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.
Example:
Input: weight[] = {4, 8, 1, 4, 2, 1}
Bin Capacity c = 10
Output: 2
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}
Bin Capacity c = 10
Output: 4
We need minimum 4 bins to accommodate all items.
Input: weight[] = {2, 5, 4, 7, 1, 3, 8};
Bin Capacity c = 10
Output: 3
Lower Bound
We can always find a lower bound on minimum number of bins required. The lower bound can be given 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.
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.
Applications
- Loading of containers like trucks.
- Placing data on multiple disks.
- Job scheduling.
- Packing advertisements in fixed length radio/TV station breaks.
- Storing a large collection of music onto tapes/CD’s, etc.
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 algorithmint 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 programint 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 |
Python3
# Python3 implementation for above approachdef 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 Codeweight = [2, 5, 4, 7, 1, 3, 8]c = 10print("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 algorithmint 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 programint 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 algorithmstatic 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 programpublic 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 |
Python3
# Python program to find number of bins required using# First Fit algorithm.# Returns number of bins required using first fit# online algorithmdef 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 programweight = [2, 5, 4, 7, 1, 3, 8]c = 10n = 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 algorithmstatic 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 codepublic 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 algorithmfunction 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 programlet 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 algorithmint 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 programint 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 algorithmstatic 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 codepublic 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 |
Python3
# Python3 program to find number # of bins required using# First Fit algorithm.# Returns number of bins required # using first fit# online algorithmdef 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 codeif __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 algorithmint 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 programint 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 algorithmstatic 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 programpublic 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 |
Python3
# Python program to find number of bins required using# Worst fit algorithm.# Returns number of bins required using worst fit# online algorithmdef 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 programweight = [ 2, 5, 4, 7, 1, 3, 8 ]c = 10n = 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 algorithmstatic 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 programpublic 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 algorithmfunction 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.
4. 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 algorithmint 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 programint 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 |
Python3
# Python program to find number of bins required using# First Fit Decreasing algorithm.# Returns number of bins required using first fit# online algorithmdef 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 algorithmdef 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 programweight = [ 2, 5, 4, 7, 1, 3, 8 ]c = 10n = 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)
This article is contributed by Dheeraj Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
