Coin Change | BFS Approach

31 May, 2022
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: 2 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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to find the minimum number// of integers requiredint 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 codeint main(){    int arr[] = { 3, 3, 4 };    int n = sizeof(arr) / sizeof(int);    int x = 7;     cout << minNumbers(x, arr, n);     return 0;}
// Java implementation of the approachimport java.util.*; class GFG{ // Function to find the minimum number// of integers requiredstatic 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 codepublic 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 implementation of the approach # Function to find the minimum number# of integers requireddef 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 = 20print(minNumbers(x, arr, n)) # This code is contributed by divyeshrabadiya07# Improved by nishant.k108
// C# implementation of the approachusing System;using System.Collections.Generic;     class GFG{ // Function to find the minimum number// of integers requiredstatic 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 codepublic 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
<script> // Javascript implementation of the approach // Function to find the minimum number// of integers requiredfunction minNumbers(x, arr, n){    // Queue for BFS    var q = [];     // Base value in queue    q.push(x);     // Boolean array to check if a number has been    // visited before    var v = new Set();     // Variable to store depth of BFS    var d = 0;     // BFS algorithm    while (q.length!=0) {         // Size of queue        var s = q.length;        while (s--) {             // Front most element of the queue            var c = q[0];             // Base case            if (!c)                return d;            q.shift();            if (v.has(c) || c < 0)                continue;             // Setting current state as visited            v.add(c);             // Pushing the required states in queue            for (var i = 0; i < n; i++)                q.push(c - arr[i]);        }         d++;    }     // If no possible solution    return -1;} // Driver codevar arr = [3, 3, 4];var n = arr.length;var x = 7;document.write(minNumbers(x, arr, n)); // This code is contributed by rrrtnx.</script>
2
 
Time complexity: O(N * X)Auxiliary Space: O(N), since N extra space has been taken.
Rajput-Ji
divyeshrabadiya07
NishantKumar37
pankajsharmagfg
rrrtnx
rishavnitro
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