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 requiredint 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 Codeint main(){ int N = 10; cout << findCount(N) << endl; return 0;}// This code is contributed by kanugargng |
Java
// Java implementation of the approachimport 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 requireddef 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 codeN = 10print(findCount(N)) |
C#
// C# implementation of the above approachusing 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> |
Output
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 coinsint findCount(int n){ return log(n)/log(2)+1;}// Driver codeint main() { int N = 10; cout << findCount(N) << endl; return 0;} |
Java
// Java program to find minimum number of coinsclass GFG{ // Function to find minimum number of coinsstatic int findCount(int n){ return (int)(Math.log(n) / Math.log(2)) + 1; }// Driver codepublic 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 coinsimport math# Function to find minimum number of coinsdef findCount(n): return int(math.log(n, 2)) + 1 # Driver codeN = 10print(findCount(N))# This code is contributed by divyesh072019 |
C#
// C# program to find minimum number of coinsusing 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> |
Output
4
Time complexity : O(log(n))
Auxiliary space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
