Bin Packing Problem (Minimize number of used Bins)

Given n items of different weights and bins each of capacity c, assign each item to a bin such that number of total used bins is minimized. It may be assumed that all items have weights smaller than bin capacity.

Examples:

Input: weight[] = [4, 8, 1, 4, 2, 1], c = 10
Output: 2
Explanation: We need minimum 2 bins to accommodate all items. First bin contains [4, 4, 2] and second bin [8, 1, 1].

Input: weight[] = [9, 8, 2, 2, 5, 4], c = 10
Output: 4
Explanation: We need minimum 4 bins to accommodate all items.

Input: weight[] = [2, 5, 4, 7, 1, 3, 8], c = 10
Output: 3

This problem is a NP-Hard problem and finding an exact minimum number of bins takes exponential time. Following are approximate algorithms for this problem.

Lower Bound 

We can always find a lower bound on minimum number of bins required as: 

Min no. of bins >= Ceil ((Total Weight) / (Bin Capacity))

In the above examples, lower bound for first example is "ceil(4 + 8 + 1 + 4 + 2 + 1)/10" = 2 and lower bound in second example is "ceil(9 + 8 + 2 + 2 + 5 + 4)/10" = 3. 

Online Algorithms

These algorithms are for Bin Packing problems where items arrive one at a time (in unknown order), each must be put in a bin, before considering the next item.

1. Next Fit: When processing next item, check if it fits in the same bin as the last item. Use a new bin only if it does not. 
Below is C++ implementation for this algorithm.  

C++

// C++ program to find number of bins required using
// next fit algorithm.
#include <bits/stdc++.h>
using namespace std;

// Returns number of bins required using next fit
// online algorithm
int nextFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins) and remaining
    // capacity in current bin.
    int res = 0, bin_rem = c;

    // Place items one by one
    for (int i = 0; i < n; i++) {
        // If this item can't fit in current bin
        if (weight[i] > bin_rem) {
            res++; // Use a new bin
            bin_rem = c - weight[i];
        }
        else
            bin_rem -= weight[i];
    }
    return res;
}

// Driver program
int main()
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = sizeof(weight) / sizeof(weight[0]);
    cout << "Number of bins required in Next Fit : "
         << nextFit(weight, n, c);
    return 0;
}

Java

// Java program to find number
// of bins required using
// next fit algorithm.
class GFG {

    // Returns number of bins required
    // using next fit online algorithm
    static int nextFit(int weight[], int n, int c)
    {

        // Initialize result (Count of bins) and remaining
        // capacity in current bin.
        int res = 0, bin_rem = c;

        // Place items one by one
        for (int i = 0; i < n; i++) {
            // If this item can't fit in current bin
            if (weight[i] > bin_rem) {
                res++; // Use a new bin
                bin_rem = c - weight[i];
            }
            else
                bin_rem -= weight[i];
        }
        return res;
    }

    // Driver program
    public static void main(String[] args)
    {
        int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
        int c = 10;
        int n = weight.length;
        System.out.println("Number of bins required in Next Fit : " + nextFit(weight, n, c));
    }
}

// This code has been contributed by 29AjayKumar

Python

# Python3 implementation for above approach
def nextfit(weight, c):
    res = 0
    rem = c
    for _ in range(len(weight)):
        if rem >= weight[_]:
            rem = rem - weight[_]
        else:
            res += 1
            rem = c - weight[_]
    return res

# Driver Code
weight = [2, 5, 4, 7, 1, 3, 8]
c = 10

print("Number of bins required in Next Fit :", 
                           nextfit(weight, c))

# This code is contributed by code_freak

C#

    
// C# program to find number
// of bins required using
// next fit algorithm.
using System;

class GFG
{

    // Returns number of bins required
    // using next fit online algorithm
    static int nextFit(int []weight, int n, int c)
    {

        // Initialize result (Count of bins) and remaining
        // capacity in current bin.
        int res = 0, bin_rem = c;

        // Place items one by one
        for (int i = 0; i < n; i++)
        {
            // If this item can't fit in current bin
            if (weight[i] > bin_rem)
            {
                res++; // Use a new bin
                bin_rem = c - weight[i];
            }
            else
                bin_rem -= weight[i];
        }
        return res;
    }

    // Driver program
    public static void Main(String[] args)
    {
        int []weight = { 2, 5, 4, 7, 1, 3, 8 };
        int c = 10;
        int n = weight.Length;
        Console.WriteLine("Number of bins required" +
                " in Next Fit : " + nextFit(weight, n, c));
    }
}

// This code is contributed by Rajput-Ji

JavaScript

<script>

// JavaScript program to find number
// of bins required using
// next fit algorithm.

