Problem Description

You are given a string s that contains only digits from 1 to 9, and an integer k.

Your task is to partition the string s into substrings such that:

• Every digit in s belongs to exactly one substring (no overlapping, no digits left out)
• The numerical value of each substring must be less than or equal to k

A partition that satisfies these conditions is called a "good" partition.

You need to find the minimum number of substrings required to create a good partition. If it's impossible to create a good partition, return -1.

For example:

• If s = "123" and k = 50, you could partition it as ["1", "23"] (values: 1 and 23, both ≤ 50) using 2 substrings
• The value of a substring is its interpretation as an integer (e.g., "23" has value 23)
• A substring is a contiguous sequence of characters from the original string

The challenge is to find the optimal way to split the string into the fewest possible valid substrings, where each substring's numerical value doesn't exceed k.

Intuition

When we look at this problem, we need to find the minimum number of substrings to partition the string. This immediately suggests a dynamic programming or recursive approach where we try different ways to split the string and choose the best one.

The key insight is that at each position in the string, we have choices: we can take just the current digit as a substring, or we can try to extend it by including more digits, as long as the resulting value doesn't exceed k.

Think of it like this: standing at position i in the string, we ask ourselves - "What are all the valid substrings I can take starting from here?" For each valid substring (one whose value is ≤ k), we would then need to solve the same problem for the remaining part of the string.

This naturally leads to a recursive solution where:

• At each position i, we try all possible valid substrings starting from i
• For a substring from position i to j, if its value is ≤ k, we recursively solve for position j+1
• We want to minimize the total number of partitions, so we take the minimum among all valid choices

The recursive nature of the problem becomes clear: minimum partitions from position i = 1 + minimum partitions from position j+1, where we try all valid j values and pick the best one.

Since we might revisit the same positions multiple times through different recursive paths, memoization helps us avoid redundant calculations by storing the results for each position we've already computed.

If at any position we can't form any valid substring (like when even a single digit exceeds k), the problem has no solution, which we handle by returning infinity during recursion and ultimately returning -1.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution implements a memoization search (top-down dynamic programming) approach using recursion with caching.

Implementation Details:

We define a recursive function dfs(i) that calculates the minimum number of partitions needed starting from index i of the string s.

Base Case:

• When i >= n (where n is the length of string), we've processed the entire string, so we return 0 (no more partitions needed).

Recursive Case: For each position i, we:

1. Initialize res = inf to track the minimum partitions and v = 0 to build the current substring's value
2. Try extending the substring by iterating j from i to n-1:
   • Update the value: v = v * 10 + int(s[j]) (this builds the integer value digit by digit)
   • If v > k, we break immediately as any further extension would also exceed k
   • Otherwise, we recursively solve for the remaining string: dfs(j + 1)
   • Update our minimum: res = min(res, dfs(j + 1))
3. Return res + 1 (adding 1 for the current partition)

Memoization: The @cache decorator is used to memorize previously computed results for each index i, preventing redundant calculations when the same subproblem is encountered again.

Final Answer:

• Call dfs(0) to get the minimum partitions starting from index 0
• If the result is inf, it means no valid partition exists, so return -1
• Otherwise, return the computed minimum number of partitions

Time Complexity: O(n²) in the worst case, where we might check all possible substrings Space Complexity: O(n) for the recursion stack and memoization cache

The algorithm efficiently explores all valid partitioning possibilities while avoiding repeated work through memoization, ensuring we find the optimal solution.

Example Walkthrough

Let's walk through the solution with s = "165" and k = 42.

We start at index 0 with dfs(0):

Step 1: From index 0

• Try substring "1" (value = 1, ≤ 42): Valid! Recursively call dfs(1)
• Try substring "16" (value = 16, ≤ 42): Valid! Recursively call dfs(2)
• Try substring "165" (value = 165, > 42): Invalid! Stop extending

Step 2: Exploring dfs(1) (from "1" partition)

• Try substring "6" (value = 6, ≤ 42): Valid! Recursively call dfs(2)
• Try substring "65" (value = 65, > 42): Invalid! Stop extending

Step 3: Exploring dfs(2) (from multiple paths)

• From "1" → "6" path: Try substring "5" (value = 5, ≤ 42): Valid! Call dfs(3)
• From "16" path: Try substring "5" (value = 5, ≤ 42): Valid! Call dfs(3)

