x was -1 (indicating no previous number), "3" is accepted as the first valid substring.
• We then call dfs(1, 3, 1) to proceed to the next character.

Action in dfs - Second Iteration from index 1:

• Now x = 3 and we start from i = 1.
• We pick s[1:2] which is "2", and convert it to number y = 2.
• We check if y is exactly 1 less than x (3 - 1 = 2), and it is.
• We then call dfs(2, 2, 2) since we have found two substrings "3" and "2".

Action in dfs - Third Iteration from index 2:

• Now x = 2, and i = 2.
• We pick s[2:3] which is "1", and convert it to number y = 1.
• Again, we check if y is exactly 1 less than x (2 - 1 = 1), which it is.
• We now call dfs(3, 1, 3) because we have found a third substring "1".

Base Case Reached:

• The call dfs(3, 1, 3) has i == len(s), meaning we have reached the end of the string.
• The count of valid splits k is 3 (greater than 1), so we return True.

Result:

• The valid sequence resulting from the above splits is ["3", "2", "1"], and therefore the initial call of splitString returns True.
• We have successfully split the string into a descending sequence where each number is exactly 1 less than the previous substring.

This simple example illustrates the DFS-based recursive process of finding valid substrings and how backtracking occurs if a substring does not fit the required criteria. The process continues until the entire string is either successfully split or determined to be unsplittable according to the conditions.

Solution Implementation

Time and Space Complexity

The given Python code is a solution to the problem of splitting a string into a sequence of decreasing consecutive numbers. It uses a depth-first search (DFS) approach to recursively try out different splits.

Time Complexity

The DFS function, dfs, is called recursively, at each step it tries to construct a new number, y, by adding digits one by one from the input string s. For each number y formed, a check is made whether the previous number x is exactly one more than y(i.e., x - y == 1). The function dfs is potentially called once for each new number y formed, which can happen for each starting position i in the string s.

The worst case time complexity is O(n * 2^n), where n is the length of the string. This is because at each step, we have two choices: either include the current digit in the current number y or start a new number beginning with the current digit. We make n choices at most once for each of the 2^n potential splits of the string.

Space Complexity

The space complexity of the code is O(n). This comes from two factors:

1. The recursive call stack, which at maximum depth could be O(n), as in the worst case the recursion could go as deep as the length of the string for a split starting at each digit of the string.
2. The variable y in the inner loop, which is re-constructed for each recursive call but doesn't add to the space complexity as it is just an integer and not a recursive structure.

Hence, the space requirement grows linearly with the input size, dominated by the depth of the recursive calls.

Learn more about how to find time and space complexity quickly using problem constraints.