This problem involves a binary array (containing only 0s and 1s). You are given a fixed window size (e.g., 3). In one operation, you can pick any subarray of that fixed size and flip all its elements (0 becomes 1, 1 becomes 0). The goal is to make all elements in the array equal to 1 using the minimum number of operations, or return -1 if it's impossible.
This is a classic "Greedy with Sliding Window" problem asked by companies like Uber and Meta. It tests your ability to recognize that the order of operations doesn't matter and that for each index, you only have one chance to decide whether to flip it or not (based on its current state as you traverse from left to right). It evaluates logical consistency and handling of boundary conditions.
The primary pattern is the Greedy approach. As you iterate through the array from left to right, if you encounter a 0 at index i, you must perform a flip starting at index i. There is no other way to turn that 0 into a 1 because any flip starting at an earlier index has already been decided, and any flip starting at a later index won't include index i. You keep doing this until you reach the last few elements where a full window can no longer fit.
Array: [0, 1, 1, 1, 0, 0], Window size: 3.
[0, 1, 2]. Array: [1, 0, 0, 1, 0, 0]. Ops = 1.[1, 2, 3]. Array: [1, 1, 1, 0, 0, 0]. Ops = 2.[3, 4, 5]. Array: [1, 1, 1, 1, 1, 1]. Ops = 3.
Result: 3.A common error is trying to find the "best" place to flip using recursion or dynamic programming, which leads to exponential time complexity. Another mistake is failing to check the remaining elements at the end of the array; if after the greedy process there are still any 0s left in the last "window-minus-one" elements, it means the task is impossible. Some also forget that flipping an element twice is the same as not flipping it at all.
Practice "Greedy on Arrays" problems where decisions are forced by the direction of traversal. This "Left-to-Right" forcing logic is very common in flipping or toggle-style problems. Also, learn to optimize range flips using a difference array or a queue to avoid O(N*K) complexity, although for smaller window sizes, a direct flip might suffice depending on the constraints.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Minimum Number of K Consecutive Bit Flips | Hard | Solve | |
| Longest Nice Subarray | Medium | Solve | |
| Maximum XOR for Each Query | Medium | Solve | |
| Shortest Subarray With OR at Least K II | Medium | Solve | |
| XOR Queries of a Subarray | Medium | Solve |