Step 4: Base case dfs(3)

• Index 3 ≥ length(3), return 0 (end of string)

Backtracking with results:

• dfs(2) = 1 + dfs(3) = 1 + 0 = 1
• dfs(1) from "6" → "5" path = 1 + dfs(2) = 1 + 1 = 2
• dfs(1) from "65" path fails (value > k)
• dfs(0) has two valid options:
  • Path 1: "1" → remaining string = 1 + dfs(1) = 1 + 2 = 3
  • Path 2: "16" → remaining string = 1 + dfs(2) = 1 + 1 = 2

The minimum is 2, achieved by partitioning as ["16", "5"].

The memoization ensures that when we compute dfs(2) once, we reuse that result (value = 1) for all future calls to dfs(2), avoiding redundant calculations. This optimization is crucial as the recursion tree can have overlapping subproblems.

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2) in the worst case, where n is the length of string s. This occurs because:

• The recursive function dfs(i) can be called for each position i from 0 to n-1, giving us n possible states
• For each state i, the inner loop iterates from position i to potentially position n-1 in the worst case, which takes O(n) time
• Due to memoization with @cache, each state is computed only once
• Therefore, the total time complexity is O(n) * O(n) = O(n^2)

However, the reference answer states O(n). This could be achievable if the value constraint k limits how far the inner loop can extend. If k has at most d digits (where d is constant), then from any position i, we can only extend the substring by at most d positions before v > k. In this case:

• Each state still computes in O(d) = O(1) time
• With n states and constant work per state, the time complexity becomes O(n)

The space complexity is O(n):

• The memoization cache stores at most n states (one for each starting position)
• The recursion depth can go up to n in the worst case when each partition contains only one character
• Therefore, the space complexity is O(n) for both the cache and the recursion stack

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

Common Pitfalls

1. Integer Overflow When Building Substring Values

One critical pitfall in this solution is the potential for integer overflow when building the value of a substring. As we iterate through the string and build current_value = current_value * 10 + int(s[end_index]), this value can grow very large, especially with long substrings.

Why it's a problem:

• In languages with fixed integer sizes, this could cause overflow
• Even in Python (which handles arbitrary precision), comparing extremely large numbers with k becomes inefficient
• The algorithm might waste time building huge numbers that will obviously exceed k

Solution: Add an early termination check before updating the value:

# Check if adding the next digit would definitely exceed k
next_digit = int(s[end_index])
if current_value > (k - next_digit) // 10:
    break
current_value = current_value * 10 + next_digit

2. Not Handling Single Digit Greater Than k

Another pitfall is not immediately checking if any single digit in the string is already greater than k. If k = 5 and the string contains '9', it's impossible to create a valid partition.

Why it's a problem:

• The algorithm will unnecessarily explore all possibilities before returning infinity
• This wastes computational resources on an impossible case

Solution: Add a preprocessing check:

def minimumPartition(self, s: str, k: int) -> int:
    # Early check: if any single digit > k, impossible to partition
    if any(int(digit) > k for digit in s):
        return -1
  
    # Continue with the rest of the implementation...

3. Inefficient String to Integer Conversion

The current implementation converts each character to an integer repeatedly using int(s[end_index]). While not a major issue, it can be optimized.

Why it's a problem:

• Repeated string-to-integer conversions add overhead
• Each digit is converted multiple times across different recursive calls

Solution: Pre-convert all digits once:

def minimumPartition(self, s: str, k: int) -> int:
    # Convert string digits to integers once
    digits = [int(ch) for ch in s]
  
    @cache
    def dfs(start_index: int) -> int:
        # ... rest of the code
        for end_index in range(start_index, string_length):
            current_value = current_value * 10 + digits[end_index]
            # ... rest of the loop

4. Cache Memory Issues with Very Long Strings

The @cache decorator creates an unbounded cache that stores results for every unique input. For very long strings, this could consume significant memory.

Why it's a problem:

• Memory usage grows linearly with string length
• In memory-constrained environments, this could cause issues

Solution: Use lru_cache with a maximum size or implement manual memoization with controlled memory usage:

from functools import lru_cache

@lru_cache(maxsize=1000)  # Limit cache size
def dfs(start_index: int) -> int:
    # ... implementation

These optimizations make the solution more robust and efficient, especially for edge cases and large inputs.