[Dynamic Programming] Minimum Steps to One - Medium

The Minimum Steps to One

The Minimum Steps to One problem is a well-known challenge in dynamic programming. It asks us to find the least number of steps to reduce a positive integer ( n ) to 1. We can do this by subtracting 1, dividing by 2, or dividing by 3. Our goal is to find the best way to do these operations with the least steps.

In this article, we will look at different ways to solve the Minimum Steps to One problem using dynamic programming. First, we will give an overview of the problem and explain dynamic programming. Next, we will discuss the recursive method in Java. After that, we will talk about the dynamic programming method in Java. We will also check optimized solutions in Python and C++. Finally, we will compare the performance of these different methods and answer some common questions.

Dynamic Programming Minimum Steps to One Problem Overview
Understanding Dynamic Programming Concepts
Recursive Approach to Minimum Steps to One in Java
Dynamic Programming Approach to Minimum Steps to One in Java
Optimized Dynamic Programming Solution for Minimum Steps to One in Python
Recursive Approach to Minimum Steps to One in Python
Dynamic Programming Implementation of Minimum Steps to One in C++
Optimized Approach for Minimum Steps to One in C++
Performance Comparison of Different Approaches
Frequently Asked Questions

If you want to read more about dynamic programming, we can check articles on Fibonacci numbers and their different forms. You may find these links helpful: Dynamic Programming Fibonacci Number and Dynamic Programming Fibonacci with Memoization.

Understanding Dynamic Programming Concepts

Dynamic Programming (DP) is a way to solve tough problems by breaking them into easier parts. It works well for optimization problems. These problems can be solved by using solutions of smaller parts. Some key ideas in dynamic programming are:

Overlapping Subproblems: We use DP when a problem can be split into smaller problems that we solve many times. For example, in the Minimum Steps to One problem, we can do the same calculations for the same values again and again.
Optimal Substructure: A problem has optimal substructure when the best solution to the problem includes the best solutions to its smaller problems. For example, to find minimum steps to reach one, the best way to lower a number depends on its smaller results.

Steps in Dynamic Programming

Define the State: We need to know what the state means. In the Minimum Steps to One problem, the state is dp[n], which is the least number of steps to change n to 1.
State Transition: We create a relation that shows how to find the state from smaller states. For the Minimum Steps to One problem:
- If n is divisible by 3, then dp[n] = min(dp[n], dp[n/3] + 1)
- If n is divisible by 2, then dp[n] = min(dp[n], dp[n/2] + 1)
- Always check dp[n] = min(dp[n], dp[n-1] + 1)
Base Case: We need to set the base case. For the Minimum Steps to One problem, dp[1] = 0. We do not need steps to change 1 to 1.
Compute Values: We use a loop to fill the DP table or array based on the state transition we made.
Return Result: We can get the result for the original problem from the DP table.

Example

For the value n = 10, we will check each number from 2 to 10. We will apply the rules we set.

public class MinStepsToOne {
    public static int minSteps(int n) {
        int[] dp = new int[n + 1];
        dp[1] = 0; // Base case

        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i - 1] + 1; // Step from i to i-1
            if (i % 2 == 0) {
                dp[i] = Math.min(dp[i], dp[i / 2] + 1);
            }
            if (i % 3 == 0) {
                dp[i] = Math.min(dp[i], dp[i / 3] + 1);
            }
        }
        return dp[n];
    }
}

This code shows how to find the minimum steps to change n to 1 using dynamic programming. It uses overlapping subproblems and optimal substructure well.

For more about dynamic programming, we can look at articles on the Fibonacci number problem or the Climbing Stairs problem.

Recursive Approach to Minimum Steps to One in Java

In the Minimum Steps to One problem, we want to find how many steps we need to take to change a number n into 1. We can do this with some allowed actions:

If n can be divided by 3, we divide it by 3.
If n can be divided by 2, we divide it by 2.
We can also subtract 1 from n.

The recursive way helps us to break this problem into smaller parts. We can find the minimum steps for n by looking at the results from the three actions we can take.

Recursive Method Implementation

Here is a simple Java code for the recursive approach:

public class MinimumStepsToOne {

    public static int minSteps(int n) {
        // Base case: when n is 1, no steps are needed
        if (n == 1) {
            return 0;
        }

        // Initialize steps to a large value
        int steps = Integer.MAX_VALUE;

        // If divisible by 3, take that path
        if (n % 3 == 0) {
            steps = Math.min(steps, minSteps(n / 3) + 1);
        }

        // If divisible by 2, take that path
        if (n % 2 == 0) {
            steps = Math.min(steps, minSteps(n / 2) + 1);
        }

        // Take the path of subtracting 1
        steps = Math.min(steps, minSteps(n - 1) + 1);

        return steps;
    }

    public static void main(String[] args) {
        int n = 10; // Example input
        System.out.println("Minimum steps to reduce " + n + " to 1: " + minSteps(n));
    }
}

Explanation of the Code

The minSteps method takes an integer n.
It first checks if n is 1. If it is, it returns 0 because we do not need any steps.
For each call, it:
- Divides n by 3 if it can and sees how many steps this takes.
- Divides n by 2 if it can and checks the steps needed.
- Subtracts 1 from n and checks the steps for this.
Finally, it gives back the smallest steps found from the three actions, adding 1 for the current operation.

