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 + 4

Input: 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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int index(vector<int> vec, int value)
{
    vector<int>::iterator it;
    it = find(vec.begin(), vec.end(), value);
    return (it - vec.begin());
}
 
// 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 += pow(2, i);
        list.push_back(sum);
    }
 
    for (i = 0; i < 20; i++) {
        if (list[i] >= N) {
            return (index(list, list[i]) + 1);
        }
    }
}
 
// Driver Code
int main()
{
    int N = 10;
    cout << findCount(N) << endl;
    return 0;
}
 
// This code is contributed by kanugargng

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output



4

Time Complexity: O(n)

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:

CPP14

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// CPP progarm 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;
}

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Output

4

Time complexity : O(log(n))
 

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