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995_Minimum_Number_of_K_Consecutive_Bit_Flips.cpp
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46 lines (46 loc) · 1.5 KB
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/*
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 <= 105
1 <= k <= nums.length
*/
class Solution {
public:
int minKBitFlips(vector<int>& nums, int k) {
int ans = 0, flips = 0;
for(int i = 0; i < nums.size(); i++) {
if((nums[i] + flips) % 2 == 0) {
if(i > nums.size() - k) {
return -1;
} else {
ans++;
flips++;
nums[i] = -1;
}
}
if(i + 1 >= k) {
flips -= (nums[i - k + 1] < 0);
}
}
return ans;
}
};