[Dynamic Programming] Non-negative Integers without Consecutive Ones - Medium

Non-negative integers without consecutive ones are numbers that do not have two ’1’s next to each other in their binary form. We can use dynamic programming to count how many of these numbers are there up to a certain non-negative integer. This method helps us use results we already found to solve bigger problems. It is much faster than using simple recursive methods.

In this article, we will look into the dynamic programming way to count non-negative integers without consecutive ones. We will go over the problem and its rules. We will explain the dynamic programming method and show how to do it in Java, Python, and C++. We will also talk about how to make it better, compare different methods, and see where we can use this idea in real life. We will cover these sections:

  • Dynamic Programming Non-negative Integers without Consecutive Ones - Medium Solution Overview
  • Understanding the Problem Statement and Constraints
  • Dynamic Programming Approach for Non-negative Integers
  • Implementing the Solution in Java
  • Implementing the Solution in Python
  • Implementing the Solution in C++
  • Optimizing the Dynamic Programming Solution
  • Comparing Different Approaches
  • Real World Applications of Non-negative Integers without Consecutive Ones
  • Frequently Asked Questions

If you want to learn more about dynamic programming, you can check this article on Fibonacci numbers. It is a great place to start. You might also like this article on the climbing stairs problem for more about dynamic programming uses.

Understanding the Problem Statement and Constraints

The problem “Non-negative Integers without Consecutive Ones” asks us to count how many non-negative integers are less than or equal to a number n that do not have consecutive ones in their binary form.

Problem Statement

Given a non-negative integer n, we need to find out how many numbers between 0 and n have binary forms without consecutive ones.

Constraints

  • 0 <= n <= 10^9
  • The binary form of n can have up to 30 bits.

Examples

  • For n = 5 (which is 101 in binary), the valid numbers are 0, 1, 2, 3, 5. This adds up to 5.
  • For n = 1, the valid numbers are 0 and 1, which adds up to 2.

Observations

  • We cannot make binary numbers with consecutive ones. This limits how many valid numbers we can have.
  • We can use Fibonacci numbers to help find a solution. The count of valid numbers is similar to Fibonacci sequences.

This helps us think about a dynamic programming way to find the answer. We can use Fibonacci properties to get the result without checking every number one by one.

Dynamic Programming Approach for Non-negative Integers

We can solve the problem of counting non-negative integers without consecutive ones using dynamic programming. The main idea is to set a state that captures the problem. We can do this by using some properties of the Fibonacci sequence.

State Definition

We define dp[n] as the count of valid integers with length n. We can form these valid integers based on the last digit: - If the last digit is 0, the digits before it can be any valid combination of length n-1. - If the last digit is 1, the second last digit must be 0. So, the digits before can be any valid combination of length n-2.

Recurrence Relation

We can write the relation like this:

dp[n] = dp[n-1] + dp[n-2]

Where: - dp[0] = 1 (base case: one valid integer, which is an empty number) - dp[1] = 2 (valid integers are: 0 and 1)

Implementation

Here are code snippets that show the dynamic programming solution in Java, Python, and C++.

Java Implementation

public class Solution {
    public int findIntegers(int n) {
        int[] dp = new int[32];
        dp[0] = 1;
        dp[1] = 2;

        for (int i = 2; i < 32; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }

        int result = 0, prevBit = 0;
        for (int i = 31; i >= 0; i--) {
            if ((n & (1 << i)) != 0) {
                result += dp[i];
                if (prevBit == 1) {
                    return result;
                }
                prevBit = 1;
            } else {
                prevBit = 0;
            }
        }
        return result + 1;
    }
}

Python Implementation

class Solution:
    def findIntegers(self, n: int) -> int:
        dp = [0] * 32
        dp[0], dp[1] = 1, 2

        for i in range(2, 32):
            dp[i] = dp[i - 1] + dp[i - 2]

        result, prev_bit = 0, 0
        for i in range(31, -1, -1):
            if (n & (1 << i)) != 0:
                result += dp[i]
                if prev_bit == 1:
                    return result
                prev_bit = 1
            else:
                prev_bit = 0
        return result + 1

C++ Implementation

class Solution {
public:
    int findIntegers(int n) {
        vector<int> dp(32);
        dp[0] = 1; dp[1] = 2;

        for (int i = 2; i < 32; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }

        int result = 0, prev_bit = 0;
        for (int i = 31; i >= 0; i--) {
            if (n & (1 << i)) {
                result += dp[i];
                if (prev_bit == 1) {
                    return result;
                }
                prev_bit = 1;
            } else {
                prev_bit = 0;
            }
        }
        return result + 1;
    }
};

The dynamic programming approach above gives us a good solution to count non-negative integers without consecutive ones. This method uses ideas from the Fibonacci sequence. For more about dynamic programming methods, we can read the article on Dynamic Programming - Fibonacci Number.

