Let's sort the jelly flavors such that their $$$a$$$ values are non-decreasing. We will call a particular way of buying the flavors at stores A and B a solution (each flavor is either bought at store A, store B, or not bought at all). A maximal solution is a solution for which the number of bought flavors is maximal. For a particular solution, let $$$j$$$ be the largest index of all flavors that were bought at store A.

You can now prove the following statement: There always exists a maximal solution such that all flavors with an index $$$ \lt j$$$ are bought at one of the two stores.

Proof: Consider any maximal solution $$$S$$$. If it satisfies the said property, we are done. If it doesn't, we will construct a new maximal solution $$$T$$$ that satisfies the property:
- We buy all flavors that are bought at store B in solution $$$S$$$ also at store B.
- We buy the first $$$u$$$ flavors with the smallest indices (that are not already bought at store B) at store A, where $$$u$$$ is the amount of flavor bought at store $$$A$$$ in solution $$$S$$$. Because all flavors bought at store A have the smallest indices available, there will be no flavor with index $$$ \lt j$$$ that is not bought. Since the flavors are sorted such that their $$$a$$$ values are non-decreasing, solution $$$T$$$ will spend not more money at store A than solution $$$S$$$ does, and therefore $$$T$$$ is a valid solution. The amount of flavors bought is the same for $$$S$$$ and $$$T$$$ and thus $$$T$$$ is also a maximal solution.
$$$T$$$ therefore satisfies the desired property.

Knowing that a maximal solution with this property always exists, we can design a DP-algorithm for solving the problem. Using DP, we can calculate two arrays of values:
- $$$DP_1[i][a]$$$: The minimal amount of dollars $$$b$$$ for which you can buy all flavors with an index $$$ \lt i$$$ using at most $$$a$$$ dollars at store A and at most $$$b$$$ dollars at store B
- $$$DP_2[i][b]$$$: The maximal amount of distinct flavors with an index $$$\geq i$$$ that you can buy using at most $$$b$$$ dollars at store B The recursive formula for these values should be easy to derive. Finally, the answer will be $$$\max\limits_{0\leq i\leq n}\max\limits_{0\leq a\leq x}(i + DP_2[i][y - DP_1[i][a]])$$$.

#include <bits/stdc++.h>
#include "jelly.h"
using namespace std;
#define ve vector
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1e9 + 10;

int find_maximum_unique(int x, int y, ve<int> A, ve<int> B) {

    int n = (int)A.size();
    ve<ve<int>> DPA(n + 1, ve<int>(x + 1, INF));
    ve<ve<int>> DPB(n + 1, ve<int>(y + 1, INF));
    ve<pii> F(n);

    for (int i = 0; i < n; i++) {
        F[i] = {A[i], B[i]};
    }
    sort(F.begin(), F.end());

    DPA[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= x; j++) {

            int a = j - F[i - 1].first;
            if (a >= 0) {
                DPA[i][j] = min(DPA[i][j], DPA[i - 1][a]);
            }

            if (DPA[i - 1][j] != INF) {
                int b = DPA[i - 1][j] + F[i - 1].second;
                if (b <= y) {
                    DPA[i][j] = min(DPA[i][j], b);
                }
            }

        }
    }

    DPB[n] = ve<int>(y + 1, 0);
    for (int i = n - 1; i >= 0; i--) {
        for (int j = 0; j <= y; j++) {
            DPB[i][j] = DPB[i + 1][j];
            int b = j - F[i].second;
            if (b >= 0) {
                DPB[i][j] = max(DPB[i][j], DPB[i + 1][b] + 1);
            }
        }
    }

    int ans = 0;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= x; j++) {
            int b = DPA[i][j];
            if (b != INF) {
                ans = max(ans, i + DPB[i][y - b]);
            }
        }
    }

    return ans;
}