Return the maximum value over all triplets of indices(i, j, k)such thati < 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].
Example 1:
Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
It can be shown that there are no ordered triplets of indices with a value greater than 77.
Example 2:
Input: nums = [1,10,3,4,19]
Output: 133
Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
It can be shown that there are no ordered triplets of indices with a value greater than 133.
Example 3:
Input: nums = [1,2,3]
Output: 0
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.
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
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;
}
};