This method takes a lot of time because it has overlapping problems. We can make it faster by using techniques like memoization or dynamic programming. If you want to learn more about dynamic programming, you can check articles like Dynamic Programming: Fibonacci Number or Dynamic Programming: Climbing Stairs.

Dynamic Programming Approach to Minimum Steps to One in Java

The Minimum Steps to One problem is a well-known dynamic programming challenge. We want to reduce a number ( n ) to 1 using the least number of steps. We can use these operations:

Subtract 1 from ( n ).
If ( n ) can be divided by 2, divide it by 2.
If ( n ) can be divided by 3, divide it by 3.

In the dynamic programming approach, we save results of smaller problems. This way, we do not have to calculate them again. We create an array called dp. In this array, dp[i] shows the minimum number of steps to reduce ( i ) to 1.

Java Implementation

public class MinimumStepsToOne {
    public static int minSteps(int n) {
        // Create a dp array to store minimum steps for each number
        int[] dp = new int[n + 1];

        // Base case
        dp[1] = 0; // Minimum steps to reduce 1 to 1 is 0

        for (int i = 2; i <= n; i++) {
            // Start with the option of subtracting 1
            dp[i] = dp[i - 1] + 1;

            // Check if divisible by 2
            if (i % 2 == 0) {
                dp[i] = Math.min(dp[i], dp[i / 2] + 1);
            }

            // Check if divisible by 3
            if (i % 3 == 0) {
                dp[i] = Math.min(dp[i], dp[i / 3] + 1);
            }
        }

        return dp[n];
    }

    public static void main(String[] args) {
        int n = 10; // Example input
        System.out.println("Minimum steps to reduce " + n + " to 1: " + minSteps(n));
    }
}

Explanation of the Code

We start by making a dp array. Its size is ( n + 1 ).
For each number ( i ) from 2 to ( n ):
- We first set the minimum steps to the result of subtracting 1 from ( i ).
- Then we check if ( i ) can be divided by 2 or 3. If it can, we find the minimum steps by dividing ( i ) by those numbers.
The final answer is in dp[n]. This shows the minimum steps to reduce ( n ) to 1.

This dynamic programming solution works in ( O(n) ) time and uses ( O(n) ) space. It’s efficient for big values of ( n . If you want to read more about similar dynamic programming problems, you can check Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs.

Optimized Dynamic Programming Solution for Minimum Steps to One in Python

We can solve the Minimum Steps to One problem using a simple dynamic programming method in Python. We will use an array to keep track of the minimum steps needed to reach each number down to 1. The main actions we can do are subtracting 1, dividing by 2, and dividing by 3.

Code Implementation

def min_steps_to_one(n):
    # We create a DP array where dp[i] shows the minimum steps to reach 1 from i
    dp = [0] * (n + 1)

    for i in range(2, n + 1):
        # We start with the option to subtract one
        dp[i] = dp[i - 1] + 1
        
        # We check for division by 2
        if i % 2 == 0:
            dp[i] = min(dp[i], dp[i // 2] + 1)
        
        # We check for division by 3
        if i % 3 == 0:
            dp[i] = min(dp[i], dp[i // 3] + 1)

    return dp[n]

# Example usage
n = 10
print(f"Minimum steps to reduce {n} to 1: {min_steps_to_one(n)}")

Explanation of the Code

Initialization: We make a list called dp with size n + 1 and fill it with zeros. dp[i] will show the minimum steps to reduce i to 1.
Dynamic Programming Loop: For every number from 2 to n, we find the minimum steps:
- We start by thinking the best option is to reduce by 1.
- If i can be divided by 2, we check if dividing by 2 gives a smaller step count.
- If i can be divided by 3, we do the same check for division by 3.
Return the Result: In the end, dp[n] has the minimum steps needed to reduce n to 1.

This method has a time complexity of (O(n)) and a space complexity of (O(n)). It works well for larger values of n. This way, we can quickly find out the minimum steps to reduce any integer to one.

Recursive Approach to Minimum Steps to One in Python

The Minimum Steps to One problem is about finding the least number of steps to change a number ( n ) to 1. We can use these operations:

Subtract 1 from ( n ).
If ( n ) can be divided by 2, divide ( n ) by 2.
If ( n ) can be divided by 3, divide ( n ) by 3.

