Minimum number of coins that can generate all the values in the given range
Given an integer N, the task is to find the minimum number of coins required to create all the values in the range [1, N].
Examples:
Input: N = 5
Output: 3
The coins {1, 2, 4} can be used to generate all the values in the range [1, 5].
1 = 1
2 = 2
3 = 1 + 2
4 = 4
5 = 1 + 4Input: N = 10
Output: 4
Approach: The problem is a variation of coin change problem, it can be solved with the help of binary numbers. In the above example, it can be seen that to create all the values between 1 to 10, denominator {1, 2, 4, 8} are required which can be rewritten as {20, 21, 22, 23}. The minimum number of coins to create all the values for a value N can be computed using the below algorithm.
// A list which contains the sum of all previous // bit values including that bit value list = [ 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023] // range = N for value in list: if(value >= N): print(list.index(value) + 1) break
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count // of minimum coins required int findCount( int N) { // To store the required sequence vector< int > list; int sum = 0; int i; // Creating list of the sum of all // previous bit values including // that bit value for (i = 0; i < 20; i++) { sum += (1<<i); list.push_back(sum); } for (i = 0; i < 20; i++) { if (list[i] >= N) { return (i + 1); } } } // Driver Code int main() { int N = 10; cout << findCount(N) << endl; return 0; } // This code is contributed by kanugargng |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the count // of minimum coins required static int findCount( int N) { Vector list = new Vector(); int sum = 0 ; int i; // Creating list of the sum of all // previous bit values including // that bit value for (i = 0 ; i < 20 ; i++) { sum += Math.pow( 2 , i); list.add(sum); } for (i = 0 ; i < 20 ; i++) { if (( int )list.get(i) >= N) return (list.indexOf(list.get(i)) + 1 ); } return 0 ; } // Driver Code public static void main(String[] args) { int N = 10 ; // Function Call to find count System.out.println(findCount(N)); } } // This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach # Function to return the count # of minimum coins required def findCount(N): # To store the required sequence list = [] sum = 0 # Creating list of the sum of all # previous bit values including # that bit value for i in range ( 0 , 20 ): sum + = 2 * * i list .append( sum ) for value in list : if (value > = N): return ( list .index(value) + 1 ) # Driver code N = 10 print (findCount(N)) |
C#
// C# implementation of the above approach using System; using System.Collections.Generic; class GFG { // Function to return the count // of minimum coins required static int findCount( int N) { List< int > list = new List< int >(); int sum = 0; int i; // Creating list of the sum of all // previous bit values including // that bit value for (i = 0; i < 20; i++) { sum += ( int )Math.Pow(2, i); list.Add(sum); } for (i = 0; i < 20; i++) { if (( int )list[i] >= N) return (i + 1); } return 0; } // Driver Code public static void Main(String[] args) { int N = 10; Console.WriteLine(findCount(N)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript implementation of the above approach // Function to return the count // of minimum coins required function findCount(N) { let list = []; let sum = 0; let i; // Creating list of the sum of all // previous bit values including // that bit value for (i = 0; i < 20; i++) { sum += parseInt(Math.pow(2, i), 10); list.push(sum); } for (i = 0; i < 20; i++) { if (list[i] >= N) return (i + 1); } return 0; } let N = 10; document.write(findCount(N)); // This code is contributed by rameshtravel07. </script> |
4
Time Complexity: O(n)
Auxiliary Space: O(1)
Efficient approach: The minimum number of coins required to create all the values in the range [1, N] will be log(N)/log(2) + 1.
Below is the implementation of the above approach:
C++14
// C++ program to find minimum number of coins #include<bits/stdc++.h> using namespace std; // Function to find minimum number of coins int findCount( int n) { return log (n)/ log (2)+1; } // Driver code int main() { int N = 10; cout << findCount(N) << endl; return 0; } |
Java
// Java program to find minimum number of coins class GFG{ // Function to find minimum number of coins static int findCount( int n) { return ( int )(Math.log(n) / Math.log( 2 )) + 1 ; } // Driver code public static void main(String[] args) { int N = 10 ; System.out.println(findCount(N)); } } // This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program to find minimum number of coins import math # Function to find minimum number of coins def findCount(n): return int (math.log(n, 2 )) + 1 # Driver code N = 10 print (findCount(N)) # This code is contributed by divyesh072019 |
C#
// C# program to find minimum number of coins using System; class GFG{ // Function to find minimum number of coins static int findCount( int n) { return ( int )(Math.Log(n) / Math.Log(2)) + 1; } // Driver code public static void Main() { int N = 10; Console.Write(findCount(N)); } } // This code is contributed by rutvik_56. |
Javascript
<script> // Javascript program to find minimum number of coins // Function to find minimum number of coins function findCount(n) { return parseInt(Math.log(n)/Math.log(2), 10)+1; } let N = 10; document.write(findCount(N)); </script> |
4
Time complexity : O(log(n))
Auxiliary space: O(1)
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