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 position of each number, and use a prefix sum array s to record the sum from the starting point to the current position. We use a variable j to record the left endpoint of the current non-repeating subarray.
We iterate through the array. For each number v, if d[v] exists, 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), where n is the length of the array nums.
PythonJavaC++GoTypeScriptRust
class Solution:
def maximumUniqueSubarray(self, nums: List[int]) -> int:
d = [0] * (max(nums) + 1)
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
impl Solution {
pub fn maximum_unique_subarray(nums: Vec<i32>) -> i32 {
let m = *nums.iter().max().unwrap() as usize;
let mut d = vec![0; m + 1];
let n = nums.len();
let mut s = vec![0; n + 1];
for i in 0..n {
s[i + 1] = s[i] + nums[i];
}
let mut ans = 0;
let mut j = 0;
for (i, &v) in nums.iter().enumerate().map(|(i, v)| (i + 1, v)) {
j = j.max(d[v as usize]);
ans = ans.max(s[i] - s[j]);
d[v as usize] = i;
}
ans
}
}(code-box)
Solution 2: Two Pointers (Sliding Window)
The problem is essentially asking us to find the longest subarray where all elements are distinct. We can use two pointers i and j to point to the left and right boundaries of the subarray, initially i = 0 and j = 0. Additionally, we use a hash table vis to record the elements in the subarray.
We iterate through the array. For each number x, if x is in vis, we continuously remove nums[i] from vis until x is no longer in vis. This way, we find a subarray that contains no duplicate elements. We add x to vis, update the subarray sum s, and then update the answer ans = max(ans, s).
After the iteration, we can get the maximum subarray sum.
The time complexity is O(n), and the space complexity is O(n), where n is the length of the array nums.
PythonJavaC++GoTypeScriptRust
class Solution:
def maximumUniqueSubarray(self, nums: List[int]) -> int:
vis = set()
ans = s = i = 0
for x in nums:
while x in vis:
y = nums[i]
s -= y
vis.remove(y)
i += 1
vis.add(x)
s += x
ans = max(ans, s)
return ans(code-box)
class Solution {
public int maximumUniqueSubarray(int[] nums) {
Set<Integer> vis = new HashSet<>();
int ans = 0, s = 0, i = 0;
for (int x : nums) {
while (vis.contains(x)) {
s -= nums[i];
vis.remove(nums[i++]);
}
vis.add(x);
s += x;
ans = Math.max(ans, s);
}
return ans;
}
}(code-box)
class Solution {
public:
int maximumUniqueSubarray(vector<int>& nums) {
unordered_set<int> vis;
int ans = 0, s = 0, i = 0;
for (int x : nums) {
while (vis.contains(x)) {
s -= nums[i];
vis.erase(nums[i++]);
}
vis.insert(x);
s += x;
ans = max(ans, s);
}
return ans;
}
};(code-box)
func maximumUniqueSubarray(nums []int) (ans int) {
vis := map[int]bool{}
var s, i int
for _, x := range nums {
for vis[x] {
s -= nums[i]
vis[nums[i]] = false
i++
}
vis[x] = true
s += x
ans = max(ans, s)
}
return
}(code-box)
function maximumUniqueSubarray(nums: number[]): number {
const vis: Set<number> = new Set();
let [ans, s, i] = [0, 0, 0];
for (const x of nums) {
while (vis.has(x)) {
s -= nums[i];
vis.delete(nums[i++]);
}
vis.add(x);
s += x;
ans = Math.max(ans, s);
}
return ans;
}(code-box)
use std::collections::HashSet;
impl Solution {
pub fn maximum_unique_subarray(nums: Vec<i32>) -> i32 {
let mut vis = HashSet::new();
let (mut ans, mut s, mut i) = (0, 0, 0);
for &x in &nums {
while vis.contains(&x) {
let y = nums[i];
s -= y;
vis.remove(&y);
i += 1;
}
vis.insert(x);
s += x;
ans = ans.max(s);
}
ans
}
}(code-box)
