Cutting Ribbons — Binary Search on Answer (Length)

# Python
def maxLength(ribbons, k):
    def feasible(length):
        return sum(r // length for r in ribbons) >= k

    lo, hi = 1, max(ribbons)
    while lo < hi:
        mid = lo + (hi-lo+1)//2  # upper mid (maximize)
        if feasible(mid): lo = mid
        else: hi = mid-1
    return lo if feasible(lo) else 0

// Java
public int maxLength(int[] ribbons, int k) {
    int lo=1, hi=0;
    for (int r:ribbons) hi=Math.max(hi,r);
    while (lo<hi) {
        int mid=lo+(hi-lo+1)/2, cnt=0;
        for (int r:ribbons) cnt+=r/mid;
        if (cnt>=k) lo=mid; else hi=mid-1;
    }
    long cnt=0; for (int r:ribbons) cnt+=r/lo;
    return cnt>=k?lo:0;
}