There are two cases in this problem:

(Sort the input in increasing order)

(Considering 1-Based indexing)

Case 1:

If Sum of elements till [n-1]th element is less than the nth element then the most optimal answer=2*(nth element).

Why?

Consider the array 1 2 3 4 11

**Not Optimal Approach:**

Suppose Two Guys are A and B

1.A will read 1 and at the same time B will read 2 (B will not waste time waiting for A to Finish 1 so he is reading 2 to save time).

```
A 1
B 2
```

2.As A will finish 1 then he will not wait for 2 because B is still reading it.So A will read 3.

```
A 1 3
B 2
```

3.When B finishes 2 then he will not wait for 3 because A is reading it so B will read 4.

```
A 1 3
B 2 4
```

4.As A Finishes 3, he will not wait for 4 as B is reading it, So A will Read 11.

```
A 1 3 11
B 2 4
```

5.Meanwhile,B Finishes 4 , Now instead of waiting for 11 he will read his pending books (1 and 3) .

```
A 1 3 11
B 2 4 1 3
```

6.As B finishes reading 1 and 3 then he is left with only one book i.e 11 which A is reading,So he **Waits** for A to finish 11 .

```
A 1 3 11 [A will finish reading 11 at 15th time]
B 2 4 1 3 [B will wait for 11 for 5 minutes because he completed remaining
books at 10th time ]
```

7.When A finishes 11 , B will read 11 with a delay of 5 minutes.Meanwhile A will finish his Pending Books (2 and 4 ) as B is not reading these books right now.

```
A 1 3 11 2 4
B 2 4 1 3 11 [So B's time of reading all books is 2+4+1+3+11+delay=26 (delay-
5 Minutes) ]
```

**So Total Time taken to read books by both A and B is 26 Minutes.**

**Optimal Approach:**

1.A will read 11,At the same time B will read 1 2 3 4.

Because 1+2+3+4=10 and 10<11

```
A 11
B 1 2 3 4
```

2.Now B will read 11 and A will read 1 2 3 4

```
A 11 1 2 3 4
B 1 2 3 4 11
```

**No delay -** **So total Time taken is 22** **=> (11*2)** **=> (maximum time** ***2)**

**Case 2:**

If Sum of elements till [n-1]th element is greater than the nth element then the most optimal answer=Sum of all elements.

Consider Case - 1 2 3 5 10

**Non Optimal Approach is same as Case 1**

**Optimal Approach:**

1.A will read 10, at the same time B will read 1 2 3.

```
A 10 [10 Minutes]
B 1 2 3 [6 Minutes]
```

2.When B finishes 1 2 3 he reads 5 and A is still reading 10.

```
A 10 [10 minutes]
B 1 2 3 5 [11 Minutes]
```

3.When A finishes 10 He reads 1 2 3 as B as already read these books and B is not reading these books right now so A can read 1 2 3.

```
A 10 1 2 3 [16 Minutes]
B 1 2 3 5 [11 Minutes]
```

4.When B finish 5 He will read his last pending book 10 , He can read it now because A is not reading 10 right now and A is busy reading 1 2 3.

```
A 10 1 2 3 [16 Minutes]
B 1 2 3 5 10 [21 Minutes]
```

5.When A finish reading 1 2 3 He will read his last pending book 5 , He can read it now because B is not reading 5 right now and B is busy reading 10.

```
A 10 1 2 3 5 Both 21 Minutes
B 1 2 3 5 10
```

**No delay- So total Time taken is 21 => (Sum of all times).**

General Strategy for Case 2:

```
A should read like this:
n 1 2 3 - - - n-1
B should read like this:
1 2 3- - - - - n
```

I Hope You got the difference between Optimal approach and Non Optimal Approach.

My AC Code - Click Here