Cut Ribbons of Same Length - Google Top Interview Questions


Problem Statement :


You are given a list of positive integers ribbons and an integer k. 
Given that you can cut the ribbons as many times as you want, return the largest r such that you can have k ribbons of length r. 
If there is no solution, return -1.

Constraints

1 ≤ n ≤ 100,000 where n is the length of ribbons

1 ≤ k

Example 1

Input

ribbons = [1, 2, 3, 4, 9]

k = 5

Output

3

Explanation

We can cut the ribbon of size 9 into 3 pieces of size 3 each. Then cut the ribbon of size 4 into size 1 
and 3. Then we can achieve 5 ribbons of size 3.



Example 2

Input

ribbons = [8]

k = 4

Output


2

Explanation

We can cut the ribbon into 4 pieces of size 2.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& ribbons, int k) {
    vector<int> nums = ribbons;
    sort(nums.begin(), nums.end());
    int n = nums.size();
    int mv = nums[n - 1];
    int sm = 0;
    int l = 1;
    int r = mv;
    int ans = 0;
    while (l <= r) {
        int mid = floor((l + r) / 2);
        sm = 0;
        for (int j = 0; j < ribbons.size(); j++) {
            sm += floor((ribbons[j] * 1.0) / mid);
            // cout<<i << " " << sm<<endl;
        }
        if (sm >= k) {
            ans = mid;
            l = mid + 1;
        } else {
            r = mid - 1;
        }
        // sm = 0;
    }
    // print(ans)
    cout << ans << endl;
    if (ans > 0) return ans;
    return -1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] A, int k) {
        Arrays.sort(A);
        int l = 1;
        int r = A[A.length - 1];
        int res = -1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (check(A, mid, k)) {
                res = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return res;
    }

    public boolean check(int A[], int mid, int k) {
        int sum = 0;
        for (int i = 0; i < A.length; i++) {
            int cnt = A[i] / mid;
            sum += cnt;
        }
        return sum >= k;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, a, k):
        def is_possible(r):
            res = 0
            for x in a:
                ans = x // r
                if ans == 0:
                    break
                else:
                    res += ans
            return res >= k

        a.sort(reverse=True)
        lo, hi = 1, max(a)
        n = len(a)
        while lo <= hi:
            mid = (lo + hi) // 2
            if is_possible(mid):
                lo = mid + 1
            else:
                hi = mid - 1
        return hi if hi > 0 else -1


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