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.

Print the minimum possible penalty after performing the operations.

### Input:

The first line of the input contains the number 𝑛(length of array) and m(maximum number of operations) The next n integers denotes the elements of the array.

### Output:

print the index of the element if it satisfies the condition else print -1

### Constraints:

1 <= nums.length <= 105 1 <= maxOperations, nums[i] <= 109

### Sample Input 1

1 2 9

### Sample Output 1

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.

### Sample Input 2

4 4 2 4 8 2

### Sample Output 2

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 an you should return 2.

## Solution of Minimum Limit of Balls in a Bag in java:–

import java.util.*; import java.io.*; public class Main { public static int findOperations(int [] a, int mid){ int n = a.length; int cnt = 0; for(int i = 0; i<n; i++) { cnt += (a[i] - 1) / mid; } return cnt; } public static void main(String args[]) { Scanner input = new Scanner(System.in); int n = input.nextInt(), k = input.nextInt(); int []a = new int[n]; int maxi = 0; for(int i = 0; i < n; i++){ a[i] = input.nextInt(); maxi = Math.max(a[i], maxi); } int l = 1, r = maxi, ans = 0; while(l<=r) { int mid = (l+r)/2; // current penality if(findOperations(a, mid) <= k) { ans = mid; r = mid -1; } else{ l = mid + 1; } } System.out.println(ans); } }

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