The recursive way looks at all the combinations of these operations to get to 1. But it can be slow because it does some calculations again. Here is how we can do it in Python:

def min_steps_to_one(n):
    # Base case
    if n == 1:
        return 0

    # Recursive calls for the three operations
    steps = 1 + min_steps_to_one(n - 1)  # Subtract 1
    if n % 2 == 0:
        steps = min(steps, 1 + min_steps_to_one(n // 2))  # Divide by 2
    if n % 3 == 0:
        steps = min(steps, 1 + min_steps_to_one(n // 3))  # Divide by 3

    return steps

# Example usage
n = 10
print(f"Minimum steps to reduce {n} to 1: {min_steps_to_one(n)}")

Explanation of the Code

The function min_steps_to_one(n) finds the minimum steps needed for ( n ) using recursion.
The base case checks if ( n ) is 1. If yes, it returns 0 because we do not need any steps.
It looks at the minimum steps for each operation and returns the smallest value.
The example shows us how to call the function.

This way has an exponential time complexity. This means it can be slow when ( n ) is bigger. For better speed, we can use memoization or dynamic programming.

If we want to learn more about dynamic programming solutions, we can check other articles like Dynamic Programming Coin Change and Dynamic Programming Fibonacci with Memoization.

Dynamic Programming Implementation of Minimum Steps to One in C++

The Minimum Steps to One problem is about finding how many steps we need to take to change a number n into 1. We can do this by using these operations:

If n can be divided by 3, we divide it by 3.
If n can be divided by 2, we divide it by 2.
We can also subtract 1 from n.

We can solve this problem in an easy way by using dynamic programming in C++. Let’s see how we can do it.

C++ Code Implementation

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int minSteps(int n) {
    vector<int> dp(n + 1, 0);

    // Base case
    dp[1] = 0;  // Minimum steps to reduce 1 to 1 is 0

    for (int i = 2; i <= n; i++) {
        // Start with the operation of subtracting 1
        dp[i] = dp[i - 1] + 1;

        // Check if divisible by 2
        if (i % 2 == 0) {
            dp[i] = min(dp[i], dp[i / 2] + 1);
        }

        // Check if divisible by 3
        if (i % 3 == 0) {
            dp[i] = min(dp[i], dp[i / 3] + 1);
        }
    }

    return dp[n];
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;
    cout << "Minimum steps to reduce " << n << " to 1: " << minSteps(n) << endl;
    return 0;
}

Explanation of the Code

We create a vector called dp that is size n + 1 to keep the minimum steps for each number from 1 to n.
The base case is dp[1] which is 0.
For each number i from 2 to n, we find the minimum steps:
- We start by assuming we subtract 1 from i.
- If i can be divided by 2, we check if dividing by 2 gives us fewer steps.
- If i can be divided by 3, we check if dividing by 3 gives us fewer steps.
At the end, we return dp[n], which tells us the minimum steps to change n to 1.

This solution using dynamic programming is good because it takes O(n) time and uses O(n) space. It works well for big numbers n.

If you want to learn more about dynamic programming and other problems, you can check articles on Dynamic Programming: Coin Change and Dynamic Programming: Fibonacci with Memoization.

Optimized Approach for Minimum Steps to One in C++

We can solve the Minimum Steps to One problem well with dynamic programming in C++. We want to use an array to keep track of the minimum steps needed to reduce each number from 1 to n to 1.

Dynamic Programming Approach

We can create a dynamic programming array called dp. Here, dp[i] shows the minimum steps needed to reduce i to 1. We can write the rules like this:

If i is even, then dp[i] = dp[i/2] + 1
If i is odd, then dp[i] = dp[i-1] + 1
Also, if i can be divided by 3, we check dp[i] = min(dp[i], dp[i/3] + 1)

C++ Implementation

Here is the C++ code that shows the optimized way:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int minStepsToOne(int n) {
    vector<int> dp(n + 1, INT_MAX);
    dp[1] = 0; // Base case: 0 steps to reduce 1 to 1

    for (int i = 2; i <= n; ++i) {
        // We can always subtract 1
        dp[i] = dp[i - 1] + 1; 
        
        // Check if even
        if (i % 2 == 0) {
            dp[i] = min(dp[i], dp[i / 2] + 1);
        }
        
        // Check if divisible by 3
        if (i % 3 == 0) {
            dp[i] = min(dp[i], dp[i / 3] + 1);
        }
    }
    
    return dp[n];
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;
    cout << "Minimum steps to reduce " << n << " to 1: " << minStepsToOne(n) << endl;
    return 0;
}

Complexity Analysis

Time Complexity: O(n) because we look at all numbers from 2 to n.
Space Complexity: O(n) because we store the dp array.

