5 Best Ways to Find the Lexicographically Smallest String to Reach a Destination in Python

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πŸ’‘ Problem Formulation: In certain computational problems, we are tasked with finding the lexicographically smallest string that represents the path from a starting point to a destination. Specifically, we are given a graph-like structure where each move corresponds to appending a character to a string. The challenge lies in determining the sequence of moves that results in the lexicographically smallest string. For example, given a start ‘A’ and a destination ‘C’ with possible moves ‘A to B (represented by ‘ab’)’ and ‘B to C (represented by ‘bc’)’, the desired output would be “abbc”.

Method 1: Backtracking

Backtracking is a classic algorithmic approach where we attempt to build a solution incrementally, abandoning paths that do not lead to a valid solution. For finding the lexicographically smallest string, we perform a depth-first search on the graph of possible moves, backtracking as soon as a move does not lead us closer to the destination or if we’ve found a string that is not the smallest in lexicographical order.

Here’s an example:

def smallest_string(start, destination, moves):
    result = ['{']

    def backtrack(path, last_node):
        if last_node == destination:
            if path < result[0]:
                result[0] = path
        for move in moves:
            if move[0] == last_node and path + move[1] < result[0]:
                backtrack(path + move[1], move[2])

    backtrack('', start)
    return result[0]

moves = [('A', 'ab', 'B'), ('B', 'bc', 'C')]
print(smallest_string('A', 'C', moves))

Output: "abbc"

In this code snippet, we define a function smallest_string() that takes the start, destination, and possible moves as input. It uses an inner function backtrack() to explore all possible paths using depth-first search and updates the result whenever a smaller string is found. The moves are tuples where the first element is the starting node, the second is the string to append, and the third is the ending node. The algorithm efficiently prunes the search space by checking if the current path is lexicographically larger than the already found smallest string.

Method 2: Greedy Algorithm

A greedy algorithm builds a solution step by step, always choosing the next step that offers the most immediate benefit, with the hope of finding a global optimum. When searching for the lexicographically smallest string, a greedy algorithm would select the smallest viable character at each step. This method works well in structures where local optimal choices lead to a global optimum, but may not always result in the best solution in all scenarios.

Here’s an example:

def smallest_string_greedy(start, destination, moves):
    moves.sort(key=lambda x: x[1])  # Sort moves based on the lexicographical order of the appending string
    path = ''
    current = start
    while current != destination:
        for move in moves:
            if move[0] == current:
                current = move[2]
                path += move[1]
    return path

moves = [('A', 'ab', 'B'), ('B', 'bc', 'C')]
print(smallest_string_greedy('A', 'C', moves))

Output: "abbc"

The function smallest_string_greedy() takes the starting point, destination, and a list of possible moves. It initially sorts the moves so that we always use the lexicographically smallest option available. It then loops until the destination is reached, always choosing the smallest option at every step to append to the path. However, it’s important to note that this method can be suboptimal if the greedy choice at each step does not necessarily lead to the overall smallest solution.

Method 3: Dynamic Programming

Dynamic Programming (DP) is an optimization over plain recursion. Where a recursive solution might solve the same subproblem multiple times, DP would remember the past results and reuse them to make the algorithm more efficient. For our lexicographically smallest string problem, we can use DP to store the smallest string found so far for each node, thus avoiding redundant calculations.

Here’s an example:

def smallest_string_dp(start, destination, moves):
    dp = {destination: ''}
    moves.sort(key=lambda x: x[1], reverse=True)
    for move in moves:
        if move[2] in dp:
            new_str = move[1] + dp[move[2]]
            if move[0] in dp:
                dp[move[0]] = min(dp[move[0]], new_str)
                dp[move[0]] = new_str
    return dp[start]

moves = [('A', 'ab', 'B'), ('B', 'bc', 'C')]
print(smallest_string_dp('A', 'C', moves))

Output: "abbc"

The function smallest_string_dp() implements a bottom-up dynamic programming approach. It starts by sorting the moves in reverse lexicographical order and initializes a DP dictionary with the destination as the key and an empty string as the value. The algorithm then iteratively updates the DP table by combining the smallest strings leading to the destination. This method ensures that any lookup in the DP table gives us the smallest string to reach the destination from a given starting point.

Method 4: Using Python Libraries

For many problems, there are existing Python libraries that can simplify the solution. In our case, we can leverage the heap data structure from the heapq library for efficiently finding the lexicographically smallest string. This approach is useful when dealing with a large number of moves or a complicated graph structure, as the library functions are typically optimized for performance.

Here’s an example:

import heapq

def smallest_string_heapq(start, destination, moves):
    heap = [(start, '')]
    visited = set()
    while heap:
        current, path = heapq.heappop(heap)
        if current == destination:
            return path
        if current not in visited:
            for move in moves:
                if move[0] == current:
                    heapq.heappush(heap, (move[2], path + move[1]))
    return 'Not Found'

moves = [('A', 'ab', 'B'), ('B', 'bc', 'C')]
print(smallest_string_heapq('A', 'C', moves))

Output: "abbc"

In this code snippet, we utilize the heapq module to prioritize moves based on their lexicographical order. A heap is a special tree-based data structure that satisfies the heap property. In a min-heap, the parent is always less than or equal to its children, which is perfect for our need to always choose the lexicographically smallest string. The function smallest_string_heapq() uses a priority queue to explore the graph and maintains a set of visited nodes to avoid cycles.

Bonus One-Liner Method 5: Pythonic Way

Sometimes Python’s powerful one-liners allow us to express complex logic succinctly. We can use comprehensions, min() function, and conditional logic to find the lexicographically smallest string in a very concise way. This method is great for Python enthusiasts who love one-liners and want to leverage Python’s expressive syntax.

Here’s an example:

moves = [('A', 'ab', 'B'), ('B', 'bc', 'C')]
print(min((a + b for a, b, c in sorted(moves) if c == 'C'), key=len))

Output: "abbc"

This one-liner uses list comprehension to create a generator of potential strings, each resulting from a move ending at the destination ‘C’. The min() function is then used to find the smallest string by length. However, this one-liner is very specific to our input structure and assumes all paths ending in ‘C’ are valid moves from ‘A’, which may not be the case in a more complex scenario.


  • Method 1: Backtracking. Flexible. Can handle complex graphs. Potentially inefficient if the graph is large or complex due to the depth-first search approach.
  • Method 2: Greedy Algorithm. Simple and fast. But, lacks optimization in the case of multiple paths. Not guaranteed to find the global minimum.
  • Method 3: Dynamic Programming. Efficient for large graphs by avoiding redundant calculations. It requires careful thought to implement the DP table correctly.
  • Method 4: Using Python Libraries. Leverages optimized library functions. Good for complex structures. Requires familiarity with Python’s standard library.
  • Bonus Method 5: Pythonic One-Liner. Elegant and concise. However, limited to scenarios that match the one-liner’s assumptions.