Problem Description

You need to create a string of digits based on a pattern of 'I' (increasing) and 'D' (decreasing) characters.

Given a pattern string of length n containing only 'I' and 'D' characters, you must construct a string num of length n + 1 following these rules:

1. The string num can only use digits '1' through '9', and each digit can be used at most once
2. When pattern[i] == 'I', the digit at position i in num must be less than the digit at position i + 1 (increasing relationship)
3. When pattern[i] == 'D', the digit at position i in num must be greater than the digit at position i + 1 (decreasing relationship)

The goal is to find the lexicographically smallest string that satisfies all these conditions.

For example:

• If pattern = "ID", you need 3 digits where the first is less than the second (due to 'I'), and the second is greater than the third (due to 'D'). A valid answer would be "132".
• If pattern = "DDD", you need 4 digits in strictly decreasing order. The lexicographically smallest would be "4321".

The solution uses a depth-first search (DFS) approach with backtracking. It tries different digit combinations, checking at each step whether the current digit satisfies the increasing or decreasing constraint based on the pattern. The search continues until a valid arrangement is found, which will be the lexicographically smallest due to trying digits in ascending order from 1 to 9.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem doesn't involve nodes and edges or graph traversal. We're constructing a string based on a pattern.

Need to solve for kth smallest/largest?

• No: While we need the lexicographically smallest string, we're not finding the kth element from a collection.

Involves Linked Lists?

• No: The problem deals with string construction, not linked list operations.

Does the problem have small constraints?

• Yes: The constraints are quite small - we can only use digits 1-9 (at most 9 digits), and each digit can be used at most once. This means the maximum length of our result string is 10.

Brute force / Backtracking?

• Yes: Given the small constraints and the need to find a valid arrangement that satisfies multiple conditions (increasing/decreasing patterns), backtracking is ideal. We need to:
  • Try different digit combinations
  • Check if each placement satisfies the I/D constraints
  • Backtrack when we hit an invalid state
  • Continue until we find the lexicographically smallest valid solution

Conclusion: The flowchart correctly leads us to use a Backtracking approach. The small constraint size (max 9 digits) makes it feasible to explore different permutations, and the need to satisfy specific ordering conditions while finding the optimal solution makes backtracking the perfect fit for this problem.

Intuition

When we need the lexicographically smallest string, our first instinct should be to place the smallest available digits as early as possible. However, we're constrained by the pattern - we can't just place digits 1, 2, 3... in order because we must respect the 'I' and 'D' relationships.

The key insight is that with only 9 possible digits and each used at most once, we're dealing with a very limited search space. This makes backtracking feasible - we can systematically try placing each available digit at each position and see if it works.

Why backtracking specifically? Consider what happens when we try to build the string greedily from left to right:

• At position 0, we'd want to place '1' for lexicographic minimality
• But what if the pattern starts with "DDD"? We'd need digits that decrease from '1', which is impossible since we can't use digits less than 1

This shows we can't always make locally optimal choices. Sometimes we need to "reserve" smaller digits for later positions. When we hit a dead end (like trying to find a digit smaller than what we've already placed for a 'D' pattern), we need to backtrack and try a different digit at an earlier position.

The backtracking approach naturally handles this by:

1. Trying digits in ascending order (1, 2, 3...) at each position for lexicographic minimality
2. Checking if the current digit satisfies the pattern constraint with the previous digit
3. If we can't place any valid digit at a position, we backtrack and try the next available digit at the previous position
4. The first complete valid string we find will be lexicographically smallest because we always try smaller digits first

This systematic exploration with the ability to undo choices (backtrack) when we hit constraints is exactly what makes backtracking the natural solution for this problem.

Learn more about Stack, Greedy and Backtracking patterns.

Solution Approach

The implementation uses a depth-first search (DFS) with backtracking to systematically explore all possible digit arrangements. Let's walk through the key components:

Data Structures:

• vis: A boolean array of size 10 to track which digits (1-9) have been used
• t: A temporary list to build the current string candidate
• ans: Stores the first valid solution found (which will be lexicographically smallest)

DFS Function Logic:

The recursive dfs(u) function works as follows, where u represents the current position we're trying to fill:

1. Base Case: When u == len(pattern) + 1, we've successfully placed all required digits. Since we try digits in ascending order, the first valid solution is the lexicographically smallest, so we save it and return.

2. Recursive Case: For each digit from 1 to 9:

   • Check if the digit is already used (vis[i])
   • If we're not at position 0, validate the pattern constraint:
     • For 'I' pattern: The previous digit must be less than current digit (int(t[-1]) < i)
     • For 'D' pattern: The previous digit must be greater than current digit (int(t[-1]) > i)
   • If constraints are satisfied:
     • Mark digit as used: vis[i] = True
     • Add digit to current string: t.append(str(i))
     • Recursively try to fill the next position: dfs(u + 1)
     • Backtrack by unmarking the digit and removing it from the string

Early Termination:

The code includes an optimization with if ans: return at the beginning of the DFS function. Once we find the first valid solution, we stop exploring further branches since:

• We try digits in ascending order (1, 2, 3...)
• The first valid arrangement found will be lexicographically smallest
• No need to explore other possibilities

Example Walkthrough:

For pattern = "ID":

1. Start at position 0, try digit 1
2. Move to position 1, need a digit > 1 (due to 'I'), try 2
3. Move to position 2, need a digit < 2 (due to 'D'), can't use 1 (already used)
4. Backtrack to position 1, try 3
5. Move to position 2, need digit < 3, try 2
6. Success! Return "132"

The backtracking ensures we systematically explore all possibilities while the digit ordering (trying 1 before 2, 2 before 3, etc.) guarantees the first valid solution is lexicographically smallest.

