LeetCode in Kotlin
995. Minimum Number of K Consecutive Bit Flips
Hard
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
Solution
class Solution {
fun minKBitFlips(nums: IntArray, k: Int): Int {
val n = nums.size
val pref = IntArray(n)
for (i in 0 until n) {
if (i == 0) {
if (nums[i] == 0) {
pref[i]++
}
} else {
pref[i] = pref[i - 1]
val flips = pref[i] - if (i - k >= 0) pref[i - k] else 0
if (flips % 2 == nums[i]) {
if (i + k > n) {
return -1
}
pref[i]++
}
}
}
return pref[n - 1]
}
}