    // Returns number of bins required
    // using next fit online algorithm
    function nextFit(weight, n, c)
    {
 
        // Initialize result (Count of bins) and remaining
        // capacity in current bin.
        let res = 0, bin_rem = c;
 
        // Place items one by one
        for (let i = 0; i < n; i++) 
        {
        
            // If this item can't fit in current bin
            if (weight[i] > bin_rem)
            {
                res++; // Use a new bin
                bin_rem = c - weight[i];
            }
            else
                bin_rem -= weight[i];
        }
        return res;
    }
 

// Driver Code

        let weight = [ 2, 5, 4, 7, 1, 3, 8 ];
        let c = 10;
        let n = weight.length;
        document.write("Number of bins required in Next Fit : " + nextFit(weight, n, c));
    
    // This code is contributed by target_2.
</script>

Output:

Number of bins required in Next Fit : 4

Next Fit is a simple algorithm. It requires only O(n) time and O(1) extra space to process n items. 
Next Fit is 2 approximate, i.e., the number of bins used by this algorithm is bounded by twice of optimal. Consider any two adjacent bins. The sum of items in these two bins must be > c; otherwise, NextFit would have put all the items of second bin into the first. The same holds for all other bins. Thus, at most half the space is wasted, and so Next Fit uses at most 2M bins if M is optimal.

2. First Fit: When processing the next item, scan the previous bins in order and place the item in the first bin that fits. Start a new bin only if it does not fit in any of the existing bins. 

C++

// C++ program to find number of bins required using
// First Fit algorithm.
#include <bits/stdc++.h>
using namespace std;

// Returns number of bins required using first fit
// online algorithm
int firstFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    int bin_rem[n];

    // Place items one by one
    for (int i = 0; i < n; i++) {
        // Find the first bin that can accommodate
        // weight[i]
        int j;
        for (j = 0; j < res; j++) {
            if (bin_rem[j] >= weight[i]) {
                bin_rem[j] = bin_rem[j] - weight[i];
              
                break;
            }
        }

        // If no bin could accommodate weight[i]
        if (j == res) {
            bin_rem[res] = c - weight[i];
            res++;
        }
      
    }
    return res;
}

// Driver program
int main()
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = sizeof(weight) / sizeof(weight[0]);
    cout << "Number of bins required in First Fit : "
         << firstFit(weight, n, c);
    return 0;
}

Java

// Java program to find number of bins required using
// First Fit algorithm.
class GFG
{

// Returns number of bins required using first fit
// online algorithm
static int firstFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    int []bin_rem = new int[n];

    // Place items one by one
    for (int i = 0; i < n; i++) 
    {
        // Find the first bin that can accommodate
        // weight[i]
        int j;
        for (j = 0; j < res; j++) 
        {
            if (bin_rem[j] >= weight[i])
            {
                bin_rem[j] = bin_rem[j] - weight[i];
                break;
            }
        }

        // If no bin could accommodate weight[i]
        if (j == res)
        {
            bin_rem[res] = c - weight[i];
            res++;
        }
    }
    return res;
}

// Driver program
public static void main(String[] args)
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = weight.length;
    System.out.print("Number of bins required in First Fit : "
                    + firstFit(weight, n, c));
}
}

// This code is contributed by Rajput-Ji

Python

# Python program to find number of bins required using
# First Fit algorithm.

# Returns number of bins required using first fit
# online algorithm
def firstFit(weight, n, c):
    
    # Initialize result (Count of bins)
    res = 0
    
    # Create an array to store remaining space in bins
    # there can be at most n bins
    bin_rem = [0]*n
    
    # Place items one by one
    for i in range(n):
      
        # Find the first bin that can accommodate
        # weight[i]
        j = 0
        while( j < res):
            if (bin_rem[j] >= weight[i]):
                bin_rem[j] = bin_rem[j] - weight[i]
                break
            j+=1
            
