1695. Maximum Erasure Value

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 <= 105
• 1 <= 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 $nums$.

Python3JavaC++GoTypeScript

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

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;
    }
}

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;
    }
};

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
}

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;
}

Solution 2: Two Pointers

The problem is actually asking us to find the longest subarray in which all elements are distinct. We can use two pointers $i$ and $j$ to point to the left and right boundaries of the subarray, initially $i = 0$, $j = 0$. In addition, we use a hash table $vis$ to record the elements in the subarray.

We traverse the array, for each number $x$, if $x$ is in $vis$, then we continuously remove $nums[i]$ from $vis$, until $x$ is not in $vis$. In this way, we find a subarray without duplicate elements. We add $x$ to $vis$, update the sum of the subarray $s$, and then update the answer $ans = \max(ans, s)$.

After the traversal, we can get the maximum sum of the subarray.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Python3JavaC++GoTypeScript

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

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;
    }
}

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;
    }
};

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
}

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;
}

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