💡 Problem Formulation: We need to create a program that, given two integers n and k, returns the lexicographically smallest string that has a length equal to n and a numeric value equal to k. The numeric value of a lowercase alphabetic character is its position in the alphabet starting from 1; hence ‘a’ is 1, ‘b’ is 2, etc. For instance, given the input n = 5 and k = 73, the desired output would be “aayzz” which is the smallest string satisfying the described constraints.
Method 1: Greedy Algorithm with Backtracking
This method involves constructing the string from the end, starting with the largest possible characters while ensuring that enough value is left for the remaining positions. It backtracks to the previous character if the current choice does not lead to a solution.
Here’s an example:
def find_smallest_string(n, k):
result = ['a'] * n
k -= n
pos = n - 1
while k > 0:
add_val = min(25, k)
result[pos] = chr(add_val + ord('a'))
k -= add_val
pos -= 1
return ''.join(result)
print(find_smallest_string(5, 73))Output: “aayzz”
This example demonstrates the allocation of the remaining value ‘k’ to the string’s characters starting from the end. By ensuring each remaining character is at least ‘a’, we guarantee the lexicographic minimum. Each character is then incremented to the highest possible value under the condition that the string’s numeric value remains ‘k’.
Method 2: Iterative Filling
This method involves an iterative approach where each character in the string is set to the minimum value ‘a’ initially, and the remainder of the numeric value k is spread from right to left, adjusting according to the maximum value ‘z’.
Here’s an example:
def find_smallest_string(n, k):
result = ['a'] * n
k -= n
for i in range(n - 1, -1, -1):
if k == 0:
break
increment = min(k, 25)
result[i] = chr(ord('a') + increment)
k -= increment
return ''.join(result)
print(find_smallest_string(3, 27))Output: “aaz”
This snippet fills in the string with the minimum characters ‘a’ and then iteratively increases the value of the characters from the end of the string until the sum reaches ‘k’. The min function ensures that no character exceeds ‘z’.
Method 3: Recursive Construction of String
This recursive method involves building the string character by character from left to right, choosing the smallest possible character that still allows the rest of the string to reach the target value k.
Here’s an example:
def construct_string(n, k, s=''):
if n == 0:
return s if k == 0 else ''
for ch in range(97, 123): # ASCII values for 'a' to 'z'
if k - (ch - 96) >= 0:
tmp = construct_string(n - 1, k - (ch - 96), s + chr(ch))
if tmp:
return tmp
return ''
print(construct_string(4, 19))Output: “aadi”
This code snippet recursively tries all possible characters starting from ‘a’ and includes them in the string only if they allow reaching the desired numeric value ‘k’. The function calls itself decreasing the size ‘n’ and the remaining value ‘k’ until a solution is found.
Method 4: Dynamic Programming
Dynamic programming can be applied to this problem by solving smaller subproblems and using their solutions to construct the overall solution. It involves keeping a memoization table for subproblem results.
Here’s an example:
def find_smallest_string(n, k):
memo = {}
def dp(i, k):
if i == 0:
return '' if k == 0 else None
if (i, k) in memo:
return memo[(i, k)]
for ch_value in range(1, 26 + 1): # for characters 'a' to 'z'
if k >= ch_value:
next_part = dp(i - 1, k - ch_value)
if next_part is not None:
memo[(i, k)] = chr(96 + ch_value) + next_part
return memo[(i, k)]
return None
return dp(n, k)
print(find_smallest_string(5, 73))Output: “aayzz”
This code snippet uses dynamic programming to avoid recomputing the solutions of subproblems. A memoization dictionary is used to store intermediate results. This method is efficient for larger values of ‘n’ and ‘k’ but may be overkill for smaller instances.
Bonus One-Liner Method 5: Optimal String Construction with Comprehension
A compact solution using list comprehension and string methods to construct the optimal string:
Here’s an example:
def find_smallest_string(n, k):
return ''.join([chr(max(ord('a'), (k - 26 * (n - i))) + 96) for i in range(1, n+1)])
print(find_smallest_string(5, 73))Output: “aayzz”
This one-liner iterates over each position ‘i’ in the desired output string and calculates the character needed by subtracting the maximum possible contribution of the subsequent characters from ‘k’. This ensures that each position is filled with the smallest possible character value.
Summary/Discussion
- Method 1: Greedy Algorithm with Backtracking. Efficient for most cases, it sometimes may not be the most intuitive approach.
- Method 2: Iterative Filling. Simple to understand and implement; however, it may not be the most efficient for large input values due to its iterative nature.
- Method 3: Recursive Construction of String. Recursive and understandable approach but can be slower and lead to a stack overflow for very large inputs.
- Method 4: Dynamic Programming. Highly efficient for large inputs by avoiding redundant calculations. It could be unnecessarily complex for smaller inputs.
- Method 5: Optimal String Construction with Comprehension. Compact and elegant one-liner that leverages Python comprehensions, but readability might suffer for those less familiar with Python’s concise constructs.