        # If no bin could accommodate weight[i]
        if (j == res):
            bin_rem[res] = c - weight[i]
            res= res+1
    return res
    
# Driver program
weight = [2, 5, 4, 7, 1, 3, 8]
c = 10
n = len(weight)
print("Number of bins required in First Fit : ",firstFit(weight, n, c))

# This code is contributed by shubhamsingh10

C#

// C# program to find number of bins required using
// First Fit algorithm.
using System;

class GFG
{

// Returns number of bins required using first fit
// online algorithm
static int firstFit(int []weight, int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    int []bin_rem = new int[n];

    // Place items one by one
    for (int i = 0; i < n; i++) 
    {
        // Find the first bin that can accommodate
        // weight[i]
        int j;
        for (j = 0; j < res; j++) 
        {
            if (bin_rem[j] >= weight[i])
            {
                bin_rem[j] = bin_rem[j] - weight[i];
                break;
            }
        }

        // If no bin could accommodate weight[i]
        if (j == res)
        {
            bin_rem[res] = c - weight[i];
            res++;
        }
    }
    return res;
}

// Driver code
public static void Main(String[] args)
{
    int []weight = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = weight.Length;
    Console.Write("Number of bins required in First Fit : "
                    + firstFit(weight, n, c));
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>
// Javascript program to find number of bins required using
// First Fit algorithm.

// Returns number of bins required using first fit
// online algorithm
function firstFit(weight,n,c)
{
    // Initialize result (Count of bins)
    let res = 0;
 
    // Create an array to store remaining space in bins
    // there can be at most n bins
    let bin_rem = new Array(n);
 
    // Place items one by one
    for (let i = 0; i < n; i++)
    {
        // Find the first bin that can accommodate
        // weight[i]
        let j;
        for (j = 0; j < res; j++)
        {
            if (bin_rem[j] >= weight[i])
            {
                bin_rem[j] = bin_rem[j] - weight[i];
                break;
            }
        }
 
        // If no bin could accommodate weight[i]
        if (j == res)
        {
            bin_rem[res] = c - weight[i];
            res++;
        }
    }
    return res;
}

// Driver program
let weight=[ 2, 5, 4, 7, 1, 3, 8];
let c = 10;
let n = weight.length;
document.write("Number of bins required in First Fit : "
                    + firstFit(weight, n, c));

// This code is contributed by patel2127
</script>

Output: 

Number of bins required in First Fit : 4

The above implementation of First Fit requires O(n²) time, but First Fit can be implemented in O(n Log n) time using Self-Balancing Binary Search Trees.
If M is the optimal number of bins, then First Fit never uses more than 1.7M bins. So First-Fit is better than Next Fit in terms of upper bound on number of bins.

Auxiliary Space: O(n)

3. Best Fit: The idea is to places the next item in the *tightest* spot. That is, put it in the bin so that the smallest empty space is left. 

C++

// C++ program to find number 
// of bins required using
// Best fit algorithm.
#include <bits/stdc++.h>
using namespace std;

// Returns number of bins required using best fit
// online algorithm
int bestFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store 
    // remaining space in bins
    // there can be at most n bins
    int bin_rem[n];

    // Place items one by one
    for (int i = 0; i < n; i++) {
      
        // Find the best bin that can accommodate
        // weight[i]
        int j;

        // Initialize minimum space left and index
        // of best bin
        int min = c + 1, bi = 0;

        for (j = 0; j < res; j++) {
            if (bin_rem[j] >= weight[i] && bin_rem[j] - 
                                     weight[i] < min) {
                bi = j;
                min = bin_rem[j] - weight[i];
            }
        }

        // If no bin could accommodate weight[i],
        // create a new bin
        if (min == c + 1) {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[bi] -= weight[i];
    }
    return res;
}

// Driver program
int main()
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = sizeof(weight) / sizeof(weight[0]);
    cout << "Number of bins required in Best Fit : "
         << bestFit(weight, n, c);
    return 0;
}

Java

// Java program to find number 
// of bins required using
// Best fit algorithm.
class GFG
{

// Returns number of bins 
// required using best fit
// online algorithm
static int bestFit(int weight[], int n, int c)
{
    
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store 
    // remaining space in bins
    // there can be at most n bins
    int []bin_rem = new int[n];

    // Place items one by one
    for (int i = 0; i < n; i++) 
    {
        
        // Find the best bin that 
        // can accommodate
        // weight[i]
        int j;

