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# 2841. Maximum Sum of Almost Unique Subarray

## Problem Statement

<br>

You are given an integer array `nums` and two positive integers `m` and `k`.

Return *the **maximum sum** out of all **almost unique** subarrays of length* `k` *of* `nums`. If no such subarray exists, return `0`.

A subarray of `nums` is **almost unique** if it contains at least `m` distinct elements.

A subarray is a contiguous **non-empty** sequence of elements within an array.

&#x20;

**Example 1:**

<pre><code><strong>Input: nums = [2,6,7,3,1,7], m = 3, k = 4
</strong><strong>Output: 18
</strong><strong>Explanation: There are 3 almost unique subarrays of size k = 4. These subarrays are [2, 6, 7, 3], [6, 7, 3, 1], and [7, 3, 1, 7]. Among these subarrays, the one with the maximum sum is [2, 6, 7, 3] which has a sum of 18.
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [5,9,9,2,4,5,4], m = 1, k = 3
</strong><strong>Output: 23
</strong><strong>Explanation: There are 5 almost unique subarrays of size k. These subarrays are [5, 9, 9], [9, 9, 2], [9, 2, 4], [2, 4, 5], and [4, 5, 4]. Among these subarrays, the one with the maximum sum is [5, 9, 9] which has a sum of 23.
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: nums = [1,2,1,2,1,2,1], m = 3, k = 3
</strong><strong>Output: 0
</strong><strong>Explanation: There are no subarrays of size k = 3 that contain at least m = 3 distinct elements in the given array [1,2,1,2,1,2,1]. Therefore, no almost unique subarrays exist, and the maximum sum is 0.
</strong></code></pre>

&#x20;

**Constraints:**

* `1 <= nums.length <= 2 * 104`
* `1 <= m <= k <= nums.length`
* `1 <= nums[i] <= 109`

## Intuition

```
Approach:
Apply sliding window for a window and find ans
```

### Links

<https://leetcode.com/problems/maximum-sum-of-almost-unique-subarray/>

### Video Links

### Approach 1:

```
Sliding window
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    long long maxSum(vector<int>& arr, int m, int k) {
        long long ans = 0, window_sum=0;
        unordered_map<int,int> mp;
        int low=0, high;
        
        for(high=0; high<k; high++){
            mp[arr[high]]++;
            window_sum += arr[high];
        }

        if(mp.size() >= m)
            ans = max(ans, window_sum);

        while (high<arr.size()) {
            if(mp[arr[low]] == 1)
                mp.erase(arr[low]);
            else
                mp[arr[low]]--;
            window_sum -= arr[low++];

            mp[arr[high]]++;
            window_sum += arr[high++];

            if(mp.size() >= m)
                ans = max(ans, window_sum);
        }

        return ans;  
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###