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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 <= 105
  • 1 <= k <= nums.length

Companies:
Google, Amazon, Adobe

Related Topics:
Array, Bit Manipulation, Sliding Window, Prefix Sum

Similar Questions:

Solution 1. Queue

// OJ: https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(K)
class Solution {
public:
    int minKBitFlips(vector<int>& A, int k) {
        queue<int> q;
        int ans = 0;
        for (int i = 0, N = A.size(); i < N; ++i) {
            if (q.size() && q.front() <= i) q.pop();
            int flip = q.size() % 2;
            if ((A[i] ^ flip) == 0) {
                if (i + k > N) return -1;
                ++ans;
                q.push(i + k);
            }
        }
        return ans;
    }
};

TODO

Checkout https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/discuss/238609/JavaC%2B%2BPython-One-Pass-and-O(1)-Space