1138HardArrayBit ManipulationQueueSliding WindowPrefix SumBrute-Force Search

Flipping Windows to Raise Every Switch

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Trains the technique from

LeetCode 995Minimum Number of K Consecutive Bit Flips

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A row of switches holds only 0 and 1. One move picks exactly k neighbouring switches and flips every one of them, so each 0 becomes a 1 and each 1 becomes a 0.

Return the fewest moves that leave every switch showing 1, or -1 when no number of moves can do it.

Examples

Example 1

Input: switches = [0, 0, 0, 0], k = 2
Output: 2

Flip the first two switches and then the last two.

Example 2

Input: switches = [0, 1], k = 2
Output: -1

Only one window exists and it covers both switches, so flipping it turns the 1 off while turning the 0 on, and the row can never come out all ones.

Example 3

Input: switches = [1, 0, 1, 0, 1, 0, 1], k = 1
Output: 3

A window of one switch flips a single switch, so each of the three zeros costs its own move.

Constraints

1 <= k <= switches.length <= 10^5
0 <= switches[i] <= 1

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def min_k_bit_flips(switches: list[int], k: int) -> int:

Java

public int minKBitFlips(int[] switches, int k)

The intended approach

Time O(n) over the row.Space O(n) for the retirement list.

Only the leftmost switch of the row sits in a single possible window, the one starting at it, so whether that window is used is forced by whether the switch shows `0`. Once that is settled the same argument applies to the next switch, and so on: at every position the decision is forced by what the switch shows after the windows already opened have had their say. What a switch shows after those windows is decided by how many of them cover it, and only whether that count is odd or even matters. So walk the row carrying the number of open windows covering the current position. A window opened at position `i` stops covering from `i + k` onwards, so note that retirement in a list and subtract it on arrival. When the switch and the running count together come out even, the switch is showing `0` and a window has to open here. If there is no room for a full window before the row ends, nothing can ever fix that switch and the row is impossible. Otherwise count the move and carry on. Each position costs a fixed amount of work.

Pythonentry point: min_k_bit_flips

def min_k_bit_flips(switches, k):
    n = len(switches)
    retire = [0] * (n + 1)
    moves = 0
    live = 0
    for i in range(n):
        live -= retire[i]
        if (switches[i] + live) % 2 == 0:
            if i + k > n:
                return -1
            moves += 1
            live += 1
            retire[i + k] += 1
    return moves

The obvious version

Pack the row into a bit pattern and walk outwards from it one move at a time, breadth first, trying every window position from each pattern reached, until the all-ones pattern turns up. The first time it appears is the fewest moves, and a search that runs dry means the row is impossible. Costs two to the power of the row length.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_min_k_bit_flips(switches, k):
    from collections import deque
    n = len(switches)
    start = 0
    for i in range(n):
        if switches[i]:
            start |= 1 << i
    goal = (1 << n) - 1
    if start == goal:
        return 0
    band = (1 << k) - 1
    seen = {start: 0}
    queue = deque([start])
    while queue:
        state = queue.popleft()
        step = seen[state]
        for i in range(n - k + 1):
            nxt = state ^ (band << i)
            if nxt == goal:
                return step + 1
            if nxt not in seen:
                seen[nxt] = step + 1
                queue.append(nxt)
    return -1

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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