1760. Minimum Limit of Balls in a Bag

1760. Minimum Limit of Balls in a Bag

Difficulty: Medium

Topics: Array, Binary Search

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 <= 105
• 1 <= maxOperations, nums[i] <= 109

Hint:

1. Let's change the question if we know the maximum size of a bag what is the minimum number of bags you can make
2. note that as the maximum size increases the minimum number of bags decreases so we can binary search the maximum size

Solution:

We can use binary search to find the minimum possible penalty. The key insight is that if we can determine whether a given penalty is achievable, we can narrow down the search range using binary search.

Steps to Solve:

1. Binary Search Setup:

   • The minimum penalty is 1 (all balls are split into single-ball bags).
   • The maximum penalty is the largest number in the nums array.

2. Feasibility Check:

   • For a given penalty mid, check if it's possible to achieve it with at most maxOperations splits.
   • To do this, for each bag size in nums, calculate the number of splits required to make all bags have mid balls or fewer. If the total splits exceed maxOperations, the penalty mid is not feasible.

3. Iterate:

   • Use binary search to adjust the range [low, high] based on whether a penalty mid is feasible.

Let's implement this solution in PHP: 1760. Minimum Limit of Balls in a Bag

<?php
/**
 * @param Integer[] $nums
 * @param Integer $maxOperations
 * @return Integer
 */
function minimumSize($nums, $maxOperations) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

/**
 * Helper function to check if a penalty is feasible
 *
 * @param $nums
 * @param $maxOperations
 * @param $penalty
 * @return bool
 */
function canAchievePenalty($nums, $maxOperations, $penalty) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example 1
$nums = [9];
$maxOperations = 2;
echo minimumSize($nums, $maxOperations); // Output: 3

// Example 2
$nums = [2, 4, 8, 2];
$maxOperations = 4;
echo minimumSize($nums, $maxOperations); // Output: 2
?>

Explanation:

1. Binary Search:

   • The search space is between 1 and the maximum number in the nums array.
   • The midpoint mid represents the current penalty we are testing.

2. Feasibility Check (canAchievePenalty):

   • For each bag, calculate the required splits to ensure all bags have mid balls or fewer:
     • ceil(balls / mid) - 1 gives the number of splits needed.
   • If the total splits exceed maxOperations, the penalty is not feasible.

3. Adjust Search Space:

   • If the penalty is feasible, reduce the upper bound (high = mid).
   • If not, increase the lower bound (low = mid + 1).

4. Result:

   • When the loop exits, low contains the smallest feasible penalty.

Complexity:

• Time Complexity: O(n . log(max(nums)))
  • Binary search runs in O(log(max(nums))), and the feasibility check for each midpoint takes O(n).
• Space Complexity: O(1), as we only use constant extra space.

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