Implementing the Solution in Java

We want to count non-negative integers without consecutive ones using dynamic programming. We can do this in Java. The main idea is to see that we can use binary numbers. Also, we can use Fibonacci numbers to help us count.

Java Implementation

public class NonConsecutiveOnes {
    public int findIntegers(int n) {
        // Create an array to store Fibonacci numbers
        int[] fib = new int[32];
        fib[0] = 1; // F(0)
        fib[1] = 2; // F(1)

        // Fill the Fibonacci array
        for (int i = 2; i < 32; i++) {
            fib[i] = fib[i - 1] + fib[i - 2];
        }

        int prevBit = 0, result = 0;
        for (int i = 31; i >= 0; i--) {
            // Check if the ith bit is set
            if ((n & (1 << i)) != 0) {
                result += fib[i]; // Add the count of valid numbers with this bit not set
                if (prevBit == 1) {
                    // If the previous bit was also set, break
                    return result;
                }
                prevBit = 1; // Set the previous bit flag
            } else {
                prevBit = 0; // Reset the previous bit flag if the current bit is not set
            }
        }
        // Include the last valid number (n itself)
        return result + 1;
    }

    public static void main(String[] args) {
        NonConsecutiveOnes solution = new NonConsecutiveOnes();
        int n = 5; // Example input
        System.out.println("Count of non-negative integers without consecutive ones: " + solution.findIntegers(n));
    }
}

Explanation of the Code

  • Fibonacci Array: We start with an array called fib. It stores Fibonacci numbers up to F(31). We need these numbers to count valid bit setups.
  • Bit Manipulation: We go through each bit of the number n. If a bit is set (1), we add counts from the Fibonacci array.
  • Consecutive Ones Check: The prevBit helps us check if two bits are set one after another. If they are, we stop and return the result since we cannot have consecutive ones.
  • Final Count: After checking all bits, we add one to the result. This includes the number n itself if it does not have consecutive ones.

This way, we can count non-negative integers without consecutive ones up to a number n using dynamic programming and bit manipulation.

Implementing the Solution in Python

We want to count non-negative integers that do not have consecutive ones. We can do this using dynamic programming in Python. The main idea is to use Fibonacci-like sequences. We can find the number of valid integers of length n by looking at the lengths n-1 and n-2.

Approach

  1. Understanding the Pattern:
    • We define f(n) to be the count of valid integers of length n.
    • We can write the relation as:
      • f(n) = f(n-1) + f(n-2)
    • This works because if we add a digit:
      • If we add 0, the digits before can be any valid setup of n-1.
      • If we add 1, the digit before must be 0, so the digits before can be any valid setup of n-2.
  2. Base Cases:
    • f(1) = 2 (0, 1)
    • f(2) = 3 (00, 01, 10)
  3. Implementation:

Here is the Python code to put this logic into action:

def findIntegers(n: int) -> int:
    # Change n to binary to find its length
    binary = bin(n)[2:]
    length = len(binary)
    
    # Create arrays for dp
    dp = [0] * (length + 1)
    dp[1] = 2  # f(1) = 2
    dp[2] = 3  # f(2) = 3
    
    # Fill the dp array
    for i in range(3, length + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    
    # Count the valid integers up to n
    count = 0
    prev_bit = 0
    
    for i in range(length):
        if binary[i] == '1':
            count += dp[length - i]  # Add counts for smaller bits
            if prev_bit == 1:  # If we find two consecutive 1s
                return count
            prev_bit = 1
        else:
            prev_bit = 0
    
    return count + 1  # Include the number itself

Example Usage

n = 5
result = findIntegers(n)
print(f"The count of non-negative integers without consecutive ones up to {n} is: {result}")

Explanation of the Code

  • The function findIntegers finds how many valid integers we can make without consecutive ones up to a number n.
  • It first changes n to binary to see its length.
  • It creates a dynamic programming array dp to keep counts for each length.
  • It goes through each bit of n, looking for consecutive ones and counting accordingly.

This Python code counts the non-negative integers without consecutive ones using dynamic programming. It works well and is fast for the problem we have.

Implementing the Solution in C++

We will implement the solution for counting non-negative integers without consecutive ones in C++. We can use dynamic programming. The main idea is to use properties of the Fibonacci sequence to count valid integers based on their binary form.