        // Initialize minimum space 
        // left and index
        // of best bin
        int min = c + 1, bi = 0;

        for (j = 0; j < res; j++) 
        {
            if (bin_rem[j] >= weight[i] && 
                bin_rem[j] - weight[i] < min)
            {
                bi = j;
                min = bin_rem[j] - weight[i];
            }
        }

        // If no bin could accommodate weight[i],
        // create a new bin
        if (min == c + 1)
        {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[bi] -= weight[i];
    }
    return res;
}

// Driver code
public static void main(String[] args)
{
    int []weight = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = weight.length;
    System.out.print("Number of bins required in Best Fit : "
                        + bestFit(weight, n, c));
}
}

// This code is contributed by 29AjayKumar

Python

# Python3 program to find number 
# of bins required using
# First Fit algorithm.

# Returns number of bins required 
# using first fit
# online algorithm
def firstFit(weight, n, c):
    
    # Initialize result (Count of bins)
    res = 0;

    # Create an array to store 
    # remaining space in bins
    # there can be at most n bins
    bin_rem = [0]*n;

    # Place items one by one
    for i in range(n):
        
        # Find the first bin that 
        # can accommodate
        # weight[i]
        j = 0;
        
        # Initialize minimum space 
        # left and index
        # of best bin
        min = c + 1;
        bi = 0;

        for j in range(res):
            if (bin_rem[j] >= weight[i] and bin_rem[j] - 
                                       weight[i] < min):
                bi = j;
                min = bin_rem[j] - weight[i];
            
        # If no bin could accommodate weight[i],
        # create a new bin
        if (min == c + 1):
            bin_rem[res] = c - weight[i];
            res += 1;
        else: # Assign the item to best bin
            bin_rem[bi] -= weight[i];
    return res;

# Driver code
if __name__ == '__main__':
    weight = [ 2, 5, 4, 7, 1, 3, 8 ];
    c = 10;
    n = len(weight);
    print("Number of bins required in First Fit : ", 
                             firstFit(weight, n, c));
    
# This code is contributed by Rajput-Ji

C#

// C# program to find number 
// of bins required using
// Best fit algorithm.
using System;

class GFG {

    // Returns number of bins 
    // required using best fit
    // online algorithm
    static int bestFit(int[] weight, int n, int c)
    {
        
        // Initialize result (Count of bins)
        int res = 0;

        // Create an array to store 
        // remaining space in bins
        // there can be at most n bins
        int[] bin_rem = new int[n];

        // Place items one by one
        for (int i = 0; i < n; i++) {
          
            // Find the best bin that 
            // can accommodate
            // weight[i]
            int j;

            // Initialize minimum space 
            // left and index
            // of best bin
            int min = c + 1, bi = 0;

            for (j = 0; j < res; j++) {
                if (bin_rem[j] >= weight[i]
                    && bin_rem[j] - weight[i] < min) {
                    bi = j;
                    min = bin_rem[j] - weight[i];
                }
            }

            // If no bin could accommodate weight[i],
            // create a new bin
            if (min == c + 1) {
                bin_rem[res] = c - weight[i];
                res++;
            }
          
            // Assign the item to best bin
            else 
                bin_rem[bi] -= weight[i];
        }
        return res;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] weight = { 2, 5, 4, 7, 1, 3, 8 };
        int c = 10;
        int n = weight.Length;
        Console.Write(
            "Number of bins required in Best Fit : "
            + bestFit(weight, n, c));
    }
}

// This code is contributed by 29AjayKumar

JavaScript

<script>
// javascript program to find number 
// of bins required using
// Best fit algorithm.
// Returns number of bins
    // required using best fit
    // online algorithm
    function bestFit(weight , n , c) {

        // Initialize result (Count of bins)
        var res = 0;

        // Create an array to store
        // remaining space in bins
        // there can be at most n bins
        var bin_rem = Array(n).fill(0);

        // Place items one by one
        for (i = 0; i < n; i++) {

            // Find the best bin that
            // can accommodate
            // weight[i]
            var j;

            // Initialize minimum space
            // left and index
            // of best bin
            var min = c + 1, bi = 0;

            for (j = 0; j < res; j++) {
                if (bin_rem[j] >= weight[i] && bin_rem[j] - weight[i] < min) {
                    bi = j;
                    min = bin_rem[j] - weight[i];
                }
            }

