Minimum Suffix Flips in Java

The problem of finding minimum suffix flips involves working with two binary strings: initial string s and target string target. Here, both the strings are of the length n and the initial string s is an entire string of zeros (i. e., s = "000. .. 0"). The aim is to ensure that the string s is transformed to the target string by making as few flip operations as possible.

Example 1:

Input

Output:

 
Minimum operations needed: 3

Explanation

Example 2:

Input

Output:

 
Minimum operations needed: 3    

Explanation

Approach 1: Using Recursion

The core idea of the Recursive Approach is to use recursion to handle the transformation. At each step, we check if the current character matches the expected character (based on the number of flips so far). If not, we perform a flip and make a recursive call to handle the rest of the string.

Algorithm

Step 1: Define minOperations method which calls the helper method minOperationsHelper with the initial index 0 and flipped set to false.

Step 2: The minOperationsHelper method checks if the index has reached the end of the string. If so, it returns 0 because no more operations are needed.

Step 3: Determine the expected character at the current index based on the flip status ('1' if flipped is true, '0' if false).

Step 4: Character Comparison:If the current character in target does not match the expected character, it means a flip is needed: Increment the operation count by 1. Make a recursive call with the next index and the flip status toggled.

Step 5: If the characters match, no flip is needed: Make a recursive call with the next index without changing the flip status.

Filename: MinimumSuffixFlips.java

Output:

 
Minimum operations needed: 3   

Time Complexity

The overall time complexity is O(n), where n is the length of the target string. The recursion processes each character in the string once.

Space Complexity

The Overall space complexity is O(n), due to the recursion call stack, which can go as deep as the length of the string in the worst case.

Approach 2: Using Stack

The core idea of the approach using a stack is to track the transitions in the target string where the bit changes from '0' to '1' or from '1' to '0'. Each transition represents a point where a flip is necessary. By pushing the characters of the target string onto a stack and counting the number of changes, we can determine the number of required flips.

Algorithm

Step 1: Initialization: Create an empty stack to keep track of the transitions in the target string. Initialize an operations counter to 0, which will keep track of the number of flip operations needed.

Step 2: Iterate Through the Target String: Loop through each character in the target string using a for-loop. For each character, check if the stack is empty or if the top element of the stack is different from the current character.

Step 3: If either condition is true, push the current character onto the stack and increment the operations counter, this signifies a new segment that needs flipping.

Step 4: The operations counter will contain the minimum number of flip operations needed to transform s into target and then return the minimum flips.

Filename: MinimumSuffixFlips1.java

Output:

 
Minimum operations needed: 3    

Time Complexity

The time complexity of the stack-based approach is O(n), where n is the length of the target string. The algorithm iterates through the target string once. Each character in the target string is processed in constant time.

Space Complexity

The space complexity of the stack-based approach is O(n). In the worst case, the stack can hold all characters of the target string if each character is different from its predecessor. Therefore, the space used by the stack grows linearly with the length of the target string.

Approach 3: Direct Transition Counting

The problem of transforming an initial binary string s (all zeros) into a target binary string target using the minimum number of flip operations can be efficiently solved using the Direct Transition Counting approach.

Algorithm

Step 1: Start by setting the operations counter to 0, which tracks the number of flip operations performed. Initialize a variable currentBit to '0', representing the initial state of the bits in the string s.

Step 2: Iterate Through the Target String: Loop through each character in the target string target. For each character, compare it with currentBit.

Step 3: If the current character in target is different from currentBit, it means a transition has occurred:

Step 3.1: Increment the operations counter by 1. Update currentBit to the current character in target.

Step 4: After iterating through the entire target string, the operations counter will hold the minimum number of flips required to transform s into target.

Filename: MinimumSuffixFlips2.java

Output:

 
Minimum operations needed: 3    

Time Complexity

The time complexity of the Direct Transition Counting approach is O(n), where n is the length of the target string. It is because the algorithm iterates through the target string once.

Space Complexity

The space complexity of the Direct Transition Counting approach is O(1). The algorithm uses a constant amount of extra space regardless of the input size. Only a few variables (operations, currentBit) are used to keep track of the current state and count of operations.

Approach 4: Greedy Flip Counting

Greedy Flip Counting is a straightforward method to determine the minimum number of flip operations required to transform an initial binary string s (initially all zeros) into a target binary string target, this approach relies on a greedy strategy to count transitions directly, resulting in an efficient solution.

Algorithm

Step 1: Set the operations counter to 0 to count the number of flip operations needed. Begin by comparing each bit in the target string with its previous bit to detect changes.

Step 2: Iterate through the target string: For each character, compare it with the previous character (or '0' for the first character).

Step 3: Whenever the current character differs from the previous character, increment the operations counter because it indicates a flip operation is required to match that segment of the string.

Step 4: After processing the entire string, return the operations counter, which now contains the minimum number of flips required.

Filename: MinimumSuffixFlips3.java

Output:

 
Minimum operations needed: 3    

Time Complexity

The time complexity of the Greedy Flip Counting approach is O(n), where n is the length of the target string. The algorithm iterates through the target string once. Each character in the target string is processed in constant time.

Space Complexity

The space complexity of the Greedy Flip Counting approach is O(1). The algorithm uses a constant amount of extra space regardless of the input size. Only a few variables (operations) are used to keep track of the number of flip operations.

Approach 5: Two-Pointer Technique

The Two-Pointer Technique provides an efficient way to solve the problem of transforming an initial binary string s (all zeros) into a target binary string target, this technique uses two pointers to track transitions in the target string and count the number of segments that need flipping.

Algorithm

Step 1: Initialize two pointers, current and next. Set the current pointer to the first character of the target string. Set the next pointer to the second character. Initialize the operations counter to 0.

Step 2: Iterate Through the Target String: Loop through the target string with the next pointer. Compare the character at the next pointer with the character at the current pointer.

Step 3: Whenever the character at the next pointer differs from the character at the current pointer, it indicates a transition:

Step 3.1: Increment the operations counter. Move the current pointer to the next pointer. Continue this process until the end of the string.

Step 4: The operations counter will hold the minimum number of flips required to transform s into target and then return the result.

Filename: MinimumSuffixFlips4.java

Output:

 
Minimum operations needed: 3    

Time Complexity

The time complexity of the Two-Pointer Technique is O(n), where n is the length of the target string. It is because, the algorithm iterates through the target string once.

Space Complexity

The space complexity of the Two-Pointer Technique is O(1). The algorithm uses a constant amount of extra space regardless of the input size.

Conclusion

In conclusion, the problem of Minimum Suffix Flips can be approached in various ways, each with its own strengths in terms of readability, efficiency, and space utilization. The choice of approach can depend on specific constraints and preferences.