Minimum Limit of Balls in a Bag - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [9], maxOperations = 2

› Output: 3

💡 Note: Split bag of 9 balls: 9 → 6,3 → 3,3,3. Penalty is max(3,3,3) = 3.

Example 2 — Multiple Bags

$ Input: nums = [2,4,8,2], maxOperations = 4

› Output: 2

💡 Note: Split 8 into 4,4, then each 4 into 2,2. Result: [2,4,2,2,2,2,2]. Max penalty = 4, but we can do better by splitting the 4 too for penalty 2.

Example 3 — No Operations Needed

$ Input: nums = [1,2,3], maxOperations = 5

› Output: 3

💡 Note: Already optimal - max is 3 and we don't need any operations.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ maxOperations ≤ 10⁹
1 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

a Amazon 23 G Google 15 M Microsoft 12

The key insight is that this is a binary search on answer problem. If we can achieve penalty P with the given operations, we can also achieve any penalty > P. The optimal approach uses binary search to find the minimum achievable penalty. Time: O(n × log(max(nums))), Space: O(1)

Common Approaches

✓ Binary Search on Answer

⏱️ Time: O(n × log(max(nums))) Space: O(1)

Instead of checking penalties linearly, use binary search since if penalty P is achievable, then any penalty > P is also achievable. This monotonic property allows efficient searching.

Linear Search on Answer

⏱️ Time: O(max(nums) × n) Space: O(1)

For each possible penalty value, count operations needed to achieve it. The first value that requires ≤ maxOperations is our answer.

Binary Search on Answer — Algorithm Steps

Step 1: Set search range [1, max(nums)]
Step 2: Binary search on penalty values
Step 3: For each mid, check if achievable with helper function
Step 4: Narrow search based on feasibility

Visualization

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Step-by-Step Walkthrough

Initial range

Search between 1 and max(nums)

Check middle

Test if mid penalty is achievable

Narrow range

Continue until left==right

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int canAchievePenalty(int* nums, int numsSize, int penalty, int maxOperations) {
    long long operations = 0;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] > penalty) {
            operations += (nums[i] - 1) / penalty;
        }
    }
    return operations <= maxOperations;
}

int solution(int* nums, int numsSize, int maxOperations) {
    int left = 1;
    int right = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] > right) {
            right = nums[i];
        }
    }
    
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (canAchievePenalty(nums, numsSize, mid, maxOperations)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    
    return left;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        if (num != 0 || token[0] == '0') {
            nums[numsSize++] = num;
        }
        token = strtok(NULL, "[,]");
    }
    
    int maxOperations;
    scanf("%d", &maxOperations);
    
    printf("%d\n", solution(nums, numsSize, maxOperations));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × log(max(nums)))

Binary search log(max) iterations × checking all n bags

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Minimum Limit of Balls in a Bag - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Binary Search on Answer — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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