Overview

This problem asks us to find the minimum number of operations to turn all lights on by repeatedly toggling the states of lights. Each operation flips the states of \(K\) consecutive lights.

Analysis

A naive approach would be a greedy method where we scan lights from left to right, and whenever we find a light that is off, we flip \(K\) consecutive lights starting from that position. However, since the number of operations can be up to \(O(N)\) and each operation flips \(K\) lights, the time complexity becomes \(O(NK)\), which is too slow for this problem where \(N\) can be up to \(500,000\).

Therefore, we need a way to efficiently manage the effects of flip operations. Since flip operations have linearity, we can use lazy evaluation to manage the number of flips for each light using a difference array. This allows us to compute the current state of each light in \(O(1)\).

Algorithm

1. Create the initial state array arr (on: 1, off: 0).
2. Prepare a difference array diff to manage the effects of flip operations.
3. Prepare a variable flip_count to keep track of the cumulative number of flips up to the current position.
4. Loop through each light \(i\) from left to right:
   • Add the value from the difference array at position \(i\) to flip_count to get the cumulative flip count up to position \(i\).
   • Compute the current state of the light as (arr[i] + flip_count) % 2.
   • If the current state is off (0):
     • An operation is needed. However, since the operation range is from \(i\) to \(i+K-1\), if \(i+K-1 \geq N\), the range goes out of bounds and the operation is impossible, so output -1.
     • Increment the operation count and record the effect in the difference array (diff[i] += 1, diff[i+K] -= 1).
     • Increment flip_count to reflect the current state as on.
5. Output the minimum number of operations.

With this method, the state of each light can be computed in constant time, and the entire process runs in \(O(N)\) time.

Complexity

• Time complexity: \(O(N)\)
• Space complexity: \(O(N)\)

Implementation Notes

• Allocate the difference array with size \(N+2\) to prevent out-of-bounds access.

• Before performing an operation, check that the operation range fits within the array.

• The current state is computed as (initial value + flip count) % 2. Since performing a flip operation twice returns the light to its original state, taking mod 2 is sufficient.

  Source Code

import sys

def main():
    data = sys.stdin.read().splitlines()
    if not data:
        print(-1)
        return
    nk = data[0].split()
    if len(nk) < 2:
        print(-1)
        return
    N = int(nk[0])
    K = int(nk[1])
    S = data[1].strip()
    
    arr = [1 if c == '1' else 0 for c in S]
    diff = [0] * (N + 2)
    flip_count = 0
    res = 0
    
    for i in range(N):
        flip_count += diff[i]
        current = (arr[i] + flip_count) % 2
        if current == 0:
            if i + K - 1 >= N:
                print(-1)
                return
            res += 1
            flip_count += 1
            diff[i] += 1
            diff[i + K] -= 1
            current = 1
    
    print(res)

if __name__ == "__main__":
    main()

This editorial was generated by deepseekv3.