transmitters at positions 3, 7, and 11. - Base at position 3 with a cost of 1 covers house 1 and house 5. - Base at position 11 with a cost of 3 covers house 10. - Total cost = 1 + 3 = **3**.

Solution

Method 1 - Using Dynamic Programming

The optimal solution can be derived recursively by addressing one house at a time. A prerequisite for the algorithm is to ensure that the houses, bases, and their corresponding costs are sorted.

For the first house, the minimum cost is straightforward: since no previous solution exists with zero houses, we simply select t