C++ Implementation

Here is a simple implementation in C++:

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    int findIntegers(int n) {
        vector<int> dp(32);
        dp[0] = 1; // 0
        dp[1] = 2; // 0, 1

        // Fill the dp array
        for (int i = 2; i < 32; i++) {
            dp[i] = dp[i - 1] + dp[i - 2]; // Fibonacci relation
        }

        int prev_bit = 0, result = 0;
        for (int i = 31; i >= 0; i--) {
            if (n & (1 << i)) {
                result += dp[i]; // Add valid integers for this bit position
                if (prev_bit) {
                    return result; // If previous bit was 1, break (invalid)
                }
                prev_bit = 1; // Set prev_bit since current bit is 1
            } else {
                prev_bit = 0; // Reset prev_bit since current bit is 0
            }
        }

        return result + 1; // Include the number itself
    }
};

int main() {
    Solution solution;
    int n = 5; // Example input
    cout << "Count of non-negative integers without consecutive ones: " << solution.findIntegers(n) << endl;
    return 0;
}

Explanation of the Code

  • Dynamic Programming Array (dp): We use dp[i] to store the count of valid integers that we can form with i bits.
  • Fibonacci Relation: The relation dp[i] = dp[i - 1] + dp[i - 2] comes from the fact that if the current bit is 1, the previous bit must be 0. If the current bit is 0, the previous bit can be either 0 or 1.
  • Bitwise Operations: We go through the bits of n from the highest to the lowest. For each bit, we check if it is set (1) or not (0) and update our result.
  • Final Count: We add 1 to the total count to include the number itself if it is valid.

This implementation counts non-negative integers without consecutive ones up to a given n using dynamic programming and bit manipulation.

Optimizing the Dynamic Programming Solution

We can optimize the dynamic programming solution for counting non-negative integers that do not have consecutive ones. We will use the properties of the Fibonacci sequence. The main idea is that the number of valid integers of length n without consecutive ones matches the Fibonacci sequence.

Fibonacci Relation

Let f(n) be the count of valid integers of length n. We can define the relationship like this:

  • If the last digit is 0, the first n-1 digits can be any valid combination. This gives us f(n-1).
  • If the last digit is 1, the previous digit must be 0. So, the first n-2 digits can be any valid combination, giving f(n-2).

This leads us to the formula:

f(n) = f(n-1) + f(n-2)

with base cases: - f(0) = 1 (the empty number) - f(1) = 2 (numbers: 0, 1)

Space Optimization

Instead of keeping an entire array for f(n), we can save space by storing only the last two values we computed. This is because the current value only depends on the last two values.

Java Implementation

public class NonConsecutiveOnes {
    public int findIntegers(int n) {
        int[] dp = new int[32];
        dp[0] = 1; // f(0)
        dp[1] = 2; // f(1)
        
        for (int i = 2; i < 32; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        
        int prevBit = 0, result = 0;
        for (int i = 31; i >= 0; i--) {
            if ((n & (1 << i)) != 0) {
                result += dp[i];
                if (prevBit == 1) {
                    return result; // Invalid case, stop here
                }
                prevBit = 1; // Current bit is 1
            } else {
                prevBit = 0; // Current bit is 0
            }
        }
        return result + 1; // Count the number itself
    }
}

Python Implementation

class NonConsecutiveOnes:
    def findIntegers(self, n: int) -> int:
        dp = [0] * 32
        dp[0], dp[1] = 1, 2
        
        for i in range(2, 32):
            dp[i] = dp[i - 1] + dp[i - 2]
        
        prev_bit, result = 0, 0
        
        for i in range(31, -1, -1):
            if (n & (1 << i)) != 0:
                result += dp[i]
                if prev_bit == 1:
                    return result  # Invalid case, stop here
                prev_bit = 1  # Current bit is 1
            else:
                prev_bit = 0  # Current bit is 0
        
        return result + 1  # Count the number itself

C++ Implementation

class Solution {
public:
    int findIntegers(int n) {
        vector<int> dp(32);
        dp[0] = 1; // f(0)
        dp[1] = 2; // f(1)
        
        for (int i = 2; i < 32; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        
        int prev_bit = 0, result = 0;
        
        for (int i = 31; i >= 0; i--) {
            if (n & (1 << i)) {
                result += dp[i];
                if (prev_bit == 1) {
                    return result; // Invalid case, stop here
                }
                prev_bit = 1; // Current bit is 1
            } else {
                prev_bit = 0; // Current bit is 0
            }
        }
        return result + 1; // Count the number itself
    }
};

This optimization makes the time complexity O(log n) and space complexity O(1). It gives a good solution for counting non-negative integers without consecutive ones. For more on dynamic programming and its uses, you can look at tutorials on Fibonacci numbers and climbing stairs.

Comparative Analysis of Different Approaches

When we solve the problem of counting non-negative integers without consecutive ones, we see that different ways can give us different results. Here, we look at some methods.

  1. Brute Force:
    • We generate all non-negative integers up to n and check for consecutive ones.
    • The complexity is O(2^n) because there are many combinations.
    • This method is not good for large n.
  2. Dynamic Programming:
    • We use the Fibonacci sequence method where:

      • Let dp[i] be the count of valid integers of length i.

