one simple solution is:

sort the array.

then,just add numbers from the end of array at indexes which are multiples of 3.

e.g.

1,2,3,4,5,6,7,8

Now,answer=6+3=9

as on 1st day,1,6,7,8 and on 2nd day 2,3,4,5.

Since you are required to maximize the second min element, first sort the array in descending and simply add the elements in the indices: 2,5,8,11 .

## Code for your reference

```
int findSum(int A[], int n){
int sum = 0;
sort(A,A+N, greater<int>());
for(int i = 0; i<N-N/4;i++){
sum = (sum+A[i])%mod;
}
return sum;
```

Can you please post the second question too ?

I would like to know the solution for 2nd one as I was not able to solve it during the contest

The second question was given a binary string : consisting of only zeroes and ones

At each step we can remove alternating subsequence from the string

Alternating subsequence is defined as “01010…” , “101…” , “01…” , “10…”

Note that “100” and “0111” are not alternating subsequences.

We have to minimize the number of steps to reduce the binray string to empty string.

Test Case : " 0100100111"

Output : 3