Problem Brief
Coloring Houses
INTERNOA
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The city of Hackerland can be represented with an even number n houses arranged in a row. A painter must paint the houses using at most three colors. The following conditions must hold true:
n=6 then houses will be [1,2,3,4,5,6], so the houses at the same distance from both the ends will be [1,6], [2,5], [3,4].
The task is to find the number of ways to paint the houses using at most three colors such that both the above conditions hold true. Since the answer can be large, report it modulo 10^9 + 7. Two ways are considered different if at least one house is colored differently.
Function Description
Complete the countWaysToColorHouses function in the editor.
countWaysToColorHouses takes in a single parameter:
int n, the number of houses
1Example 1
Input
n = 4
Output
18
Explanation
For (color1, color2, color3, color2)
(color1, color3, color1, color3)
The number of ways to paint 4 houses using three colors is 18. Return 18 modulo
n = 4, some of the possible valid arrangements are:
(10^9 + 7) which is 18.2Example 2
Input
n = 2
Output
6
Explanation
The valid arrangements for 2 houses are:
(color1, color2)
(color1, color3)
(color2, color1)
(color3, color1)
(color2, color3)
(color3, color2)
3Example 3
Input
n = 4
Output
18
Explanation
Total valid arrangements for 4 hourses are 18.
Some of the valid arrangements are:
{color1, color2, color1, color2}
{color1, color3, color1, color3}
{color2, color1, color2, color1}
{color3, color1, color3, color1}
{color2, color3, color2, color3}
{color3, color2, color3, color2}
Constraints
Limits and guarantees your solution can rely on.