LeetCode [995] Minimum Number of K Consecutive Bit Flips

 995. Minimum Number of K Consecutive Bit Flips

Hard

In an array A containing only 0s and 1s, a K-bit flip consists of choosing a (contiguous) subarray of length K 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.

 

Example 1:

Input: A = [0,1,0], K = 1
Output: 2
Explanation: Flip A[0], then flip A[2].

Example 2:

Input: A = [1,1,0], K = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].

Example 3:

Input: A = [0,0,0,1,0,1,1,0], K = 3
Output: 3
Explanation:
Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]

 

Note:

1. 1 <= A.length <= 30000
2. 1 <= K <= A.length

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class Solution { public int minKBitFlips(int[] A, int K) { int n = A.length; boolean[] everFlipped = new boolean[n]; int windowFlipCnt = 0, totalCnt = 0; for(int i=0; i<n; ++i){ if(i-K>=0 && everFlipped[i-K]==true){ windowFlipCnt--; } //1. windowFlipCnt%2==0 && A[i]==0: A[i] has been flipped even times. still 0 //2. windowFlipCnt%2==1 && A[i]==1: A[i] has been flipped odd times. became 0 //in both cases we need to flip it again if(windowFlipCnt%2==A[i]){ if(i+K>n) return -1; windowFlipCnt++; totalCnt++; everFlipped[i] = true; } } return totalCnt; } }