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++ 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 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 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# 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 |
<script> // Javascript implementation of the approach // Function to find the minimum number // of integers required function 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 code var 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.