i am not understanding the approach of the mentor …can u explain me the concept in better way why we choose arr[i]*i + dp[i-2] ,dp[i-1]
Neighbouring enemy problem
@neha_153
This problem is similar to Alice and Bob. Let’s consider a simple case to begin with, if the sequence is 2 2 2. Then, clearly, Alex has only one choice - to pick the number 2. In that case, his score would be 2*3 = 6.
Moving forward, if the sequence was 2 2 3. Then Alex has two choices - to choose 2, or 3. Now, on choosing 2, his score would be 2 2 = 4. But (2 + 1) = 3 would be eliminated from the sequence, so 4 will be his final score. On the other hand, if he chooses 3, his score would be 3 1 = 3, and yet again, (3 - 1) = 2 will be eliminated from the array. Clearly, the maximal score is 4.
As it is clearly visible, it is the frequency of the number appearing, as well as the number itself that is significant. Hence, we initiate an array, count[i], storing the number of times the element i appears in the sequence.
Hence for the above sequence 2 2 3, count[2] = 2, and count[3] = 3.
Finally, to find the optimal score, it would be easy to first break the problem down into smaller problem. In this case, we break the sequence into smaller sequence and find optimal solution for it.
For the sequence of number containing only 0, the answer would be 0. Similarly, if a sequence contains only the number 0 and 1, then the solution would be count[1]*1.
Now we are ready to build the recursive solution for this problem. For a sequence of number containing only the numbers, 0 to n, we can either pick the nth element, or not. Hence,
dp[i] =
max(
dp[i - 1], // Not picking the ith number, then the number before it can be considered
dp[i - 2] + i*a[i] // Picking the ith number, then the number before it is eliminated, hence the number before that is considered
);
hope its cleared if yes dont forget to hit like and mark resolved 
is it arr[i]-1,arr[i]+1 or adjacent elements in the question?
@neha_153 arr[i]-1 and +1 it is.
like if a[i]=2 then delete 1 and 3 if exists
so iterate from i=1 to max element
ok got it …i was thinking i need to delete adjacent elements