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Minimum Operations to Make Binary Array Elements Equal to One II

Difficulty: Medium

Problem Description

You are given a binary array nums. You can do the following operation on the array any number of times (possibly zero): Choose any index i from the array and flip all the elements from index i to the end of the array. Flipping an element means changing its value from 0 to 1, and from 1 to 0. Return the minimum number of operations required to make all elements in nums equal to 1.

Key Insights

  • Flipping elements from a certain index affects all subsequent elements.
  • The goal is to minimize the number of flips to ensure all elements are 1.
  • A flip is only necessary when encountering a 0, which can lead to a consecutive series of flips.
  • The solution involves counting transitions between 0 and 1 to determine required operations.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve the problem, we can iterate through the array and count the number of transitions between 0 and 1. Whenever we encounter a 0 following a 1 (or vice versa), it indicates a point where a flip is needed. We can keep track of these transitions to calculate the minimum number of operations required to make the entire array equal to 1.

The main data structure used is a simple integer counter to track flips, as we need to count transitions in a single pass through the array.

Code Solutions

def minOperations(nums):
    n = len(nums)
    flips = 0
    # Initial state
    current = 1
    
    for i in range(n):
        if nums[i] != current:
            flips += 1
            current = nums[i]
    
    return flips + (1 if nums[-1] == 0 else 0)

# Example usage
print(minOperations([0, 1, 1, 0, 1]))  # Output: 4
print(minOperations([1, 0, 0, 0]))      # Output: 1
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