Example Walkthrough

Let's trace through the solution for pattern = "DI" step by step:

Goal: Create a 3-digit string where digit[0] > digit[1] (D) and digit[1] < digit[2] (I)

Step 1: Position 0 (u=0)

• Try digit 1: Place it temporarily → t = ["1"]
• Mark vis[1] = True

Step 2: Position 1 (u=1)

• Pattern[0] = 'D', so we need digit < 1
• Try digits 1-9: None work! (1 is used, and 2-9 are all ≥ 1)
• Backtrack to position 0

Step 3: Back at Position 0

• Remove 1, unmark vis[1] = False
• Try digit 2: Place it → t = ["2"]
• Mark vis[2] = True

Step 4: Position 1 (u=1)

• Pattern[0] = 'D', so we need digit < 2
• Try digit 1: It's available and 1 < 2 ✓
• Place it → t = ["2", "1"]
• Mark vis[1] = True

Step 5: Position 2 (u=2)

• Pattern[1] = 'I', so we need digit > 1
• Try digit 3: It's available and 3 > 1 ✓
• Place it → t = ["2", "1", "3"]
• Mark vis[3] = True

Step 6: Base Case (u=3)

• We've placed all 3 digits successfully
• Save ans = "213" and return

Final Answer: "213"

This example demonstrates:

• How we start with the smallest available digit (1) but backtrack when it doesn't work
• How the 'D' pattern forces us to start with a larger digit to leave room for decrease
• How we systematically try digits in ascending order, ensuring lexicographic minimality
• The first complete valid string we find ("213") is our answer

Solution Implementation

Time and Space Complexity

Time Complexity: O(9! × n) where n is the length of the pattern string.

The algorithm uses backtracking to explore all possible permutations of digits 1-9 that satisfy the pattern constraints. In the worst case:

• We try up to 9 choices for the first position
• Up to 8 choices for the second position
• Up to 7 choices for the third position, and so on
• This gives us at most 9! permutations to explore
• For each valid permutation attempt, we perform O(n) work to build the string of length n+1
• The pruning conditions (checking 'I' and 'D' constraints) help reduce the search space, but don't change the worst-case complexity

Space Complexity: O(n) where n is the length of the pattern string.

The space usage includes:

• vis array: O(10) = O(1) for tracking used digits
• t list: O(n+1) for storing the current permutation being built
• Recursion call stack: O(n+1) maximum depth since we recurse up to len(pattern) + 1 times
• ans variable: O(n+1) for storing the final result string

The dominant space factor is O(n) from the recursion stack and temporary storage.

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

Common Pitfalls

1. Off-by-One Error in Pattern Indexing

The Pitfall: A very common mistake is incorrectly mapping between the position in the result string and the index in the pattern string. Since the result has n+1 digits for a pattern of length n, developers often confuse:

• Which pattern character to check when placing a digit
• Whether to check pattern[position] or pattern[position-1]

For example, when placing the digit at position 1 in the result, you need to check pattern[0] (not pattern[1]) to determine the relationship with the digit at position 0.

Wrong Implementation:

# Incorrect: checking pattern[position] instead of pattern[position-1]
if position < len(pattern):
    if pattern[position] == 'I' and last_digit >= digit:
        continue
    if pattern[position] == 'D' and last_digit <= digit:
        continue

Correct Implementation:

# Correct: pattern[position-1] determines the relationship 
# between digit at position-1 and position
if position > 0:
    last_digit = int(current_number[-1])
    if pattern[position - 1] == 'I' and last_digit >= digit:
        continue
    if pattern[position - 1] == 'D' and last_digit <= digit:
        continue

2. Not Enforcing Strict Inequality

The Pitfall: Using <= or >= comparisons instead of strict < or > comparisons. The pattern requires strictly increasing or decreasing relationships, not equal values.

Wrong Implementation:

# Incorrect: allows equal digits which violates uniqueness
if pattern[position - 1] == 'I' and last_digit > digit:  # Should be >=
    continue
if pattern[position - 1] == 'D' and last_digit < digit:  # Should be <=
    continue

Why This Matters: Even though we're using each digit only once (tracked by used_digits), the logic error above would cause us to skip valid digits. For pattern "I", if the last digit is 3, we should accept 4, 5, 6, etc., but the wrong comparison (last_digit > digit) would incorrectly reject all digits greater than 3.

3. Forgetting Early Termination Optimization

The Pitfall: Without early termination, the algorithm continues searching even after finding the first (and lexicographically smallest) valid solution, causing unnecessary computation.

Inefficient Version:

def backtrack(position):
    # Missing early termination check
    if position == len(pattern) + 1:
        result = ''.join(current_number)
        return  # This doesn't stop the entire search!

Optimized Version:

def backtrack(position):
    if result:  # Early termination - stop once we find first valid solution
        return
  
    if position == len(pattern) + 1:
        result = ''.join(current_number)
        return

4. Using Digit 0 in the Search

The Pitfall: The problem specifies using digits 1-9 only, but developers might accidentally include 0 in their search range.

Wrong Implementation:

# Incorrect: includes 0 which is not allowed
for digit in range(0, 10):  # Should start from 1
    if used_digits[digit]:
        continue

Correct Implementation:

# Correct: only use digits 1 through 9
for digit in range(1, 10):
    if used_digits[digit]:
        continue

Why This Matters: Including 0 would create invalid solutions and could also lead to lexicographically incorrect answers. For example, "012" would appear smaller than "123" but contains an invalid digit 0.