Problem 219: Skew-cost Coding

Mathematical Development

Definition. A prefix-free code corresponds to a binary tree whose leaves are the codewords. The cost of a leaf is the weighted path length from the root, where moving left adds 1 and moving right adds 4.

Theorem 1 (Greedy optimality). An optimal code is obtained by repeatedly splitting a leaf of minimum current cost.

Proof. Suppose some optimal tree splits a leaf of cost $c_2$ while leaving another leaf of smaller cost $c_1 < c_2$ unsplit. Swapping those choices decreases the total cost by

contradicting optimality. Therefore the next split must always come from a minimum-cost leaf. $\square$

Lemma 1 (Cost accounting). Splitting a leaf of cost $c$ removes that leaf and creates two leaves of costs $c+1$ and $c+4$. The net increase in total cost is

The number of leaves increases by 1.

Theorem 2 (Bulk processing of equal-cost leaves). If the current minimum cost is $c$ and there are $k$ leaves of that cost, then processing them all together is equivalent to $k$ consecutive greedy splits. The histogram update is:

• remove $k$ leaves of cost $c$
• add $k$ leaves of cost $c+1$
• add $k$ leaves of cost $c+4$

Proof. All $k$ leaves are tied for minimum cost, so the greedy rule may split them in any order. Each split has the same local effect, and the $k$ effects add linearly. $\square$

Editorial

The tree interpretation makes the greedy step completely local. Splitting a leaf of cost $c$ replaces it by two children of costs $c+1$ and $c+4$, so the total cost rises by $c+5$. The exchange argument shows that delaying a cheaper leaf in favor of a more expensive one can only make the final total worse.

That means we never need the full tree structure. A histogram of how many leaves exist at each cost is enough. Repeatedly take the smallest cost bucket, split as many leaves from that bucket as are still needed, update the counts of the $c+1$ and $c+4$ buckets, and continue until the tree has exactly $10^9$ leaves.

Pseudocode

Create a sorted map frequency with one entry:
    cost 0 appears once.

leaf_count = 1
total_cost = 0

While leaf_count < N:
    Let c be the smallest cost present in the map.
    Let available be the number of leaves of cost c.

    need = N - leaf_count
    splits = min(available, need)

    total_cost += splits * (c + 5)
    leaf_count += splits

    Add splits leaves to cost c + 1.
    Add splits leaves to cost c + 4.

    If not all leaves of cost c were split:
        put the unused remainder back into cost c.

Return total_cost

Complexity Analysis

• Time: $O(M \log M)$ where $M$ is the number of distinct cost buckets that ever become active in the histogram.
• Space: $O(M)$ for the histogram.

Answer

#include <bits/stdc++.h>
using namespace std;

/*
 * Problem 219: Skew-cost Coding
 *
 * Build a prefix-free code for N = 10^9 symbols.
 * Bit '0' costs 1, bit '1' costs 4.
 * Minimize total cost of all codewords.
 *
 * Greedy: always split the cheapest codeword.
 * Splitting cost c -> produces c+1 and c+4.
 * Use histogram for bulk processing.
 *
 * Answer: 64564225042
 */

int main() {
    const long long N = 1000000000LL;

    // Histogram: cost -> count
    // Use a sorted map for efficient min access
    map<long long, long long> freq;
    freq[0] = 1;
    long long total_cost = 0;
    long long num_codes = 1;

    while (num_codes < N) {
        auto it = freq.begin();
        long long cost = it->first;
        long long count = it->second;
        freq.erase(it);

        // How many can we split? Each split adds 1 codeword.
        long long need = N - num_codes;
        long long splits = min(count, need);

        // Split 'splits' codewords of this cost
        total_cost += splits * (cost + 5);
        num_codes += splits;

        // Add children
        freq[cost + 1] += splits;
        freq[cost + 4] += splits;

        // If we didn't split all, put the rest back
        if (splits < count) {
            freq[cost] += (count - splits);
        }
    }

    cout << total_cost << endl;
    return 0;
}

"""
Problem 219: Skew-cost Coding

Build a prefix-free code for N = 10^9 symbols.
Bit '0' costs 1, bit '1' costs 4.
Minimize total cost of all codewords.

Greedy approach: always split the cheapest codeword.
Splitting a codeword of cost c removes it and creates c+1 and c+4.
Net cost increase per split of cost c: c + 5.

Use histogram (sorted dict) for bulk processing.

Answer: 64564225042
"""

def solve_simple():
    """Alternative using plain dict with manual min tracking."""
    N = 10**9

    freq = {0: 1}
    total_cost = 0
    num_codes = 1
    min_cost = 0

    while num_codes < N:
        # Find minimum cost
        while min_cost not in freq or freq[min_cost] == 0:
            if min_cost in freq and freq[min_cost] == 0:
                del freq[min_cost]
            min_cost += 1

        cost = min_cost
        count = freq[cost]
        del freq[cost]

        need = N - num_codes
        splits = min(count, need)

        total_cost += splits * (cost + 5)
        num_codes += splits

        freq[cost + 1] = freq.get(cost + 1, 0) + splits
        freq[cost + 4] = freq.get(cost + 4, 0) + splits

        if splits < count:
            freq[cost] = freq.get(cost, 0) + (count - splits)

    print(total_cost)

if __name__ == "__main__":
    solve_simple()