 Solution

Remove any plane types with zero count, retaining the increasing order of the remaining lengths; if none remain, return the empty allocation. For the retained types put $s_0=0$ and $N_i={\sum }_{\ell =i}^k{n}_{\ell }$. Then every plane of type $j$ has Shapley valueCompute suffix counts $N_i$ in a backward pass. In a forward pass, accumulate $A\leftarrow A+\lambda (s_i-s_{i-1})/N_i$ and output $A$ as the common payment for type $i$.

For correctness, layer $i$ costs $\lambda (s_i-s_{i-1})$ and is required by exactly $N_i$ planes. A particular requiring plane is earliest among these in a random ordering with probability $1/N_i$. Its expected marginal contribution to the layer is therefore that layer's cost divided by $N_i$. Summing the layers it requires proves the formula. This is the Shapley value of an airport game.

The two passes use $O(k)$ arithmetic operations. If every individual payment must be explicitly listed, add the unavoidable $O(|N|)$ output work. For counts given in binary, the compact per-type output is polynomial in the input size; expanding an exponentially large number of identical entries need not be. With integer lengths and rational $\lambda$, all fractions can be computed exactly with polynomial bit complexity: a common denominator divides a product of at most $k$ suffix counts, whose bit length is $O(k\log |N|)$, in addition to the input denominator of $\lambda$. For arbitrary real $\lambda$, the same rational factors give symbolic multiples of it.