Description#
You are given an array of positive integers nums and want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements.
Return the maximum score you can get by erasing exactly one subarray.
An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).
Example 1:
Input: nums = [4,2,4,5,6]
Output: 17
Explanation: The optimal subarray here is [2,4,5,6].
Example 2:
Input: nums = [5,2,1,2,5,2,1,2,5]
Output: 8
Explanation: The optimal subarray here is [5,2,1] or [1,2,5].
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 104
Solutions#
Solution 1: Array or Hash Table + Prefix Sum#
We use an array or hash table $d$ to record the last occurrence of each number, use $s$ to record the prefix sum, and use $j$ to record the left endpoint of the current non-repeating subarray.
We traverse the array, for each number $v$, if $d[v]$ exists, then we update $j$ to $max(j, d[v])$, which ensures that the current non-repeating subarray does not contain $v$. Then we update the answer to $max(ans, s[i] - s[j])$, and finally update $d[v]$ to $i$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
1
2
3
4
5
6
7
8
9
10
| class Solution:
def maximumUniqueSubarray(self, nums: List[int]) -> int:
d = defaultdict(int)
s = list(accumulate(nums, initial=0))
ans = j = 0
for i, v in enumerate(nums, 1):
j = max(j, d[v])
ans = max(ans, s[i] - s[j])
d[v] = i
return ans
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
| class Solution {
public int maximumUniqueSubarray(int[] nums) {
int[] d = new int[10001];
int n = nums.length;
int[] s = new int[n + 1];
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
int ans = 0, j = 0;
for (int i = 1; i <= n; ++i) {
int v = nums[i - 1];
j = Math.max(j, d[v]);
ans = Math.max(ans, s[i] - s[j]);
d[v] = i;
}
return ans;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| class Solution {
public:
int maximumUniqueSubarray(vector<int>& nums) {
int d[10001]{};
int n = nums.size();
int s[n + 1];
s[0] = 0;
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
int ans = 0, j = 0;
for (int i = 1; i <= n; ++i) {
int v = nums[i - 1];
j = max(j, d[v]);
ans = max(ans, s[i] - s[j]);
d[v] = i;
}
return ans;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
| func maximumUniqueSubarray(nums []int) (ans int) {
d := [10001]int{}
n := len(nums)
s := make([]int, n+1)
for i, v := range nums {
s[i+1] = s[i] + v
}
for i, j := 1, 0; i <= n; i++ {
v := nums[i-1]
j = max(j, d[v])
ans = max(ans, s[i]-s[j])
d[v] = i
}
return
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
| function maximumUniqueSubarray(nums: number[]): number {
const m = Math.max(...nums);
const n = nums.length;
const s: number[] = Array.from({ length: n + 1 }, () => 0);
for (let i = 1; i <= n; ++i) {
s[i] = s[i - 1] + nums[i - 1];
}
const d = Array.from({ length: m + 1 }, () => 0);
let [ans, j] = [0, 0];
for (let i = 1; i <= n; ++i) {
j = Math.max(j, d[nums[i - 1]]);
ans = Math.max(ans, s[i] - s[j]);
d[nums[i - 1]] = i;
}
return ans;
}
|
