Minimum Time to Finish the Race — C++

// Time:  O((n + l) * logc)
// Space: O(n + l + logc)

// greedy, dp
class Solution {
public:
    int minimumFinishTime(vector<vector<int>>& tires, int changeTime, int numLaps) {
        static const int INF = numeric_limits<int>::max();

        const auto& ceil_log2 = [](const auto& x) {
            return std::__lg(x - 1) + 1;
        };
        vector<int64_t> dp(ceil_log2(changeTime + 1), INF);  // dp[i]: min time to complete i+1 laps without changing a tire
        for (int i = 0; i < size(tires); ++i) {
            const int f = tires[i][0], r = tires[i][1];
            for (int64_t curr = f, total = f, cnt = 0;
                 curr < changeTime + f;
                 curr *= r, total += curr, ++cnt) {  // at worst (f, r) = (1, 2) => 2^(cnt-1) < changeTime+1 => cnt < ceil(log2(changeTime+1))
                dp[cnt] = min(dp[cnt], total);
            }
        }
        vector<int64_t> dp2(numLaps, INF);  // dp2[i]: min time to complete i+1 laps with changing zero or more tires
        for (int i = 0; i < numLaps; ++i) {
            for (int j = 0; j < min(i + 1, static_cast<int>(size(dp))); ++j) {
                dp2[i] = min(dp2[i], (i - j - 1 >= 0 ? dp2[i - j - 1] + changeTime : 0) + dp[j]);
            }
        }
        return dp2.back();
    }
};