Formatted question description: https://leetcode.ca/all/1493.html
1493. Longest Subarray of 1’s After Deleting One Element (Medium)
Given a binary array nums, you should delete one element from it.
Return the size of the longest non-empty subarray containing only 1's in the resulting array.
Return 0 if there is no such subarray.
Example 1:
Input: nums = [1,1,0,1] Output: 3 Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.
Example 2:
Input: nums = [0,1,1,1,0,1,1,0,1] Output: 5 Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].
Example 3:
Input: nums = [1,1,1] Output: 2 Explanation: You must delete one element.
Example 4:
Input: nums = [1,1,0,0,1,1,1,0,1] Output: 4
Example 5:
Input: nums = [0,0,0] Output: 0
Constraints:
1 <= nums.length <= 10^5nums[i]is either0or1.
Related Topics:
Array
Solution 1.
prev2 and prev are the indexes of the non-one values we’ve seen most recently during scanning.
prev2 prev i
0 1 1 1 0 1 1 1 0
If the array only contains 1, then return N - 1.
Otherwise, the answer is the maximum of i - prev2 - 2.
// OJ: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestSubarray(vector<int>& A) {
int N = A.size(), prev2 = -1, prev = -1, ans = 0;
for (int i = 0; i <= N; ++i) {
if (i < N && A[i] == 1) continue;
if (i == N && prev == -1) return N - 1;
if (prev != -1) ans = max(ans, i - prev2 - 2);
prev2 = prev;
prev = i;
}
return ans;
}
};
Solution 2. Sliding Window
Sliding window [i, j) with at most 2 zeros.
- Keep extending the right edge until the window becomes invalid, i.e. more than one zero, or can’t be extended any more.
- Now
cntcan be either0,1or2. The corresponding size after deleting one element isj - i - max(cnt, 1). - Then we shrink the left edge until
cntbecomes valid, i.e.cnt <= 1. - Back to step 1 until the window reaches the end.
// OJ: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestSubarray(vector<int>& A) {
int i = 0, j = 0, N = A.size(), cnt = 0, ans = 0;
while (j < N) {
while (j < N && cnt <= 1) cnt += A[j++] == 0;
ans = max(ans, j - i - max(cnt, 1));
while (cnt > 1) cnt -= A[i++] == 0;
}
return ans;
}
};
Solution 3. Sliding Window
Sliding window [i, j) with at most 1 zero.
- Keep extending the right edge until it reaches the end or one more extension will include 2 zeros.
j - i - 1is the corresponding answer for this window.- Shrink the left edge until
cnt == 0. - Back to step 1 until the window reaches the end.
// OJ: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestSubarray(vector<int>& A) {
int i = 0, j = 0, N = A.size(), cnt = 0, ans = 0;
while (j < N) {
while (j < N && cnt + (A[j] == 0) <= 1) cnt += A[j++] == 0;
ans = max(ans, j - i - 1);
while (cnt == 1) cnt -= A[i++] == 0;
}
return ans;
}
};
Solution 4. Sliding Window
Sliding window [i, j] with at most 1 zero.
- Keep extending the right edge.
- Once the
cnt > 1, shrink the left edge. - The size for this window with one deletion is
j - i. - Back to step 1 until it reaches the end.
This solution is simpler than the previous one because we don’t need to think about when to stop the extension of the right edge.
// OJ: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestSubarray(vector<int>& A) {
int i = 0, j = 0, N = A.size(), cnt = 0, ans = 0;
for (; j < N; ++j) {
cnt += A[j] == 0;
while (cnt > 1) cnt -= A[i++] == 0;
ans = max(ans, j - i);
}
return ans;
}
};
Solution 5. Sliding Window
Sliding window [i - ans + 1, i] with at most 1 zero that never shrinks.
// OJ: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestSubarray(vector<int>& A) {
int N = A.size(), cnt = 0, ans = 0;
for (int i = 0; i < N; ++i) {
cnt += A[i] == 0;
if (cnt > 1) cnt -= A[i - ans] == 0;
else ++ans;
}
return ans - 1;
}
};
Sliding window [i, j] with at most 1 zero taht never shrinks.
// OJ: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestSubarray(vector<int>& A) {
int i = 0, j = 0, N = A.size(), cnt = 0;
for (; j < N; ++j) {
cnt += A[j] == 0;
if (cnt > 1) cnt -= A[i++] == 0;
}
return j - i - 1;
}
};
Java
class Solution {
public int longestSubarray(int[] nums) {
int length = nums.length;
int index = 0;
while (index < length && nums[index] == 0)
index++;
if (index == length)
return 0;
List<Integer> ones = new ArrayList<Integer>();
List<Integer> zeros = new ArrayList<Integer>();
int prev = 1;
int count = 1;
for (int i = index + 1; i < length; i++) {
int num = nums[i];
if (num == prev)
count++;
else {
if (prev == 1)
ones.add(count);
else
zeros.add(count);
prev = num;
count = 1;
}
}
if (prev == 1)
ones.add(count);
else
zeros.add(count);
if (index == 0 && zeros.size() == 0)
return ones.get(0) - 1;
int maxLength = ones.get(0);
int size = ones.size();
for (int i = 1; i < size; i++) {
int zero = zeros.get(i - 1);
int prevOne = ones.get(i - 1);
int one = ones.get(i);
maxLength = Math.max(maxLength, one);
if (zero == 1)
maxLength = Math.max(maxLength, prevOne + one);
}
return maxLength;
}
}