#1625

Medium

auto_awesomeString plus Depth-First Search

LeetCode Problem Workspace

Lexicographically Smallest String After Applying Operations

Optimize a string through rotations and additions to get the lexicographically smallest possible result using string manipulation and depth-first search.

string depth first search breadth first search enumeration

Problem Statement

You are given a string s of even length consisting of digits from 0 to 9, and two integers a and b.

You can apply two operations: rotate the string and add a number to a specific digit. Return the lexicographically smallest string you can obtain by applying these operations any number of times.

Examples

Example 1

Input: s = "5525", a = 9, b = 2

Output: "2050"

We can apply the following operations: Start: "5525" Rotate: "2555" Add: "2454" Add: "2353" Rotate: "5323" Add: "5222" Add: "5121" Rotate: "2151" Add: "2050" There is no way to obtain a string that is lexicographically smaller than "2050".

Example 2

Input: s = "74", a = 5, b = 1

Output: "24"

We can apply the following operations: Start: "74" Rotate: "47" Add: "42" Rotate: "24" There is no way to obtain a string that is lexicographically smaller than "24".

Example 3

Input: s = "0011", a = 4, b = 2

Output: "0011"

There are no sequence of operations that will give us a lexicographically smaller string than "0011".

Constraints

2 <= s.length <= 100
s.length is even.
s consists of digits from 0 to 9 only.
1 <= a <= 9
1 <= b <= s.length - 1

Solution Approach

Breadth-First Search (BFS)

One approach is to use BFS to explore all possible strings generated by applying the two operations in different sequences. By performing rotations and additions, we can find the lexicographically smallest string efficiently.

Depth-First Search (DFS)

DFS can be used to explore every possible operation combination recursively. Each operation sequence leads to a new string, and by pruning unpromising paths, DFS can help us find the smallest string.

Optimization with Memoization

Since the operations can repeat, memoization helps by storing previously computed results. This eliminates redundant computations and speeds up the algorithm.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen. For BFS or DFS, the number of possible operations grows exponentially, but memoization or pruning can reduce this. The space complexity will vary based on the data structure used for state storage, such as queues in BFS or recursion stack in DFS.

What Interviewers Usually Probe

The candidate's understanding of depth-first search (DFS) and breadth-first search (BFS) will be crucial to solving the problem.
The candidate should highlight ways to optimize the solution using memoization or pruning techniques to handle large input sizes.
Look for the candidate's ability to balance both rotation and addition operations effectively within the string manipulation strategy.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle the pruning of unpromising paths in DFS can lead to unnecessary computations.
Incorrectly applying rotation and addition operations could result in non-optimal solutions.
Overcomplicating the algorithm by missing opportunities for memoization to optimize repeated states.

Follow-up variants

Increase the number of operations allowed to test optimization techniques.
Consider a variation where the string length is odd, requiring different traversal strategies.
Modify the problem to handle larger digit sets or constraints to test scalability.

FAQ

How do rotation and addition operations work in this problem?

Rotation shifts the digits of the string, while addition modifies a specific digit by adding a number between 0 and 9. Both operations can be applied any number of times to optimize the string.

What is the role of depth-first search in this problem?

Depth-first search is used to explore all possible states generated by the operations. It systematically checks different operation sequences and finds the lexicographically smallest string.

What is the time complexity of this problem?

The time complexity depends on the final approach but can grow exponentially due to the large number of possible operation sequences. Optimization through memoization can help reduce this.

How can I optimize my solution for larger inputs?

Optimizing the solution can be done by using memoization or pruning to avoid recalculating the same string states. This can greatly reduce time complexity in larger inputs.

What is the primary pattern to solve this problem?

The primary pattern is string manipulation combined with depth-first search (DFS), which allows us to explore all possible operations and find the lexicographically smallest result.

terminal

Solution

Solution 1: BFS

Since the data scale of this problem is relatively small, we can use BFS to brute-force search all possible states and then take the lexicographically smallest state.

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class Solution:
    def findLexSmallestString(self, s: str, a: int, b: int) -> str:
        q = deque([s])
        vis = {s}
        ans = s
        while q:
            s = q.popleft()
            if ans > s:
                ans = s
            t1 = ''.join(
                [str((int(c) + a) % 10) if i & 1 else c for i, c in enumerate(s)]
            )
            t2 = s[-b:] + s[:-b]
            for t in (t1, t2):
                if t not in vis:
                    vis.add(t)
                    q.append(t)
        return ans

Solution 2: Enumeration

We observe that for the addition operation, a digit will return to its original state after at most $10$ additions; for the rotation operation, the string will also return to its original state after at most $n$ rotations.

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class Solution:
    def findLexSmallestString(self, s: str, a: int, b: int) -> str:
        q = deque([s])
        vis = {s}
        ans = s
        while q:
            s = q.popleft()
            if ans > s:
                ans = s
            t1 = ''.join(
                [str((int(c) + a) % 10) if i & 1 else c for i, c in enumerate(s)]
            )
            t2 = s[-b:] + s[:-b]
            for t in (t1, t2):
                if t not in vis:
                    vis.add(t)
                    q.append(t)
        return ans

Continue Practicing

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