Implement Masked Cross Entropy Loss with Label Smoothing

Problem: Implement Masked Cross Entropy Loss with Label Smoothing Given a batch of classification model outputs logits, ground-truth labels labels, a sample mask mask, and a label smoothing parameter epsilon, implement the cross entropy loss using NumPy/Python. Definition logits is a 2D array of shape (N, C), where: N is the number of samples; C is the number of classes. labels[i] is the ground-truth class of sample i, satisfying 0 <= labels[i] < C. mask[i] indicates whether sample i participates in the loss computation: mask[i] = 1: include this sample; mask[i] = 0: ignore this sample. epsilon is the label smoothing coefficient, satisfying 0 <= epsilon < 1. For sample i, define the softmax probability as: p[i][j] = exp(logits[i][j]) / sum_k exp(logits[i][k]) Use the following label smoothing rule to construct the target distribution: target[i][j] = epsilon / C if j != labels[i] target[i][labels[i]] = 1 - epsilon + epsilon / C The per-sample loss is: loss[i] = - sum_j target[i][j] * log(p[i][j]) Return the masked mean loss: sum_i mask[i] * loss[i] / sum_i mask[i] If all mask[i] are 0, return 0.0. Your implementation should be numerically stable and should avoid directly applying exp to very large logits. Input Format N C logits[0][0] logits[0][1] ... logits[0][C-1] ... logits[N-1][0] ... logits[N-1][C-1] labels[0] labels[1] ... labels[N-1] mask[0] mask[1] ... mask[N-1] epsilon Output Format Print one floating-point number, the final loss, rounded to 6 decimal places. Constraints 1 <= N <= 100000 2 <= C <= 1000 N * C <= 200000 -1000 <= logits[i][j] <= 1000 0 <= labels[i] < C mask[i] is either 0 or 1 0 <= epsilon < 1 Example 1 Input: 2 3 1 2 3 1 1 1 2 0 1 1 0 Output: 0.753109 Example 2 Input: 3 2 0 0 2 0 0 2 0 0 1 1 0 1 0 Output: 0.410038 Example Input 2 3 1 2 3 1 1 1 2 0 1 1 0 Output 0.753109