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Minimum time required to transport all the boxes from source to the destination under the given constraints

  • Difficulty Level : Hard
  • Last Updated : 05 Aug, 2021

Given two arrays, box[] and truck[], where box[i] represents the weight of the ith box and truck[i] represents the maximum load that the ith truck can carry. Now each truck takes 1 hour to transport a box from source to destination and another one hour to come back. Now, given that all the boxes are kept at the source, the task is to find the minimum time required to transport all the boxes from the source to the destination. 

Note that there will always be some time in which the boxes can be transported and only a single box can be carried by truck at any instance of time.

Examples:  

Input: box[] = {7, 6, 5, 4, 3}, truck[] = {10, 3} 
Output:
1st hour: truck[0] carries box[0] and truck[1] carries box[4] 
2nd hour: Both trucks are back at the source location. 
Now, truck[1] cannot carry anymore boxes as all the remaining boxes 
have weights more than the capacity of a truck[1]. 
So, truck[0] will carry box[1] and box[2] 
in a total of four hours. (source-destination and then destination-source) 
And finally, box[3] will take another hour to reach the destination. 
So, total time taken = 2 + 4 + 1 = 7

Input: box[] = {10, 2, 16, 19}, truck[] = {29, 25} 
Output: 3  

Approach: The idea is to use binary search and sort the two arrays. Here the lower bound will be 0 and the upper bound will be 2 * size of box[] because in the worst case, the amount of time required to transport all the boxes will be 2 * size of box array. Now compute the mid-value, and for each mid-value check if all the boxes can be transported by the loaders in time = mid. If yes, then update the upper bound as mid – 1 and if not, then update the lower bound as mid + 1.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if it is
// possible to transport all the boxes
// in the given amount of time
bool isPossible(int box[], int truck[],
                int n, int m, int min_time)
{
    int temp = 0;
    int count = 0;
 
    while (count < m) {
        for (int j = 0; j < min_time
                        && temp < n
                        && truck[count] >= box[temp];
             j += 2)
            temp++;
 
        count++;
    }
 
    // If all the boxes can be
    // transported in the given time
    if (temp == n)
        return true;
 
    // If all the boxes can't be
    // transported in the given time
    return false;
}
 
// Function to return the minimum time required
int minTime(int box[], int truck[], int n, int m)
{
 
    // Sort the two arrays
    sort(box, box + n);
    sort(truck, truck + m);
 
    int l = 0;
    int h = 2 * n;
 
    // Stores minimum time in which
    // all the boxes can be transported
    int min_time = 0;
 
    // Check for the minimum time in which
    // all the boxes can be transported
    while (l <= h) {
        int mid = (l + h) / 2;
 
        // If it is possible to transport all
        // the boxes in mid amount of time
        if (isPossible(box, truck, n, m, mid)) {
            min_time = mid;
            h = mid - 1;
        }
        else
            l = mid + 1;
    }
 
    return min_time;
}
 
// Driver code
int main()
{
    int box[] = { 10, 2, 16, 19 };
    int truck[] = { 29, 25 };
 
    int n = sizeof(box) / sizeof(int);
    int m = sizeof(truck) / sizeof(int);
 
    printf("%d", minTime(box, truck, n, m));
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.Arrays;
 
class GFG
{
 
// Function that returns true if it is
// possible to transport all the boxes
// in the given amount of time
static boolean isPossible(int box[], int truck[],
                int n, int m, int min_time)
{
    int temp = 0;
    int count = 0;
 
    while (count < m)
    {
        for (int j = 0; j < min_time
                        && temp < n
                        && truck[count] >= box[temp];
            j += 2)
            temp++;
 
        count++;
    }
 
    // If all the boxes can be
    // transported in the given time
    if (temp == n)
        return true;
 
    // If all the boxes can't be
    // transported in the given time
    return false;
}
 
// Function to return the minimum time required
static int minTime(int box[], int truck[], int n, int m)
{
 
    // Sort the two arrays
    Arrays.sort(box);
    Arrays.sort(truck);
 
    int l = 0;
    int h = 2 * n;
 
    // Stores minimum time in which
    // all the boxes can be transported
    int min_time = 0;
 
    // Check for the minimum time in which
    // all the boxes can be transported
    while (l <= h) {
        int mid = (l + h) / 2;
 
        // If it is possible to transport all
        // the boxes in mid amount of time
        if (isPossible(box, truck, n, m, mid))
        {
            min_time = mid;
            h = mid - 1;
        }
        else
            l = mid + 1;
    }
 
    return min_time;
}
 
// Driver code
public static void main(String[] args)
{
    int box[] = { 10, 2, 16, 19 };
    int truck[] = { 29, 25 };
 
    int n = box.length;
    int m = truck.length;
 
    System.out.printf("%d", minTime(box, truck, n, m));
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 implementation of the approach
 
# Function that returns true if it is
# possible to transport all the boxes
# in the given amount of time
def isPossible(box, truck, n, m, min_time) :
     
    temp = 0
    count = 0
 
    while (count < m) :
        j = 0
        while (j < min_time and temp < n and
                    truck[count] >= box[temp] ):
            temp +=1
            j += 2
 
        count += 1
 
    # If all the boxes can be
    # transported in the given time
    if (temp == n) :
        return True
 
    # If all the boxes can't be
    # transported in the given time
    return False
 
# Function to return the minimum time required
def minTime(box, truck, n, m) :
 
    # Sort the two arrays
    box.sort();
    truck.sort();
 
    l = 0
    h = 2 * n
 
    # Stores minimum time in which
    # all the boxes can be transported
    min_time = 0
 
    # Check for the minimum time in which
    # all the boxes can be transported
    while (l <= h) :
        mid = (l + h) // 2
 
