Resource assignment (Greedy Sub-pattern)

// Each need gets the smallest resource that satisfies it; count the matches.
int matchSmallestFit(vector<int> need, vector<int> res) {
    sort(need.begin(), need.end());
    sort(res.begin(), res.end());
    int i = 0, j = 0;
    while (i < (int)need.size() && j < (int)res.size()) {
        if (res[j] >= need[i]) i++;                          // match: this need is served
        j++;                                                 // too small now = too small for all later needs
    }
    return i;
}

// Beat b[] in as many positions as possible by permuting a[] (Tian Ji):
// strongest opponent first; win with our strongest, or sacrifice our weakest.
vector<int> advantage(vector<int> a, const vector<int>& b) {
    int n = a.size();
    sort(a.begin(), a.end());
    vector<int> order(n);
    iota(order.begin(), order.end(), 0);
    sort(order.begin(), order.end(), [&](int x, int y) { return b[x] > b[y]; });
    vector<int> res(n);
    int lo = 0, hi = n - 1;
    for (int k : order) res[k] = a[hi] > b[k] ? a[hi--] : a[lo++];
    return res;
}