# 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 :

```                        ```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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