Today I have done Preimage Size of Factorial Zeroes Function and leetcode's June LeetCoding Challenge with cpp.

Preimage Size of Factorial Zeroes Function

Description

Let f(x) be the number of zeroes at the end of x!. (Recall that x! = 1 * 2 * 3 * ... * x, and by convention, 0! = 1.)

For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end. Given k, find how many non-negative integers x have the property that f(x) = k.

Example 1:
Input: k = 0
Output: 5
Explanation: 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.

Example 2:
Input: k = 5
Output: 0
Explanation: There is no x such that x! ends in k = 5 zeroes.

Note:

• k will be an integer in the range [0, 10^9].

Solution

constexpr long long zeros(long long x) {
  long long result = 0;
  while(x) {
    result += x / 5;
    x /= 5;
  }
  return result;
}
class Solution {
public:
  int preimageSizeFZF(int k) {
    long long start = 0;
    long long end = LONG_LONG_MAX;
    while(start < end) {
      auto mid = (start + end) >> 1;
      if(zeros(mid) < k) {
        start = mid + 1;
      } else {
        end = mid;
      }
    }
    int begin = start;
    if(zeros(start) != k) return 0;
    start = 0;
    end = LONG_LONG_MAX - 1;
    while(start < end) {
      auto mid = (start + end + 1) >> 1;
      if(zeros(mid) > k) {
        end = mid - 1;
      } else {
        start = mid;
      }
    }
    return start - begin + 1;
  }
};

June LeetCoding Challenge 14

Description

Maximum Units on a Truck

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]:

• numberOfBoxesi is the number of boxes of type i.
• numberOfUnitsPerBoxi is the number of units in each box of the type i.

You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.

Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91

Constraints:

• 1 <= boxTypes.length <= 1000
• 1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
• 1 <= truckSize <= 10^6

Solution

class Solution {
public:
  int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
    sort(boxTypes.begin(), boxTypes.end(), [](const vector<int> &a, const vector<int> &b) {
      return a[1] > b[1];
    });
    int answer = 0;
    for(auto &boxType : boxTypes) {
      int n = min(truckSize, boxType[0]);
      answer += n * boxType[1];
      truckSize -= n;
      if(!truckSize) break;
    }
    return answer;
  }
};