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Minimum Operations to Make Binary Array Elements Equal to One I

Number: 3475

Difficulty: Medium

Paid? No

Companies: Google, Amazon, Uber

Problem Description

Given a binary array nums, you can perform an operation any number of times where you choose any 3 consecutive elements and flip all of them (i.e. change 0 to 1 and 1 to 0). Your task is to determine the minimum number of operations needed to make every element in nums equal to 1. If it is impossible to do so, return -1.

Key Insights

  • This problem is similar to the "Minimum K Bit Flips" question where we need to flip consecutive segments to achieve a target state.
  • A greedy approach is used: traverse the array and, when encountering an element that is not 1 (after considering previous flips), flip the current triple.
  • Use a difference array (or sliding window counter) to efficiently maintain the cumulative influence of prior flips.
  • Only positions where a triple flip is possible (i.e. i to i+2) can be flipped; any failure in the tail end means the transformation is impossible.

Space and Time Complexity

Time Complexity: O(n) - We traverse the array once. Space Complexity: O(n) - We use an auxiliary array (or diff array) to keep track of flip effects.

Solution

We use a greedy algorithm combined with a difference array technique. As we traverse the array, we keep track of how many flips have affected the current index using a variable currentFlip. At each index, we determine the effective value by considering the parity of currentFlip. If the effective value isn’t 1, we flip the next 3 consecutive elements and update our difference array to mark where this flip effect ends. If at any position (especially towards the end) we cannot flip due to boundary constraints, we return -1. This approach ensures that we perform the minimum number of flips required.

Code Solutions

# Python solution using a difference array to track flip effects
def minOperations(nums):
    n = len(nums)
    k = 3     # window size is fixed to 3
    ops = 0
    currentFlip = 0  # total flips affecting the current index
    diff = [0] * (n + 1)  # difference array to efficiently add and remove flip effects
    
    # Traverse through the array
    for i in range(n):
        # Update the current flip count with the current difference
        currentFlip += diff[i]
        
        # Determine the effective value at index i after previous flips
        # If currentFlip is odd, the bit is flipped from its original state
        effective = nums[i] ^ (currentFlip & 1)
        
        # If effective value is not 1, we perform a flip operation
        if effective == 0:
            # If there aren't enough elements remaining to flip, return -1
            if i + k > n:
                return -1
            ops += 1          # increment the operation counter
            currentFlip += 1  # apply the flip effect to the current index
            diff[i + k] -= 1  # schedule removal of this flip effect after i+k index
            
    return ops

# Example usage:
example1 = [0,1,1,1,0,0]
example2 = [0,1,1,1]
print(minOperations(example1))  # Output: 3
print(minOperations(example2))  # Output: -1
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