# 5 Best Ways to Program to Find Minimum Number of Subsequences Whose Concatenation Is Same as Target in Python

Rate this post

π‘ Problem Formulation: We are tasked with finding the smallest number of subsequences from a given set of strings that can be concatenated together to form a target string. For example, given subsequences [‘a’, ‘ab’, ‘ac’] and a target string ‘abac’, the minimum number of subsequences needed would be 2 (‘ab’ and ‘ac’).

## Method 1: Recursive Backtracking

This approach uses recursive backtracking to explore all possible combinations of the subsequences to match the target string. The algorithm recursively tries to match the target string with a subsequence, and on success, it moves on to the remainder of the target string until the whole string is matched or all possibilities are exhausted.

Here’s an example:

```def min_subsequences(subseqs, target):
if not target:
return 0
min_len = float('inf')
for seq in subseqs:
if target.startswith(seq):
remaining_target = target[len(seq):]
count = min_subsequences(subseqs, remaining_target)
if count != float('inf'):
min_len = min(min_len, 1 + count)
return min_len

subseqs = ['a', 'ab', 'ac']
target = 'abac'
print(min_subsequences(subseqs, target))```

Output:

`2`

This code snippet defines a function `min_subsequences()` which takes a list of possible subsequences and a target string, then calculates the minimum number of subsequences required through recursive calls. It outputs the answer when the target string is fully matched or returns infinity if it is not possible.

## Method 2: Dynamic Programming

Dynamic programming can be used to solve the problem efficiently by breaking it down into smaller subproblems, storing the results to avoid redundant calculations. It iteratively builds up a table that records the minimum number of subsequences needed to form each prefix of the target string.

Here’s an example:

```def min_subsequences_dp(subseqs, target):
dp = [float('inf')] * (len(target) + 1)
dp[0] = 0
for i in range(1, len(target) + 1):
for seq in subseqs:
if target[:i].endswith(seq):
dp[i] = min(dp[i], dp[i - len(seq)] + 1)
return dp[-1] if dp[-1] != float('inf') else -1

subseqs = ['a', 'ab', 'ac']
target = 'abac'
print(min_subsequences_dp(subseqs, target))```

Output:

`2`

The `min_subsequences_dp()` function initializes a dynamic programming table `dp` to store the minimum number of subsequences needed for each prefix of the target. For each position i in the target string, it updates the table with the minimal value if a subsequence ends at that position and adds one to the count due to the match.

## Method 3: Greedy Approach

A greedy algorithm can be applied to this problem by always choosing the longest subsequence that matches the target’s current prefix. Note that this method does not guarantee an optimal solution for all cases but can be a quick heuristic.

Here’s an example:

```def min_subsequences_greedy(subseqs, target):
subseqs.sort(key=len, reverse=True)
count = 0
while target:
for seq in subseqs:
if target.startswith(seq):
count += 1
target = target[len(seq):]
break
else:
return -1
return count

subseqs = ['a', 'ab', 'ac']
target = 'abac'
print(min_subsequences_greedy(subseqs, target))```

Output:

`2`

The `min_subsequences_greedy()` function first sorts the subsequences in descending order by length to try the longest ones first. The target is reduced by stripping the matching prefixes, and if no subsequence matches, it breaks and returns -1, indicating no solution.

## Method 4: Breadth-First Search (BFS)

This method uses BFS to traverse the “state-space tree”, where each state represents a prefix of the target string. In the breadth-first search traversal, states are generated by appending available subsequences and their depth (or level) in the tree corresponds to the number of subsequences used.

Here’s an example:

```from collections import deque

def min_subsequences_bfs(subseqs, target):
queue = deque([(0, '')])  # (number of subsequences, current prefix)
visited = {''}
while queue:
count, current = queue.popleft()
if current == target:
return count
for seq in subseqs:
if not current.endswith(seq):
next_prefix = current + seq
if next_prefix in visited:
continue
if next_prefix == target[:len(next_prefix)]:
queue.append((count + 1, next_prefix))
return -1

subseqs = ['a', 'ab', 'ac']
target = 'abac'
print(min_subsequences_bfs(subseqs, target))```

Output:

`2`

The code represents the BFS algorithm using a queue and a visited set to track already explored prefixes. Once a state representing the full target string is reached, the number of subsequences used to get there is returned as the answer.

## Bonus One-Liner Method 5: Python One-Liner Using itertools

For a fun and experimental approach, Python’s itertools can be used to create combinations of the subsequences and check if their concatenation matches the target string. This one-liner solution is not efficient and is meant for small inputs or educational purposes.

Here’s an example:

```from itertools import product

subseqs = ['a', 'ab', 'ac']
target = 'abac'
print(min(len(comb) for i in range(1, len(target)+1) for comb in product(subseqs, repeat=i) if ''.join(comb) == target))```

Output:

`2`

This one-liner uses `product()` from the itertools module to generate all possible combinations of the subsequences up to the length of the target. It then finds the shortest combination that exactly matches the target string.

## Summary/Discussion

• Method 1: Recursive Backtracking. It’s a straightforward approach that can solve the problem by exploring all possibilities. However, it may not be efficient for large datasets due to its exponential runtime complexity.
• Method 2: Dynamic Programming. It is the most efficient method for large inputs as it avoids redundant calculations, but it may require a significant amount of memory to store intermediate results.
• Method 3: Greedy Approach. It is typically fast and easy to implement but does not guarantee the optimal solution in all cases.
• Method 4: Breadth-First Search (BFS). BFS offers a correct solution by exploring the states level-by-level but may become computationally intensive with a growing size of input.
• Method 5: Python One-Liner Using itertools. While it is concise and elegant, it is impractical for actual use due to its poor time complexity.