        # If it is possible to transport all
        # the boxes in mid amount of time
        if (isPossible(box, truck, n, m, mid)) :
            min_time = mid
            h = mid - 1
     
        else :
             
            l = mid + 1
 
    return min_time
 
# Driver code
if __name__ == "__main__" :
 
    box = [ 10, 2, 16, 19 ]
    truck = [ 29, 25 ]
 
    n = len(box)
    m = len(truck)
 
    print(minTime(box, truck, n, m))
     
# This code is contributed by Ryuga

C#




// C# implementation of the approach
using System;
     
class GFG
{
 
// Function that returns true if it is
// possible to transport all the boxes
// in the given amount of time
static bool isPossible(int []box, int []truck,
                int n, int m, int min_time)
{
    int temp = 0;
    int count = 0;
 
    while (count < m)
    {
        for (int j = 0; j < min_time
                        && temp < n
                        && truck[count] >= box[temp];
            j += 2)
            temp++;
 
        count++;
    }
 
    // If all the boxes can be
    // transported in the given time
    if (temp == n)
        return true;
 
    // If all the boxes can't be
    // transported in the given time
    return false;
}
 
// Function to return the minimum time required
static int minTime(int []box, int []truck, int n, int m)
{
 
    // Sort the two arrays
    Array.Sort(box);
    Array.Sort(truck);
 
    int l = 0;
    int h = 2 * n;
 
    // Stores minimum time in which
    // all the boxes can be transported
    int min_time = 0;
 
    // Check for the minimum time in which
    // all the boxes can be transported
    while (l <= h)
    {
        int mid = (l + h) / 2;
 
        // If it is possible to transport all
        // the boxes in mid amount of time
        if (isPossible(box, truck, n, m, mid))
        {
            min_time = mid;
            h = mid - 1;
        }
        else
            l = mid + 1;
    }
 
    return min_time;
}
 
// Driver code
public static void Main(String[] args)
{
    int[] box = { 10, 2, 16, 19 };
    int []truck = { 29, 25 };
 
    int n = box.Length;
    int m = truck.Length;
 
    Console.WriteLine("{0}", minTime(box, truck, n, m));
}
}
 
/* This code contributed by PrinciRaj1992 */

PHP




<?php
 
// PHP implementation of the approach
 
// Function that returns true if it is
// possible to transport all the boxes
// in the given amount of time
function isPossible($box, $truck,
                $n, $m, $min_time)
{
    $temp = 0;
    $count = 0;
 
    while ($count < $m)
    {
        for ( $j = 0; $j < $min_time
                        && $temp < $n
                        && $truck[$count] >= $box[$temp];
            $j += 2)
            $temp++;
 
        $count++;
    }
 
    // If all the boxes can be
    // transported in the given time
    if ($temp == $n)
        return true;
 
    // If all the boxes can't be
    // transported in the given time
    return false;
}
 
// Function to return the minimum time required
function minTime( $box, $truck, $n, $m)
{
 
    // Sort the two arrays
    sort($box);
    sort($truck);
 
    $l = 0;
    $h = 2 * $n;
 
    // Stores minimum time in which
    // all the boxes can be transported
    $min_time = 0;
 
    // Check for the minimum time in which
    // all the boxes can be transported
    while ($l <= $h) {
        $mid = intdiv(($l + $h) , 2);
 
        // If it is possible to transport all
        // the boxes in mid amount of time
        if (isPossible($box, $truck, $n, $m, $mid))
        {
            $min_time = $mid;
            $h = $mid - 1;
        }
        else
            $l = $mid + 1;
    }
 
    return $min_time;
}
 
// Driver code
$box = array( 10, 2, 16, 19 );
$truck = array( 29, 25 );
 
$n = sizeof($box);
$m = sizeof($truck);
 
echo minTime($box, $truck, $n, $m);
 
 
// This code is contributed by ihritik
 
?>

Javascript




<script>
 
// Js implementation of the approach
 
// Function that returns true if it is
// possible to transport all the boxes
// in the given amount of time
function isPossible( box, truck,
                 n, m, min_time)
{
    let temp = 0;
    let count = 0;
 
    while (count < m) {
        for (let j = 0; j < min_time
                        && temp < n
                        && truck[count] >= box[temp];
             j += 2)
            temp++;
 
        count++;
    }
 
    // If all the boxes can be
    // transported in the given time
    if (temp == n)
        return true;
 
    // If all the boxes can't be
    // transported in the given time
    return false;
}
 
// Function to return the minimum time required
function minTime(box, truck, n, m)
{
 
    // Sort the two arrays
    box.sort(function(a,b){return a-b });
    truck.sort(function(a,b){return a-b });
 
    let l = 0;
    let h = 2 * n;
 
    // Stores minimum time in which
    // all the boxes can be transported
    let min_time = 0;
 
    // Check for the minimum time in which
    // all the boxes can be transported
    while (l <= h) {
        let mid = Math.floor((l + h) / 2);
 
        // If it is possible to transport all
        // the boxes in mid amount of time
        if (isPossible(box, truck, n, m, mid)) {
            min_time = mid;
            h = mid - 1;
        }
        else
            l = mid + 1;
    }
 
    return min_time;
}
 
// Driver code
let box = [ 10, 2, 16, 19 ];
let truck = [ 29, 25 ];
let n = box.length;
let m = truck.length;
document.write(minTime(box, truck, n, m));
 
 
</script>
Output: 
3

 

Time Complexity: O(N * log(N))
Auxiliary Space: O(1) 


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