Use the binary search template on Koko's eating speed k. The key is to define a correct can_finish_eating(piles, h, k) function.

For one pile with p bananas, Koko needs ceil(p / k) hours, because she can only work on one pile per hour and a partial final hour still counts as a full hour. Plain division p / k is wrong here: for p = 7 and k = 4, 7 / 4 = 1.75, but the real time is 2 hours.

Sum this value across all piles to get hours_used. If hours_used <= h, speed k is feasible; otherwise it is not. This creates the monotonic behavior needed for binary search: once a speed works, every larger speed also works.

In example 1 (piles = [3,6,7,11], h = 8), this feasibility check behaves as follows.

Implementation

def can_finish_eating(piles, h, k):
    hours_used = 0
    for p in piles:
        hours_used += ceil(float(p) / k)
    return hours_used <= h

def minEatingSpeed(piles, h):
    left, right = 1, 1000000000  # 10^9 max pile size
    ans = -1
    while left <= right:
        mid = (left + right) // 2
        if can_finish_eating(piles, h, mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1
    return ans