      • We can define the relation like this:

        dp[i] = dp[i-1] + dp[i-2]

      • Base cases are: dp[0] = 1, dp[1] = 2.

    • The complexity is O(n) in time and O(n) in space for storing results.
  3. Optimized Dynamic Programming:

    • We can reduce the space we need by only keeping the last two values instead of the whole array:

      def countNonConsecutive(n):
    if n == 0: return 1
    if n == 1: return 2
    a, b = 1, 2
    for i in range(2, n + 1):
        a, b = b, a + b
    return b

    • The complexity here is O(n) in time and O(1) in space.

  4. Mathematical Approach:
    • We can use Fibonacci properties to find counts without going through all integers:
      • For a given n, valid integers relate to Fibonacci numbers. We can use a formula from combinatorial arguments.
    • The complexity is O(log n) using matrix exponentiation for finding Fibonacci numbers.
  5. Bit Manipulation:
    • We think of valid integers as binary numbers and use bitwise operations to remove those with consecutive ones.
    • The complexity is O(n) for checking each integer’s bits.

Summary of Approaches:

Approach Time Complexity Space Complexity
Brute Force O(2^n) O(1)
Dynamic Programming O(n) O(n)
Optimized Dynamic Programming O(n) O(1)
Mathematical Approach O(log n) O(1)
Bit Manipulation O(n) O(1)

We find that dynamic programming is the best way for this problem because it is simple and works well. For bigger inputs, the mathematical method may be better, especially for finding Fibonacci numbers directly.

Real World Applications of Non-negative Integers without Consecutive Ones

Counting non-negative integers without consecutive ones has many uses in computer science and math. This is important in areas like combinatorial problems, data encoding, and algorithm design. Here are some main uses:

  • Data Encoding: We can use this concept in encoding schemes where we need to avoid certain patterns. For instance, in binary data transmission, not having consecutive ones can help lower error rates. This also makes the signal stronger.

  • Dynamic Programming: This problem is a good example in dynamic programming. It teaches us about optimal substructure and overlapping subproblems. These ideas are important for solving many complex computing problems.

  • Combinatorial Structures: Counting non-negative integers without consecutive ones is connected to Fibonacci numbers. We can see Fibonacci numbers in many combinatorial situations. This includes counting paths in graphs and arranging items.

  • Resource Allocation: When we give out resources over time, we should make sure that no two resources are given out one after another. This can help avoid conflicts and make things fair.

  • Computer Networking: In networking protocols, we should avoid sending consecutive packets. This can help stop congestion and make the network run better.

  • Robotics and Path Planning: When we plan paths for robots, we should make sure that certain states are not visited one after another. This can help make movement better and save energy.

  • Game Design: In games, stopping players from taking consecutive actions can make the game more fun. This makes players think carefully about their next move.

These applications show how useful the problem of non-negative integers without consecutive ones can be. It is relevant in many areas. For more information about dynamic programming concepts, we can check articles like Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs.

Frequently Asked Questions

1. What are non-negative integers without consecutive ones?

Non-negative integers without consecutive ones are numbers that do not have ‘1’s next to each other when we write them in binary. For example, the number 5 is ’101’ in binary. This is okay. But the number 6 is ‘110’, and this one is not okay. We need to know this idea to solve problems in dynamic programming that count these types of integers well.

2. How can dynamic programming help in solving this problem?

Dynamic programming is a good way to count non-negative integers without consecutive ones. We can split the problem into smaller parts that overlap. By using results we already found, we save time. This way, we can compute answers fast by keeping results in an array or table. This helps us find the best solutions.

3. What is the time complexity of the dynamic programming solution?

The time complexity for our dynamic programming solution is O(n). Here, n is the number of bits in the biggest number we look at. This works well because there is a direct link between the number of bits and how we manage state changes in our dynamic programming method.

4. Can you implement the solution in multiple programming languages?

Yes, we can write the solution for non-negative integers without consecutive ones in many programming languages. Some of them are Java, Python, and C++. Each version will have a similar logic but will use the style and rules of that language. For examples, you can look at the implementations in Java, Python, and C++.

5. What are some real-world applications of non-negative integers without consecutive ones?

Non-negative integers without consecutive ones are useful in real life too. They help in areas like combinatorial optimization, coding theory, and problems where we have to satisfy certain rules. They are important when we need to avoid choosing things next to each other. This can happen in resource management, scheduling tasks, or making good algorithms for data structures. Knowing this can help us create better algorithms.

For more information on similar dynamic programming problems, look at the Dynamic Programming Fibonacci Number and Dynamic Programming Coin Change articles to improve your knowledge and skills.