Minimizing Operations to Reach N from 0 Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a number N. Find the minimum number of operations required to reach N starting from 0. You have 2 operations available: Double the numberAdd one to the numberExamples: Input: N = 8Output: 4Explanation: 0 + 1 = 1 --> 1 + 1 = 2 --> 2 * 2 = 4 --> 4 * 2 = 8. Input: N = 7Output: 5Explanation: 0 + 1 = 1 --> 1 + 1 = 2 --> 1 + 2 = 3 --> 3 * 2 = 6 --> 6 + 1 = 7. Approach: To solve the problem follow the below idea: The idea is to use greedy algorithm. Think this problem in reverse order like, if we have to reduce given number n to 0. Now, we'll approach in a way that if n is even then divided it by 2 otherwise, decrement it by 1 and keep a count variable to keep track of operation required to do. Step-by-step approach: Keep a variable count for tracking number of operations.Loop until given number is positive number.Check if number is even, then divide it by 2.Otherwise, decrement it by 1.Increment the count by 1.Return count.Below is implementation of the above approach: C++ C++ #include <bits/stdc++.h> using namespace std; // Function to calculate the minimum number // of operations required to make the given // number n equal to 0. The operations allowed // are either dividing n by 2 (if it's // even) or subtracting 1 from n. int minOperation(int n) { int count = 0; while (n > 0) { // If n is even, divide it by 2. if (n % 2 == 0) { n /= 2; } // If n is odd, subtract 1 from it. else { n -= 1; } count++; } return count; } // Drivers code int main() { int n; // Test cases n = 8; // 8 -> 4 -> 2 -> 1 -> 0, so it takes 4 // operations. cout << "Minimum operations for n = " << n << ": " << minOperation(n) << endl; n = 15; // 15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1 -> 0, so it // takes 7 operations. cout << "Minimum operations for n = " << n << ": " << minOperation(n) << endl; n = 1; // 1 -> 0, so it takes 1 operation. cout << "Minimum operations for n = " << n << ": " << minOperation(n) << std::endl; return 0; } Java public class MinimumOperations { public static int minOperation(int n) { int count = 0; while (n > 0) { // If n is even, divide it by 2. if (n % 2 == 0) { n /= 2; } // If n is odd, subtract 1 from it. else { n -= 1; } count++; } return count; } public static void main(String[] args) { int n; // Test cases n = 8; // 8 -> 4 -> 2 -> 1 -> 0, so it takes 4 operations. System.out.println("Minimum operations for n = " + n + ": " + minOperation(n)); n = 15; // 15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1 -> 0, so it takes 7 operations. System.out.println("Minimum operations for n = " + n + ": " + minOperation(n)); n = 1; // 1 -> 0, so it takes 1 operation. System.out.println("Minimum operations for n = " + n + ": " + minOperation(n)); } } Python3 def min_operation(n): count = 0 while n > 0: # If n is even, divide it by 2. if n % 2 == 0: n //= 2 # If n is odd, subtract 1 from it. else: n -= 1 count += 1 return count # Test cases n = 8 print("Minimum operations for n =", n, ":", min_operation(n)) n = 15 print("Minimum operations for n =", n, ":", min_operation(n)) n = 1 print("Minimum operations for n =", n, ":", min_operation(n)) C# // C# Code using System; public class MinimumOperations { public static int MinOperation(int n) { int count = 0; while (n > 0) { // If n is even, divide it by 2. if (n % 2 == 0) { n /= 2; } // If n is odd, subtract 1 from it. else { n -= 1; } count++; } return count; } public static void Main(string[] args) { int n; // Test cases n = 8; // 8 -> 4 -> 2 -> 1 -> 0, so it takes 4 operations. Console.WriteLine("Minimum operations for n = " + n + ": " + MinOperation(n)); n = 15; // 15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1 -> 0, so it takes 7 operations. Console.WriteLine("Minimum operations for n = " + n + ": " + MinOperation(n)); n = 1; // 1 -> 0, so it takes 1 operation. Console.WriteLine("Minimum operations for n = " + n + ": " + MinOperation(n)); } } JavaScript // Javascript code class MinimumOperations { static minOperation(n) { let count = 0; while (n > 0) { // If n is even, divide it by 2. if (n % 2 === 0) { n /= 2; } // If n is odd, subtract 1 from it. else { n -= 1; } count++; } return count; } static main() { let n; // Test cases n = 8; // 8 -> 4 -> 2 -> 1 -> 0, so it takes 4 operations. console.log(`Minimum operations for n = ${n}: ${MinimumOperations.minOperation(n)}`); n = 15; // 15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1 -> 0, so it takes 7 operations. console.log(`Minimum operations for n = ${n}: ${MinimumOperations.minOperation(n)}`); n = 1; // 1 -> 0, so it takes 1 operation. console.log(`Minimum operations for n = ${n}: ${MinimumOperations.minOperation(n)}`); } } // Call the main method MinimumOperations.main(); OutputMinimum operations for n = 8: 4 Minimum operations for n = 15: 7 Minimum operations for n = 1: 1Time Complexity: O(log(n))Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms K krishna_g Follow Improve Article Tags : Greedy Geeks Premier League DSA Greedy Algorithms Geeks Premier League 2023 +1 More Practice Tags : Greedy Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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