Runtime 28 ms , Memory 59.62 MB

You are given an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

Take any bag of balls and divide it into two new bags with a positive number of balls.
For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.
Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

 

Example 1:

Input: nums = [9], maxOperations = 2
Output: 3
Explanation: 
- Divide the bag with 9 balls into two bags of sizes 6 and 3. [9] -> [6,3].
- Divide the bag with 6 balls into two bags of sizes 3 and 3. [6,3] -> [3,3,3].
The bag with the most number of balls has 3 balls, so your penalty is 3 and you should return 3.
Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2, and you should return 2.
 

Constraints:

1 <= nums.length <= 10^5
1 <= maxOperations, nums[i] <= 10^9

1760. Minimum Limit of Balls in a Bag

@@ -0,0 +1,36 @@
+class Solution {
+    public int minimumSize(int[] nums, int maxOperations) 
+    {
+
+        int l = 1;
+        int r = Arrays.stream(nums).max().getAsInt();
+
+        while (l < r) 
+        {
+            int m = (l + r) / 2;
+            if(numOperations(nums, m) <= maxOperations)
+            {
+                r = m;
+            }
+            else
+            {
+                l = m + 1;
+            }
+
+        }
+
+        return l;
+
+    }
+
+
+    public int numOperations(int[] nums, int m) 
+    {
+        int operations = 0;
+        for(int num : nums)
+            operations += (num - 1) / m;
+
+        return operations;
+    }
+
+}