Description#
Given a binary array nums
and an integer k
, return the maximum number of consecutive 1
's in the array if you can flip at most k
0
's.
Example 1:
Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Example 2:
Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Constraints:
1 <= nums.length <= 105
nums[i]
is either 0
or 1
.0 <= k <= nums.length
Solutions#
Solution 1#
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| class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
ans = 0
cnt = j = 0
for i, v in enumerate(nums):
if v == 0:
cnt += 1
while cnt > k:
if nums[j] == 0:
cnt -= 1
j += 1
ans = max(ans, i - j + 1)
return ans
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| class Solution {
public int longestOnes(int[] nums, int k) {
int j = 0, cnt = 0;
int ans = 0;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] == 0) {
++cnt;
}
while (cnt > k) {
if (nums[j++] == 0) {
--cnt;
}
}
ans = Math.max(ans, i - j + 1);
}
return ans;
}
}
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| class Solution {
public:
int longestOnes(vector<int>& nums, int k) {
int ans = 0;
int cnt = 0, j = 0;
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] == 0) {
++cnt;
}
while (cnt > k) {
if (nums[j++] == 0) {
--cnt;
}
}
ans = max(ans, i - j + 1);
}
return ans;
}
};
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| func longestOnes(nums []int, k int) int {
ans := 0
j, cnt := 0, 0
for i, v := range nums {
if v == 0 {
cnt++
}
for cnt > k {
if nums[j] == 0 {
cnt--
}
j++
}
ans = max(ans, i-j+1)
}
return ans
}
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| function longestOnes(nums: number[], k: number): number {
const n = nums.length;
let l = 0;
for (const num of nums) {
if (num === 0) {
k--;
}
if (k < 0 && nums[l++] === 0) {
k++;
}
}
return n - l;
}
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| impl Solution {
pub fn longest_ones(nums: Vec<i32>, mut k: i32) -> i32 {
let n = nums.len();
let mut l = 0;
for num in nums.iter() {
if num == &0 {
k -= 1;
}
if k < 0 {
if nums[l] == 0 {
k += 1;
}
l += 1;
}
}
(n - l) as i32
}
}
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Solution 2#
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| class Solution:
def longestOnes(self, nums: List[int], k: int) -> int:
l = r = -1
while r < len(nums) - 1:
r += 1
if nums[r] == 0:
k -= 1
if k < 0:
l += 1
if nums[l] == 0:
k += 1
return r - l
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| class Solution {
public int longestOnes(int[] nums, int k) {
int l = 0, r = 0;
while (r < nums.length) {
if (nums[r++] == 0) {
--k;
}
if (k < 0 && nums[l++] == 0) {
++k;
}
}
return r - l;
}
}
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| class Solution {
public:
int longestOnes(vector<int>& nums, int k) {
int l = 0, r = 0;
while (r < nums.size()) {
if (nums[r++] == 0) --k;
if (k < 0 && nums[l++] == 0) ++k;
}
return r - l;
}
};
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| func longestOnes(nums []int, k int) int {
l, r := -1, -1
for r < len(nums)-1 {
r++
if nums[r] == 0 {
k--
}
if k < 0 {
l++
if nums[l] == 0 {
k++
}
}
}
return r - l
}
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| function longestOnes(nums: number[], k: number): number {
const n = nums.length;
let l = 0;
let res = k;
const count = [0, 0];
for (let r = 0; r < n; r++) {
count[nums[r]]++;
res = Math.max(res, r - l);
while (count[0] > k) {
count[nums[l]]--;
l++;
}
}
return Math.max(res, n - l);
}
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