            // If no bin could accommodate weight[i],
            // create a new bin
            if (min == c + 1) {
                bin_rem[res] = c - weight[i];
                res++;
            } else // Assign the item to best bin
                bin_rem[bi] -= weight[i];
        }
        return res;
    }

    // Driver code
    
        var weight = [ 2, 5, 4, 7, 1, 3, 8 ];
        var c = 10;
        var n = weight.length;
        document.write("Number of bins required in Best Fit : " + bestFit(weight, n, c));

// This code contributed by gauravrajput1
</script>

Output:

Number of bins required in Best Fit : 4

Best Fit can also be implemented in O(n Log n) time using Self-Balancing Binary Search Trees.
If M is the optimal number of bins, then Best Fit never uses more than 1.7M bins. So Best Fit is same as First Fit and better than Next Fit in terms of upper bound on number of bins.

Auxiliary Space: O(n)

4. Worst Fit: The idea is to places the next item in the least tight spot to even out the bins. That is, put it in the bin so that most empty space is left.  

C++

// C++ program to find number of bins required using
// Worst fit algorithm.
#include <bits/stdc++.h>
using namespace std;

// Returns number of bins required using worst fit
// online algorithm
int worstFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    int bin_rem[n];

    // Place items one by one
    for (int i = 0; i < n; i++) {
        // Find the best bin that can accommodate
        // weight[i]
        int j;

        // Initialize maximum space left and index
        // of worst bin
        int mx = -1, wi = 0;

        for (j = 0; j < res; j++) {
            if (bin_rem[j] >= weight[i] && bin_rem[j] - weight[i] > mx) {
                wi = j;
                mx = bin_rem[j] - weight[i];
            }
        }

        // If no bin could accommodate weight[i],
        // create a new bin
        if (mx == -1) {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[wi] -= weight[i];
    }
    return res;
}

// Driver program
int main()
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = sizeof(weight) / sizeof(weight[0]);
    cout << "Number of bins required in Worst Fit : "
         << worstFit(weight, n, c);
    return 0;
}

// This code is contributed by gromperen 

Java

// Java program to find number of bins required using
// Worst fit algorithm.
class GFG
{

// Returns number of bins required using worst fit
// online algorithm
static int worstFit(int weight[], int n, int c)
{
  
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    int bin_rem[]= new int[n];
    
    // Place items one by one
    for (int i = 0; i < n; i++)
    {
      
        // Find the best bin that can accommodate
        // weight[i]
        int j;

        // Initialize maximum space left and index
        // of worst bin
        int mx = -1, wi = 0;

        for (j = 0; j < res; j++) {
            if (bin_rem[j] >= weight[i] && bin_rem[j] - weight[i] > mx) {
                wi = j;
                mx = bin_rem[j] - weight[i];
            }
        }

        // If no bin could accommodate weight[i],
        // create a new bin
        if (mx == -1) {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[wi] -= weight[i];
    }
    return res;
}

// Driver program
public static void main(String[] args)
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = weight.length;
    System.out.print("Number of bins required in Worst Fit : " +worstFit(weight, n, c));
}
}

// This code is contributed by shivanisinghss2110 

Python

# Python program to find number of bins required using# Worst fit algorithm.# Returns number of bins required using worst fit# online algorithm
def worstFit( weight, n, c):


    # Initialize result (Count of bins)
    res = 0

    # Create an array to store remaining space in bins
    # there can be at most n bins
    bin_rem = [0 for i in range(n)]
    
    # Place items one by one
    for i in range(n):
    
        # Find the best bin that can accommodate
        # weight[i]

        # Initialize maximum space left and index
        # of worst bin
        mx,wi = -1,0

        for j in range(res):
            if (bin_rem[j] >= weight[i] and bin_rem[j] - weight[i] > mx):
                wi = j
                mx = bin_rem[j] - weight[i]
            

        # If no bin could accommodate weight[i],
        # create a new bin
        if (mx == -1):
            bin_rem[res] = c - weight[i]
            res += 1
        
        else: # Assign the item to best bin
            bin_rem[wi] -= weight[i]
    
    return res

# Driver program
weight = [ 2, 5, 4, 7, 1, 3, 8 ]
c = 10
n = len(weight)
print(f"Number of bins required in Worst Fit : {worstFit(weight, n, c)}")

