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Coin Change | BFS Approach

  • Difficulty Level : Hard
  • Last Updated : 04 Aug, 2021
Geek Week

Given an integer X and an array arr[] of length N consisting of positive integers, the task is to pick minimum number of integers from the array such that they sum up to N. Any number can be chosen infinite number of times. If no answer exists then print -1.
Examples: 

Input: X = 7, arr[] = {3, 5, 4} 
Output:
The minimum number elements will be 2 as 
3 and 4 can be selected to reach 7.

Input: X = 4, arr[] = {5} 
Output: -1   

Approach: We have already seen how to solve this problem using dynamic-programming approach in this article.
Here, we will see a slightly different approach to solve this problem using BFS
Before that, let’s go ahead and define a state. A state SX can be defined as the minimum number of integers we would need to take from array to get a total of X.
Now, if we start looking at each state as a node in a graph such that each node is connected to (SX – arr[0], SX – arr[1], … SX – arr[N – 1])
Thus, we have to find the shortest path from state N to 0 in an unweighted and this can be done using BFS. BFS works here because the graph is unweighted.
Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of integers required
int minNumbers(int x, int* arr, int n)
{
    // Queue for BFS
    queue<int> q;
 
    // Base value in queue
    q.push(x);
 
    // Boolean array to check if a number has been
    // visited before
    unordered_set<int> v;
 
    // Variable to store depth of BFS
    int d = 0;
 
    // BFS algorithm
    while (q.size()) {
 
        // Size of queue
        int s = q.size();
        while (s--) {
 
            // Front most element of the queue
            int c = q.front();
 
            // Base case
            if (!c)
                return d;
            q.pop();
            if (v.find(c) != v.end() or c < 0)
                continue;
 
            // Setting current state as visited
            v.insert(c);
 
            // Pushing the required states in queue
            for (int i = 0; i < n; i++)
                q.push(c - arr[i]);
        }
 
        d++;
    }
 
    // If no possible solution
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 3, 4 };
    int n = sizeof(arr) / sizeof(int);
    int x = 7;
 
    cout << minNumbers(x, arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to find the minimum number
// of integers required
static int minNumbers(int x, int []arr, int n)
{
    // Queue for BFS
    Queue<Integer> q = new LinkedList<>();
 
    // Base value in queue
    q.add(x);
 
    // Boolean array to check if
    // a number has been visited before
    HashSet<Integer> v = new HashSet<Integer>();
 
    // Variable to store depth of BFS
    int d = 0;
 
    // BFS algorithm
    while (q.size() > 0)
    {
 
        // Size of queue
        int s = q.size();
        while (s-- > 0)
        {
 
            // Front most element of the queue
            int c = q.peek();
 
            // Base case
            if (c == 0)
                return d;
            q.remove();
            if (v.contains(c) || c < 0)
                continue;
 
            // Setting current state as visited
            v.add(c);
 
            // Pushing the required states in queue
            for (int i = 0; i < n; i++)
                q.add(c - arr[i]);
        }
        d++;
    }
 
    // If no possible solution
    return -1;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 3, 4 };
    int n = arr.length;
    int x = 7;
 
    System.out.println(minNumbers(x, arr, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to find the minimum number
# of integers required
def minNumbers(x, arr, n) :
 
    q = []
 
    # Base value in queue
    q.append(x)
 
    v = set([])
 
    d = 0
 
    while (len(q) > 0) :
 
        s = len(q)
        while (s) :
            s -= 1
            c = q[0]
            #print(q)
            if (c == 0) :
                return d
            q.pop(0)
            if ((c in v) or c < 0) :
                continue
 
            # Setting current state as visited
            v.add(c)
 
            # Pushing the required states in queue
            for i in range(n) :
                q.append(c - arr[i])            
             
        d += 1
        #print()
        #print(d,c)
 
    # If no possible solution
    return -1
 
arr = [ 1, 4,6 ]
n = len(arr)
x = 20
print(minNumbers(x, arr, n))
 
# This code is contributed by divyeshrabadiya07
# Improved by nishant.k108

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
 
// Function to find the minimum number
// of integers required
static int minNumbers(int x, int []arr, int n)
{
    // Queue for BFS
    Queue<int> q = new Queue<int>();
 
    // Base value in queue
    q.Enqueue(x);
 
    // Boolean array to check if
    // a number has been visited before
    HashSet<int> v = new HashSet<int>();
 
    // Variable to store depth of BFS
    int d = 0;
 
    // BFS algorithm
    while (q.Count > 0)
    {
 
        // Size of queue
        int s = q.Count;
        while (s-- > 0)
        {
 
            // Front most element of the queue
            int c = q.Peek();
 
            // Base case
            if (c == 0)
                return d;
            q.Dequeue();
            if (v.Contains(c) || c < 0)
                continue;
 
            // Setting current state as visited
            v.Add(c);
 
            // Pushing the required states in queue
            for (int i = 0; i < n; i++)
                q.Enqueue(c - arr[i]);
        }
        d++;
    }
 
    // If no possible solution
    return -1;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 3, 4 };
    int n = arr.Length;
    int x = 7;
 
    Console.WriteLine(minNumbers(x, arr, n));
}
}
 
// This code is contributed by Rajput-Ji
Output: 



2

 

Time complexity: O(N * X)
Auxiliary Space: O(N)

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