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Minimum Number of K Consecutive Bit Flips

Number: 1037

Difficulty: Hard

Paid? No

Companies: Amazon, Meta, Google

Problem Description

Given a binary array and an integer k, perform operations where you select a subarray of length k and flip every bit (i.e., turn 0s into 1s and 1s into 0s). Determine the minimum number of such k-bit flips required so that every element in the array becomes 1. If it is impossible, return -1.

Key Insights

  • Instead of flipping all k bits directly every time, use a sliding window to track the cumulative effect of previous flips.
  • Maintain a counter representing the number of flips affecting the current index, which helps determine the effective value of a bit (considering odd/even flips).
  • Use an auxiliary array (or similar structure like a queue) to mark where the effect of a flip expires.
  • The algorithm uses a greedy approach: if a bit is 0 after taking previous flips into account, and a flip is possible, perform a new flip.

Space and Time Complexity

Time Complexity: O(n) where n is the length of the input array. Space Complexity: O(n) due to the auxiliary array used to track the flip effects.

Solution

The solution leverages a greedy sliding window technique where, for each index, we adjust a flip counter based on earlier flips. When the current bit (modified by previous flips) remains 0, a new flip operation is initiated, marking its effect to last for k positions. If at any index a flip cannot be performed due to out-of-bound constraints, the function returns -1. This strategy minimizes redundant operations and efficiently computes the minimal number of flips.

Code Solutions

def minKBitFlips(nums, k):
    n = len(nums)
    flip_effect = [0] * n  # Marks where the effect of a flip ends
    flip_count = 0        # Number of active flips affecting the current index
    answer = 0            # Total number of flips performed
    
    for i in range(n):
        # Remove the effect of flips that are no longer in the current window.
        if i >= k:
            flip_count -= flip_effect[i - k]
        
        # Check if the current effective bit is 0 (considering flip_count).
        if (nums[i] + flip_count) % 2 == 0:
            # Not enough elements left to flip? Then it's impossible.
            if i + k > n:
                return -1
            flip_effect[i] = 1  # Mark a flip starting at index i.
            flip_count += 1
            answer += 1
            
    return answer

# Example usage:
print(minKBitFlips([0, 1, 0], 1))      # Output: 2
print(minKBitFlips([1, 1, 0], 2))      # Output: -1
print(minKBitFlips([0, 0, 0, 1, 0, 1, 1, 0], 3))  # Output: 3
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