#3191

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Minimum Operations to Make Binary Array Elements Equal to One I

Find the minimum number of operations to make all elements of a binary array equal to one using sliding window and state updates.

array bit manipulation queue sliding window prefix sum

Problem Statement

You are given a binary array nums and can perform the following operation any number of times: flipping an element means changing its value from 0 to 1, or vice versa.

Your task is to find the minimum number of operations to make all elements in the array equal to 1. If it is impossible to achieve this, return -1.

Examples

Example 1

Input: nums = [0,1,1,1,0,0]

Output: 3

We can do the following operations:

Example 2

Input: nums = [0,1,1,1]

Output: -1

It is impossible to make all elements equal to 1.

Constraints

3 <= nums.length <= 105
0 <= nums[i] <= 1

Solution Approach

Sliding Window Technique

The solution involves using a sliding window to identify the minimal operations needed. The key insight is that the window should expand to include zeros and flip them to ones, while adjusting the window size to track the number of flips.

State Updates with Prefix Sum

Track the number of flips made with a running state. As the window slides through the array, update the count of flips, ensuring that the window covers all necessary elements to convert zeros to ones.

Edge Case Handling

Handle cases where it’s impossible to turn all elements into 1s, such as when there are too few ones to flip the zeros. The algorithm needs to quickly detect and return -1 in such cases.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity is O(n), where n is the length of the array, as we are iterating through it only once. The space complexity is O(1) since we are using a fixed amount of extra space for the sliding window and state updates.

What Interviewers Usually Probe

Understanding of sliding window approach with state updates.
Ability to optimize for space and time complexity.
Clear reasoning for edge case handling, especially when no solution is possible.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle the edge case where it's impossible to flip all elements to one, returning an incorrect result.
Not properly adjusting the window size during the iteration, leading to unnecessary operations.
Misunderstanding how the sliding window updates, causing an inefficient solution.

Follow-up variants

Consider a variation where the flip operation can only be applied a limited number of times.
Handling larger input sizes while maintaining the O(n) time complexity.
Implementing the solution in a way that minimizes memory usage for large inputs.

FAQ

What is the key technique for solving the Minimum Operations to Make Binary Array Elements Equal to One I problem?

The key technique is using a sliding window approach combined with state updates to efficiently track the minimum number of operations needed.

How does GhostInterview help with understanding the sliding window pattern?

GhostInterview offers step-by-step feedback on implementing the sliding window pattern, helping you master this approach in the context of the problem.

What is the time complexity of the solution for Minimum Operations to Make Binary Array Elements Equal to One I?

The time complexity is O(n), as we only need to iterate through the array once while updating the sliding window.

What should be done if it's impossible to make all elements equal to 1 in this problem?

In such cases, return -1 to indicate that it is impossible to achieve the goal with the given array.

How can edge cases like a small array or a mostly ones array be handled in this problem?

Edge cases can be handled by adjusting the sliding window size and checking for early exits if the solution is found or if the task is impossible.

terminal

Solution

Solution 1: Sequential Traversal + Simulation

We notice that the first position in the array that is $0$ must undergo a flip operation, otherwise, it cannot be turned into $1$. Therefore, we can sequentially traverse the array, and each time we encounter $0$, we flip the next two elements and accumulate one operation count.

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class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = 0
        for i, x in enumerate(nums):
            if x == 0:
                if i + 2 >= len(nums):
                    return -1
                nums[i + 1] ^= 1
                nums[i + 2] ^= 1
                ans += 1
        return ans

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