LeetCode //C - 1283. Find the Smallest Divisor Given a Threshold

1283. Find the Smallest Divisor Given a Threshold

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor, divide all the array by it, and sum the division's result. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

The test cases are generated so that there will be an answer.

Example 1:

Input: nums = 1,2,5,9, threshold = 6

Output: 5

Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.

If the divisor is 4 we can get a sum of 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).ou can split the string into "ab" and "c", and change 1 character in "ab" to make it palindrome.

Example 2:

Input: s = "aabbc", k = 3

Output: 0

Explanation: You can split the string into "aa", "bb" and "c", all of them are palindrome.

Example 3:

Input: s = "leetcode", k = 8

Output: 0

Constraints:

• 1 < = n u m s . l e n g t h < = 5 ∗ 10 4 1 <= nums.length <= 5 * 10^4 1<=nums.length<=5∗104
• 1 < = n u m s i < = 10 6 1 <= numsi <= 10^6 1<=numsi<=106
• n u m s . l e n g t h < = t h r e s h o l d < = 10 6 nums.length <= threshold <= 10^6 nums.length<=threshold<=106

From: LeetCode

Link: 1283. Find the Smallest Divisor Given a Threshold

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Solution:

Ideas:

Use binary search on the divisor. For each divisor, calculate the rounded-up sum using (numsi + divisor - 1) / divisor.

Code:

int smallestDivisor(int* nums, int numsSize, int threshold) {
    int left = 1;
    int right = 0;

    for (int i = 0; i < numsSize; i++) {
        if (nums[i] > right) {
            right = nums[i];
        }
    }

    while (left < right) {
        int mid = left + (right - left) / 2;
        long long sum = 0;

        for (int i = 0; i < numsSize; i++) {
            sum += (nums[i] + mid - 1) / mid;

            if (sum > threshold) {
                break;
            }
        }

        if (sum <= threshold) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }

    return left;
}