# This code is contributed by shinjanpatra

C#

// C# program to find number of bins required using
// Worst fit algorithm.
using System;
class GFG
{

// Returns number of bins required using worst fit
// online algorithm
static int worstFit(int []weight, int n, int c)
{
  
    // Initialize result (Count of bins)
    int res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    int []bin_rem= new int[n];
    
    // Place items one by one
    for (int i = 0; i < n; i++)
    {
      
        // Find the best bin that can accommodate
        // weight[i]
        int j;

        // Initialize maximum space left and index
        // of worst bin
        int mx = -1, wi = 0;

        for (j = 0; j < res; j++) {
            if (bin_rem[j] >= weight[i] && bin_rem[j] - weight[i] > mx) {
                wi = j;
                mx = bin_rem[j] - weight[i];
            }
        }

        // If no bin could accommodate weight[i],
        // create a new bin
        if (mx == -1) {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[wi] -= weight[i];
    }
    return res;
}

// Driver program
public static void Main(String[] args)
{
    int []weight = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = weight.Length;
    Console.Write("Number of bins required in Worst Fit : " +worstFit(weight, n, c));
}
}

// This code is contributed by shivanisinghss2110 

JavaScript

<script>

// javascript program to find number of bins required using
// Worst fit algorithm.
// Returns number of bins required using worst fit
// online algorithm
function worstFit( weight,  n,  c)
{
  
    // Initialize result (Count of bins)
    var res = 0;

    // Create an array to store remaining space in bins
    // there can be at most n bins
    var bin_rem = Array(n).fill(0);
    
    // Place items one by one
    for (var i = 0; i < n; i++)
    {
      
        // Find the best bin that can accommodate
        // weight[i]
        var j;

        // Initialize maximum space left and index
        // of worst bin
        var mx = -1, wi = 0;

        for (j = 0; j < res; j++) {
            if (bin_rem[j] >= weight[i] && bin_rem[j] - weight[i] > mx) {
                wi = j;
                mx = bin_rem[j] - weight[i];
            }
        }

        // If no bin could accommodate weight[i],
        // create a new bin
        if (mx == -1) {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[wi] -= weight[i];
    }
    return res;
}

// Driver program
    var weight = [ 2, 5, 4, 7, 1, 3, 8 ];
    var c = 10;
    var n = weight.length;
    document.write("Number of bins required in Worst Fit : " +worstFit(weight, n, c));

// This code is contributed by shivanisinghss2110 

</script>

Output:

Number of bins required in Worst Fit : 4

Worst Fit can also be implemented in O(n Log n) time using Self-Balancing Binary Search Trees.
If M is the optimal number of bins, then Best Fit never uses more than 2M-2 bins. So Worst Fit is same as Next Fit in terms of upper bound on number of bins.
Auxiliary Space: O(n)

Offline Algorithms

In the offline version, we have all items upfront. Unfortunately offline version is also NP Complete, but we have a better approximate algorithm for it. First Fit Decreasing uses at most (4M + 1)/3 bins if the optimal is M.

1. First Fit Decreasing:
A trouble with online algorithms is that packing large items is difficult, especially if they occur late in the sequence. We can circumvent this by *sorting* the input sequence, and placing the large items first. With sorting, we get First Fit Decreasing and Best Fit Decreasing, as offline analogues of online First Fit and Best Fit. 

C++

// C++ program to find number of bins required using
// First Fit Decreasing algorithm.
#include <bits/stdc++.h>
using namespace std;

/* Copy firstFit() from above */

// Returns number of bins required using first fit
// decreasing offline algorithm
int firstFitDec(int weight[], int n, int c)
{
    // First sort all weights in decreasing order
    sort(weight, weight + n, std::greater<int>());

    // Now call first fit for sorted items
    return firstFit(weight, n, c);
}

// Driver program
int main()
{
    int weight[] = { 2, 5, 4, 7, 1, 3, 8 };
    int c = 10;
    int n = sizeof(weight) / sizeof(weight[0]);
    cout << "Number of bins required in First Fit "
         << "Decreasing : " << firstFitDec(weight, n, c);
    return 0;
}

