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# 2874. Maximum Value of an Ordered Triplet II

## Problem Statement

<br>

You are given a **0-indexed** integer array `nums`.

Return ***the maximum value over all triplets of indices*** `(i, j, k)` *such that* `i < j < k`*.* If all such triplets have a negative value, return `0`.

The **value of a triplet of indices** `(i, j, k)` is equal to `(nums[i] - nums[j]) * nums[k]`.

&#x20;

**Example 1:**

<pre><code><strong>Input: nums = [12,6,1,2,7]
</strong><strong>Output: 77
</strong><strong>Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
</strong>It can be shown that there are no ordered triplets of indices with a value greater than 77. 
</code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [1,10,3,4,19]
</strong><strong>Output: 133
</strong><strong>Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
</strong>It can be shown that there are no ordered triplets of indices with a value greater than 133.
</code></pre>

**Example 3:**

<pre><code><strong>Input: nums = [1,2,3]
</strong><strong>Output: 0
</strong><strong>Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
</strong></code></pre>

&#x20;

**Constraints:**

* `3 <= nums.length <= 105`
* `1 <= nums[i] <= 106`

## Intuition

```
Approach:
Better

At each element, Go Left and also Right to find the Max values
As (i - j) * k
Bascially we are iterating at j to find left(i) and Right(k)

Optimal:
Save the prefix_max and suffix_max at each step to avoid the extra loop
```

### Links

<https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-ii/description/>

### Video Links

### Approach 1:

```
```

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

```cpp
class Solution {
public:
    long long maximumTripletValue(vector<int>& v) {
        int n=v.size();
        long long ans=0;
        vector<int>l(n,0);
        vector<int>r(n,0);
        int s=v[0];
        for(int i=1;i<n;i++)
        {
            l[i]=s;
            s=max(s,v[i]);
        }
        s=v[n-1];
        for(int i=n-2;i>=0;i--)
        {
            r[i]=s;
            s=max(s,v[i]);
        }
        for(int i=1;i<n-1;i++)
        {
            long long k=(long long)(l[i]-v[i])*r[i];
            ans=max(ans,k);
        }
        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

###