Today I have done Capacity To Ship Packages Within D Days and leetcode's July LeetCoding Challenge with cpp.

Capacity To Ship Packages Within D Days

Description

A conveyor belt has packages that must be shipped from one port to another within days days.

The ith package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within days days.

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], days = 5
Output: 15
Explanation: A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10

Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.

Example 2:

Input: weights = [3,2,2,4,1,4], days = 3
Output: 6
Explanation: A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], days = 4
Output: 3
Explanation:
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Constraints:

• 1 <= days <= weights.length <= 5 * 104
• 1 <= weights[i] <= 500

Solution

class Solution {
  bool validate(vector<int> &weights, int days, int capacity) {
    int cur = 0;
    int ships = 1;
    for(auto w : weights) {
      if(cur + w > capacity) {
        cur = w;
        ships += 1;
      } else {
        cur += w;
      }
      if(ships > days) return false;
    }
    // cout << capacity << ' ' << days << endl;
    return true;
  }
public:
  int shipWithinDays(vector<int>& weights, int days) {
    int sum = accumulate(weights.begin(), weights.end(), 0);
    int maxW = *max_element(weights.begin(), weights.end());
    int len = weights.size();
    if(days >= len) return maxW;
    if(days == 1) return sum;
    int low = maxW;
    int high = sum;
    while(low < high) {
      int mid = (low + high) >> 1;
      if(validate(weights, days, mid)) {
        high = mid;
      } else {
        low = mid + 1;
      }
    }
    return low;
  }
};

July LeetCoding Challenge 2

Description

Find K Closest Elements

Given a sorted integer array arr, two integers k and x, return the k closest integers to x in the array. The result should also be sorted in ascending order.

An integer a is closer to x than an integer b if:

• |a - x| < |b - x|, or
• |a - x| == |b - x| and a < b

Example 1:

Input: arr = [1,2,3,4,5], k = 4, x = 3
Output: [1,2,3,4]

Example 2:

Input: arr = [1,2,3,4,5], k = 4, x = -1
Output: [1,2,3,4]

Constraints:

• 1 <= k <= arr.length
• 1 <= arr.length <= 10^4
• arr is sorted in ascending order.
• -10^4 <= arr[i], x <= 10^4

Solution

auto _ = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<int> findClosestElements(vector<int>& arr, int k, int x) {
    int len = arr.size();
    if(k == len) return arr;
    int pos = lower_bound(arr.begin(), arr.end(), x) - arr.begin();
    if(pos == 0) return vector<int>(arr.begin(), arr.begin() + k);
    if(pos == len) return vector<int>(arr.end() - k, arr.end());
    int start = pos;
    int sz = (arr[pos] == x);
    while(sz < k) {
      if(start == 0) break;
      if(start + sz == len) {
        start = len - k;
        break;
      }
      if(x - arr[start - 1] <= arr[start + sz] - x) {
        start -= 1;
      }
      sz += 1;
    }
    return vector<int>(arr.begin() + start, arr.begin() + start + k);
  }
};