With this dynamic programming method, we can find the minimum steps to reduce any integer to one fast and using less space. This method is better for larger numbers than the basic recursive way, which can take too long because it does the same work many times.

For more topics in dynamic programming, we can look at articles like Dynamic Programming: Fibonacci Number or Dynamic Programming: Climbing Stairs.

Performance Comparison of Different Approaches

When we try to solve the Minimum Steps to One problem, we can use different methods. These methods include recursive, dynamic programming, and optimized ones. We will compare how these methods perform based on time, space, and speed.

Recursive Approach

Time Complexity: O(3^n) - This method is slow because it makes many calls.
Space Complexity: O(n) - It needs space for the recursion stack.
Execution Speed: It is slow for big inputs because it does many repeated calculations.

Dynamic Programming Approach

Time Complexity: O(n) - This method is faster because it calculates each state only once.
Space Complexity: O(n) - It needs space for the DP array to keep results.
Execution Speed: This approach is much faster than the recursive one and works well for moderate input sizes.

Optimized Dynamic Programming Solution

Time Complexity: O(n) - It is still linear because it uses results we calculated before.
Space Complexity: O(1) - It needs very little space for the last few calculations.
Execution Speed: This method is the fastest among all, and it works well for big inputs.

Summary of Performance

The recursive method is not good for big numbers because of its slow time.
The standard dynamic programming method is much better but still needs linear space.
The optimized dynamic programming solution is the best for big inputs. It is quick and uses little space.

Example Code for Each Approach

Recursive Approach in Java:

public int minStepsRecursive(int n) {
    if (n == 1) return 0;
    int steps = minStepsRecursive(n - 1) + 1;
    if (n % 2 == 0) steps = Math.min(steps, minStepsRecursive(n / 2) + 1);
    if (n % 3 == 0) steps = Math.min(steps, minStepsRecursive(n / 3) + 1);
    return steps;
}

Dynamic Programming Approach in Java:

public int minStepsDP(int n) {
    int[] dp = new int[n + 1];
    for (int i = 2; i <= n; i++) {
        dp[i] = dp[i - 1] + 1;
        if (i % 2 == 0) dp[i] = Math.min(dp[i], dp[i / 2] + 1);
        if (i % 3 == 0) dp[i] = Math.min(dp[i], dp[i / 3] + 1);
    }
    return dp[n];
}

Optimized Dynamic Programming Solution in Python:

def min_steps_optimized(n):
    if n == 1:
        return 0
    steps = 0
    for i in range(2, n + 1):
        steps = i - 1  # Start with maximum steps
        if i % 2 == 0:
            steps = min(steps, steps + 1)
        if i % 3 == 0:
            steps = min(steps, steps + 1)
    return steps

In conclusion, the method we choose really matters for performance, especially when n is big. For practical use, we usually recommend the optimized dynamic programming solution. It is efficient and works well.

Frequently Asked Questions

1. What is the Minimum Steps to One problem in dynamic programming?

The Minimum Steps to One problem is a well-known challenge in dynamic programming. Here, we want to reduce a number n to 1 with the fewest operations. We can do this by subtracting one, dividing by two if it is even, or dividing by three if it is divisible by three. We must understand this problem to learn about dynamic programming. It helps us see how we can make recursive solutions better.

2. How can I solve the Minimum Steps to One problem using recursion in Java?

To solve the Minimum Steps to One problem with recursion in Java, we can create a function that takes an integer n. The function checks if n is 1 and then returns 0. For other numbers, it calls itself again. It looks at the three operations: subtracting one, dividing by two, and dividing by three. Then it gives back the smallest result plus one. While this method is simple, it can be slow without using memoization.

3. What is the dynamic programming approach for Minimum Steps to One in Python?

For the Minimum Steps to One problem in Python, we can use a dynamic programming method. We make an array to keep track of the minimum steps for each number from 1 to n. We go through the numbers and fill this array using previous values. This way, we do not have to use recursion. This method is much faster and shows how dynamic programming works.

4. Can you explain the optimized solution for Minimum Steps to One in C++?

The optimized solution for the Minimum Steps to One problem in C++ also uses a dynamic programming array. We store results for numbers up to n. We loop through the range and calculate the minimum steps based on the values we have already found. This method cuts down the time to O(n) and uses space well. It works great for larger numbers.

5. How does the performance of recursive and dynamic programming approaches compare for Minimum Steps to One?

The performance of recursive and dynamic programming methods for the Minimum Steps to One problem is very different. The recursive way can take a long time because of overlapping subproblems. On the other hand, the dynamic programming way makes it faster by keeping track of results we already calculated. So for big values of n, we should choose the dynamic programming method for better speed.

For more about dynamic programming techniques, we can read about the dynamic programming Fibonacci number and minimum path sum in a grid.