# PROBLEM LINK:

**Author:** Nishchith Shetty

**Tester:** Neel Shah

**Editorialist:** Rushang Gajjal

# DIFFICULTY:

EASY

# PREREQUISITES:

implemenatation, observation

# PROBLEM:

Given a number **X** , take each digit in turn from left to right . For each digit follow that many black arrows in a row and then follow one white arrow , initial position being point **A** . We have to find the node on which BOND will land after performing the given operations on **X**.

# EXPLANATION:

We can Solve this problem in two ways : 1. straight off Implementation and 2. Observation.

### Method 1: Implementation

Each element is the list [**A,B,C,D,E,F,G**] is denoted by its index **i** ranging from 0 to 6. So **A** -> 0, **B** -> 1 and so on.

For each node let’s store the next node it will reach if we follow the black arrow or the white arrow respectively.

l = [[1, 0], [3, 2], [4, 3], [2, 5], [6, 6], [0, 4], [5, 1]]

i.e. **A** -> l[0] = [1,0] -> [ **B** , **A** ]. From **A** if we move along black arrow we reach **B** or if we move along white arrow we reach **A**.

Starting from **A** i.e 0 (start = 0), Now for each digit **n** in the integer **X** we perform **n** moves along black arrow and then move along a white arrow. We can do this by

```
for i in range(n):
start = l[start][0] #move along black arrow
start = l[start][1] #move along white arrow
```

Value of start is initially 0.

After the above operations we just print the element denoted by the start variable.

- Time Complexity:
**O(|X|)**

### Method 2: Observation

Similarly if we perform **X** moves along black arrows and then a move along white arrow we can notice that each node can be represented by the value obtained by **X** mod 7.

**A** -> (**X** mod 7 = 0)

**C** -> (**X** mod 7 = 1)

**F** -> (**X** mod 7 = 2)

**D** -> (**X** mod 7 = 3)

**G** -> (**X** mod 7 = 4)

**B** -> (**X** mod 7 = 5)

**E** -> (**X** mod 7 = 6)

- Time Complexity:
**O(1)**

# AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here

Tester’s solution can be found here