Binary search on the answer (Binary Search Sub-pattern)

// Smallest speed k so that all piles are eaten within h hours (Koko).
// Only the range [lo, hi] and feasible() change between problems.
bool feasible(vector<int>& piles, int h, int k) {
    long long hours = 0;
    for (int p : piles) hours += (p + k - 1) / k;   // ceil(p / k)
    return hours <= h;
}

int smallestWorking(vector<int>& piles, int h) {
    int lo = 1;                                      // smallest answer that could ever work
    int hi = *max_element(piles.begin(), piles.end());  // this one surely works
    while (lo < hi) {
        int mid = lo + (hi - lo) / 2;
        if (feasible(piles, h, mid)) hi = mid;       // works: maybe something smaller works too
        else lo = mid + 1;                           // fails: everything smaller fails too
    }
    return lo;
}