Algorithm
Problem Name: 1011. Capacity To Ship Packages Within D Days
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 * 1041 <= weights[i] <= 500
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int count_days(int *weight, int sz, int k) {
int i, w = 0, d = 1;
for (i = 0; i < sz; i ++) {
if (w + weight[i] > k) {
d ++;
w = 0;
}
w += weight[i];
}
return d;
}
int shipWithinDays(int* weights, int weightsSize, int D){
int left, right, mid;
int i, w, d;
left = right = 0;
for (i = 0; i < weightsSize; i ++) {
w = weights[i];
if (left < w) left = w;
right += w;
}
while (left < right) {
mid = left + (right - left) / 2;
d = count_days(weights, weightsSize, mid);
if (d < = D) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
class Solution {
public int shipWithinDays(int[] weights, int days) {
int maximumWeight = 0;
int minimumWeight = weights[0];
for (int weight : weights) {
maximumWeight += weight;
minimumWeight = Math.max(minimumWeight, weight);
}
while (minimumWeight < = maximumWeight) {
int currCapacity = (maximumWeight + minimumWeight) / 2;
if (isShipmentPossible(weights, days, currCapacity)) {
maximumWeight = currCapacity - 1;
} else {
minimumWeight = currCapacity + 1;
}
}
return minimumWeight;
}
private boolean isShipmentPossible(int[] weights, int days, int capacity) {
int idx = 0;
while (idx < weights.length) {
int currWeight = 0;
while (idx < weights.length && currWeight + weights[idx] <= capacity) {
currWeight += weights[idx++];
}
days--;
}
return days >= 0;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const shipWithinDays = function(weights, days) {
let l = Math.max(...weights)
let r = weights.reduce((ac, e) => ac + e, 0)
while(l < r) {
const mid = Math.floor((l + r) / 2)
if(valid(mid)) {
r = mid
} else l = mid + 1
}
return l
function valid(mid) {
let res = 1, cur = 0
for(let w of weights) {
if(cur + w > mid) {
cur = 0
res++
}
cur += w
}
return res <= days
}
};
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def shipWithinDays(self, weights: List[int], D: int) -> int:
def check(mx):
days, cur = 1, 0
for w in weights:
if cur + w <= mx:
cur += w
else:
days += 1
cur = w
return days
l, r = max(weights), sum(weights)
while l < r:
mid = (l + r) // 2
days = check(mid)
if days <= D:
r = mid
else:
l = mid + 1
return r
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _1011_CapacityToShipPackagesWithinDDays
{
public int ShipWithinDays(int[] weights, int D)
{
var totalSum = 0;
var largestNumber = -1;
for (int i = 0; i < weights.Length; i++)
{
largestNumber = Math.Max(largestNumber, weights[i]);
totalSum += weights[i];
}
var l = largestNumber;
var h = totalSum;
while (l < = h)
{
var mid = l + (h - l) / 2;
var count = 1;
var currentSum = 0;
for (int i = 0; i < weights.Length; i++)
{
if (currentSum + weights[i] > mid)
{
count++;
if (count > D) break;
currentSum = weights[i];
}
else currentSum += weights[i];
}
if (count > D) l = mid + 1;
else h = mid - 1;
}
return l;
}
}
}
Input
Output