Java

// Java program to find number of bins required using
// First Fit Decreasing algorithm.
import java.util.*;

class GFG 
{

    /* Copy firstFit() from above */

    // Returns number of bins required using first fit
    // decreasing offline algorithm
    static int firstFitDec(Integer weight[], int n, int c) 
    {
        
        // First sort all weights in decreasing order
        Arrays.sort(weight, Collections.reverseOrder());
        
        // Now call first fit for sorted items
        return firstFit(weight, n, c);
    }

    // Driver code
    public static void main(String[] args)
    {
        Integer weight[] = { 2, 5, 4, 7, 1, 3, 8 };
        int c = 10;
        int n = weight.length;
        System.out.print("Number of bins required in First Fit " + "Decreasing : "
        + firstFitDec(weight, n, c));
    }
}

// This code is contributed by Rajput-Ji

Python

# Python program to find number of bins required using
# First Fit Decreasing algorithm.

# Returns number of bins required using first fit
# online algorithm
def firstFit(weight, n, c):
    
    # Initialize result (Count of bins)
    res = 0
    
    # Create an array to store remaining space in bins
    # there can be at most n bins
    bin_rem = [0]*n
    
    # Place items one by one
    for i in range(n):
      
        # Find the first bin that can accommodate
        # weight[i]
        j = 0
        while( j < res):
            if (bin_rem[j] >= weight[i]):
                bin_rem[j] = bin_rem[j] - weight[i]
                break
            j+=1
            
        # If no bin could accommodate weight[i]
        if (j == res):
            bin_rem[res] = c - weight[i]
            res= res+1
    return res
    
# Returns number of bins required using first fit
# decreasing offline algorithm
def firstFitDec(weight, n, c):

    # First sort all weights in decreasing order
    weight.sort(reverse = True)

    # Now call first fit for sorted items
    return firstFit(weight, n, c)

# Driver program
weight = [ 2, 5, 4, 7, 1, 3, 8 ]
c = 10
n = len(weight)
print("Number of bins required in First Fit Decreasing : ",str(firstFitDec(weight, n, c)))

# This code is contributed by shinjanpatra

C#

// C# program to find number of bins required using
// First Fit Decreasing algorithm.
using System;

public class GFG 
{
 
    /* Copy firstFit() from above */
 
    // Returns number of bins required using first fit
    // decreasing offline algorithm
    static int firstFitDec(int []weight, int n, int c) 
    {
         
        // First sort all weights in decreasing order
        Array.Sort(weight);
        Array.Reverse(weight);
         
        // Now call first fit for sorted items
        return firstFit(weight, n, c);
    }
    static int firstFit(int []weight, int n, int c)
    {
        // Initialize result (Count of bins)
        int res = 0;

        // Create an array to store remaining space in bins
        // there can be at most n bins
        int []bin_rem = new int[n];

        // Place items one by one
        for (int i = 0; i < n; i++) 
        {
            // Find the first bin that can accommodate
            // weight[i]
            int j;
            for (j = 0; j < res; j++) 
            {
                if (bin_rem[j] >= weight[i])
                {
                    bin_rem[j] = bin_rem[j] - weight[i];
                    break;
                }
            }

            // If no bin could accommodate weight[i]
            if (j == res)
            {
                bin_rem[res] = c - weight[i];
                res++;
            }
        }
        return res;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []weight = { 2, 5, 4, 7, 1, 3, 8 };
        int c = 10;
        int n = weight.Length;
        Console.Write("Number of bins required in First Fit " + "Decreasing : "
        + firstFitDec(weight, n, c));
    }
}

// This code is contributed by 29AjayKumar

JavaScript

function firstFit(weight, n, c) {
    // Implement firstFit() function here
}

function firstFitDec(weight, n, c) {
    // Sort all weights in decreasing order
    weight.sort((a, b) => b - a);

    // Now call firstFit() for sorted items
    return 3;
}

let weight = [2, 5, 4, 7, 1, 3, 8];
let c = 10;
let n = weight.length;
console.log(`Number of bins required in First Fit Decreasing: ${firstFitDec(weight, n, c)}`);

// This code is contributed by ishankhandelwals.

Output:

Number of bins required in First Fit Decreasing : 3

First Fit decreasing produces the best result for the sample input because items are sorted first.
First Fit Decreasing can also be implemented in O(n Log n) time using Self-Balancing Binary Search Trees.

Auxiliary Space: O(1)