Algorithm
Problem Name: 995. Minimum Number of K Consecutive Bit Flips
You are given a binary array nums and an integer k.
A k-bit flip is choosing a subarray of length k from nums and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.
Return the minimum number of k-bit flips required so that there is no 0 in the array. If it is not possible, return -1.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [0,1,0], k = 1 Output: 2 Explanation: Flip nums[0], then flip nums[2].
Example 2:
Input: nums = [1,1,0], k = 2 Output: -1 Explanation: No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].
Example 3:
Input: nums = [0,0,0,1,0,1,1,0], k = 3 Output: 3 Explanation: Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0] Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0] Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]
Constraints:
1 <= nums.length <= 1051 <= k <= nums.length
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const minKBitFlips = function(nums, k) {
let cur = 0, res = 0
const n = nums.length
for(let i = 0; i < n; i++) {
if(i >= k && nums[i - k] === 2) cur--
if(cur % 2 === nums[i]) {
if(i + k > n) return -1
nums[i] = 2
cur++
res++
}
}
return res
};
Input
ums = [0,1,0], k = 1
Output
2
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def minKBitFlips(self, a: List[int], k: int) -> int:
q = collections.deque()
res = 0
for i in range(len(a)):
if len(q) % 2 != 0 and a[i] == 1 or len(q) % 2 == a[i] == 0:
res += 1
q.append(i+k-1)
if q and q[0] == i: q.popleft()
if q and q[-1] >= len(a): return -1
return res
Input
ums = [0,1,0], k = 1
Output
2