Minimum number of double or triple flips to make a binary string

ghz 昨天 ⋅ 5 views

Minimum number of double or triple flips to make a binary string all 1s

I just completed a coding interview and I am stumped on a question in it. Although the test is already over, I still really want to know how this question can be solved.

Problem: Given a binary string (containing only "1" or "0"), find the minimum number of flips required to get a binary string of all "1"s.

  • When a flip is performed, the value at that index will go from "0" to "1" or "1" to "0".
  • The only flips you are allowed to perform are either doubles or triples. Meaning that when a flip is performed, it must be performed on either two or three adjacent indices.

Example:

  • For the binary string "1010", the minimum number of flips required is 2. "1010" => "1001" => "1111" (end)
  • For "10001", minimm number of flips required is 1. "10001" => "11111" (end)

Answer

This problem is an interesting one where you need to perform flips on adjacent indices in a binary string to transform it into a string of all "1"s. You can only flip 2 or 3 consecutive bits at a time, and the goal is to minimize the number of flips.

Approach:

The key observation is that you can only flip adjacent pairs or triples of bits. This means you can either change "01" to "10", or flip three consecutive bits like "010" to "111". The idea is to work from left to right and flip any segment of 0s that you encounter until you reach the end of the string.

Steps to Solve:

  1. Iterate over the string: Start from the leftmost index. Whenever you encounter a "0", you'll want to flip it.
  2. Choose the optimal flip: If there are at least two adjacent "0"s, flip them (i.e., apply a "double" flip). If there are at least three adjacent "0"s, flip three of them at once (a "triple" flip). Choose the largest flip possible to minimize the number of flips.
  3. Keep track of the flips: Continue this process until you've converted the entire string to all "1"s.

Greedy Solution:

We will use a greedy approach where we always try to flip the maximum number of consecutive zeros (either pairs or triples) in order to minimize the total number of flips.

Here’s how you can implement the solution:

def min_flips_to_all_ones(binary_str):
    n = len(binary_str)
    flips = 0
    i = 0
    
    while i < n:
        if binary_str[i] == '0':
            # Check if we can flip 3 bits
            if i + 2 < n and binary_str[i+1] == '0' and binary_str[i+2] == '0':
                # Perform a triple flip
                flips += 1
                i += 3
            # Check if we can flip 2 bits
            elif i + 1 < n and binary_str[i+1] == '0':
                # Perform a double flip
                flips += 1
                i += 2
            else:
                # Flip this bit if it's a lone '0' (which can't form a pair or triplet)
                flips += 1
                i += 1
        else:
            # Move to the next bit if it's already '1'
            i += 1

    return flips

Explanation:

  • Loop through the string: We iterate through the string, checking each position. If it's a "0", we try to flip it.
  • Try flipping 3 bits: First, we check if we can perform a "triple flip" (i.e., 3 adjacent 0s). If yes, we flip them and move 3 positions forward.
  • Try flipping 2 bits: If we can't perform a triple flip, we check if we can flip a pair of adjacent 0s. If yes, we perform the flip and move 2 positions forward.
  • Handle single 0s: If neither a double nor triple flip is possible, we flip the single 0 and move to the next position.
  • Continue until the end: The loop continues until we've processed all bits.

Example 1:

Input: "1010"

  • Flip first "10" => "1001", then flip the last "01" => "1111".
  • Minimum flips required: 2
print(min_flips_to_all_ones("1010"))  # Output: 2

Example 2:

Input: "10001"

  • Flip "10001" to "11111" (single flip of the entire "10001").
  • Minimum flips required: 1
print(min_flips_to_all_ones("10001"))  # Output: 1

Time Complexity:

  • The solution processes each bit in the string exactly once, so the time complexity is O(n), where n is the length of the binary string.

Conclusion:

This greedy approach ensures that you minimize the number of flips by always flipping the largest possible segments of 0s (either pairs or triples), which guarantees the solution is efficient.

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