Problem
A dense layer is a group of neurons that all receive the same inputs. Neuron j has its own weights weights[j] and bias biases[j]:
z_j = inputs · weights[j] + biases[j]
Then the layer's activation is applied to each z_j:
"linear": output z_j unchanged"relu": max(0, z_j)"sigmoid": 1 / (1 + e^(-z_j))
Implement dense_forward(inputs, weights, biases, activation).
Input
inputs: The n input values.
weights: One row per neuron; each row has n weights.
biases: One bias per neuron.
activation: "linear", "relu" or "sigmoid".
Output
Return the list of outputs, one per neuron, in order.
Constraints
1 ≤ n ≤ 100, 1 ≤ number of neurons ≤ 100
-10 ≤ each value ≤ 10
Examples
Input:
biases = [0.0, -1.0]
inputs = [1.0, 2.0]
weights = [
[0.5, -1.0],
[1.5, 0.5]
]
activation = "relu"
Expected output:
expected = [0.0, 1.5]Input:
std::vector<double> biases = {0.0, -1.0};
std::vector<double> inputs = {1.0, 2.0};
std::vector<std::vector<double>> weights = {
{0.5, -1.0},
{1.5, 0.5}
};
std::string activation = "relu";
Expected output:
std::vector<double> expected = {0.0, 1.5};Input:
double[] biases = new double[] {0.0, -1.0};
double[] inputs = new double[] {1.0, 2.0};
double[][] weights = new double[][] {
new double[] {0.5, -1.0},
new double[] {1.5, 0.5}
};
string activation = "relu";
Expected output:
double[] expected = new double[] {0.0, 1.5};Explanation
Neuron 1: 1×0.5 + 2×(-1) + 0 = -1.5, and ReLU gives 0. Neuron 2: 1×1.5 + 2×0.5 - 1 = 1.5, which ReLU keeps. Result: [0, 1.5].
Input:
biases = [1.0, 0.0, -1.5]
inputs = [3.0]
weights = [
[2.0],
[-1.0],
[0.5]
]
activation = "linear"
Expected output:
expected = [7.0, -3.0, 0.0]Input:
std::vector<double> biases = {1.0, 0.0, -1.5};
std::vector<double> inputs = {3.0};
std::vector<std::vector<double>> weights = {
{2.0},
{-1.0},
{0.5}
};
std::string activation = "linear";
Expected output:
std::vector<double> expected = {7.0, -3.0, 0.0};Input:
double[] biases = new double[] {1.0, 0.0, -1.5};
double[] inputs = new double[] {3.0};
double[][] weights = new double[][] {
new double[] {2.0},
new double[] {-1.0},
new double[] {0.5}
};
string activation = "linear";
Expected output:
double[] expected = new double[] {7.0, -3.0, 0.0};Explanation
With a linear activation the outputs are just 3×2 + 1 = 7, 3×(-1) + 0 = -3 and 3×0.5 - 1.5 = 0.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Treat each neuron separately: each one has its own weight row and bias.
Hint 2
Compute every neuron's dot product with the input vector, add its bias, then apply the chosen activation.
Hint 3
Make sure the activation choice is applied after the linear step, and preserve neuron order in the output.
Expected complexity
O(mn) time, O(m) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.