[Dynamic Programming] Maximum Sum of a Zigzag Subsequence - Medium

The Maximum Sum of a Zigzag Subsequence is a dynamic programming problem. We need to find a subsequence from a given sequence. This subsequence should switch between increasing and decreasing values. Our goal is to maximize the sum of its elements.

To solve this problem, we use two arrays. One array keeps track of the maximum sum of the subsequence that ends with an increase. The other array tracks the maximum sum that ends with a decrease. We carefully go through the sequence and update these arrays. This way, we can find the maximum sum of the zigzag subsequence.

In this article, we will look at the Maximum Sum of a Zigzag Subsequence in detail. First, we will explain the problem clearly. Then, we will break down the dynamic programming method. We will show how to implement it in Java, Python, and C++. We will also talk about common mistakes, performance analysis, and questions that people often ask about this dynamic programming challenge.

  • Dynamic Programming Maximum Sum of a Zigzag Subsequence Explained
  • Understanding the Problem Statement for Maximum Sum of a Zigzag Subsequence
  • Dynamic Programming Approach for Maximum Sum of a Zigzag Subsequence
  • Java Implementation of Maximum Sum of a Zigzag Subsequence
  • Python Code Example for Maximum Sum of a Zigzag Subsequence
  • C++ Solution for Maximum Sum of a Zigzag Subsequence
  • Optimizing the Dynamic Programming Solution for Zigzag Subsequences
  • Common Mistakes to Avoid in Maximum Sum of a Zigzag Subsequence
  • Performance Analysis of Zigzag Subsequence Solutions
  • Frequently Asked Questions

For more tips on dynamic programming, we can check out articles on Dynamic Programming Fibonacci Number and Dynamic Programming Maximum Subarray (Kadane’s Algorithm).

Understanding the Problem Statement for Maximum Sum of a Zigzag Subsequence

The Maximum Sum of a Zigzag Subsequence problem is about finding a subsequence in an array. We want to maximize the sum of the elements, and they must follow a zigzag pattern. A zigzag pattern means that each element is either bigger or smaller than the one before it.

Problem Definition

We have an array of integers. Our job is to find the maximum sum of a zigzag subsequence. A zigzag subsequence has these rules:

  • If we include an element ( A[i] ), the next element ( A[j] ) must be either bigger than ( A[i] ) or smaller than ( A[i] ) for ( i < j ).
  • The subsequence does not have to be next to each other.
  • We want to maximize the sum of the chosen elements that follow the zigzag pattern.

Example

For the array [1, 7, 4, 9, 2, 5], we can have a zigzag subsequence like [1, 7, 2, 5]. The maximum sum here is 15.

Constraints

  • The array can have both positive and negative numbers.
  • The length of the array can be from 1 to 10^5.

Input/Output Format

  • Input: An integer array.
  • Output: An integer that shows the maximum sum of the zigzag subsequence.

Key Points to Consider

  • We must keep the zigzag property in our subsequence.
  • It is important to create a good algorithm for big input sizes, ideally in ( O(n) ) time.
  • We should also think about special cases. For example, what if the array has all the same numbers or just one number.

We can solve this problem well using dynamic programming. This method helps us keep track of the maximum sums while following the zigzag rules.

Dynamic Programming Approach for Maximum Sum of a Zigzag Subsequence

The maximum sum of a zigzag subsequence problem is about finding a subsequence from a given sequence. We want to maximize the sum of its elements. The subsequence must follow a zigzag pattern. A zigzag pattern means the elements go up and down alternately.

Dynamic Programming Solution

  1. State Definition:
    • We define up[i] as the maximum sum of a zigzag subsequence ending at index i with the last move going up.
    • We define down[i] as the maximum sum of a zigzag subsequence ending at index i with the last move going down.
  2. Transition:
    • For each element arr[i] (where i goes from 1 to n-1):

      • If arr[i] > arr[j] for all j < i, then we can use this rule:

        up[i] = max(up[i], down[j] + arr[i])

      • If arr[i] < arr[j] for all j < i, then we can use this rule:

        down[i] = max(down[i], up[j] + arr[i])
    • We start by setting both up[i] and down[i] to arr[i]. This is because the smallest zigzag subsequence at each index is just the element itself.
  3. Final Result:
    • The result will be the maximum value between the last elements of up and down arrays.

Complexity Analysis

  • Time Complexity: O(n²) because of nested loops.
  • Space Complexity: O(n) for the up and down arrays.

Example Code Implementation

public class ZigzagSubsequence {
    public int maxZigzagSum(int[] arr) {
        if (arr.length == 0) return 0;

        int n = arr.length;
        int[] up = new int[n];
        int[] down = new int[n];

        for (int i = 0; i < n; i++) {
            up[i] = arr[i];
            down[i] = arr[i];
        }

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (arr[i] > arr[j]) {
                    up[i] = Math.max(up[i], down[j] + arr[i]);
                } else if (arr[i] < arr[j]) {
                    down[i] = Math.max(down[i], up[j] + arr[i]);
                }
            }
        }

        return Math.max(up[n - 1], down[n - 1]);
    }
}

Python Code Example

def max_zigzag_sum(arr):
    n = len(arr)
    if n == 0:
        return 0

    up = [0] * n
    down = [0] * n

    for i in range(n):
        up[i] = arr[i]
        down[i] = arr[i]

    for i in range(1, n):
        for j in range(i):
            if arr[i] > arr[j]:
                up[i] = max(up[i], down[j] + arr[i])
            elif arr[i] < arr[j]:
                down[i] = max(down[i], up[j] + arr[i])

    return max(up[n - 1], down[n - 1])

C++ Solution for Maximum Sum of a Zigzag Subsequence

class Solution {
public:
    int maxZigzagSum(vector<int>& arr) {
        int n = arr.size();
        if (n == 0) return 0;

        vector<int> up(n, 0);
        vector<int> down(n, 0);

        for (int i = 0; i < n; ++i) {
            up[i] = arr[i];
            down[i] = arr[i];
        }

        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (arr[i] > arr[j]) {
                    up[i] = max(up[i], down[j] + arr[i]);
                } else if (arr[i] < arr[j]) {
                    down[i] = max(down[i], up[j] + arr[i]);
                }
            }
        }

        return max(up[n - 1], down[n - 1]);
    }
};

This dynamic programming approach helps us find the maximum sum of a zigzag subsequence by using the rules of subsequences and zigzag patterns. For more on dynamic programming, we can read about Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Java Implementation of Maximum Sum of a Zigzag Subsequence

To solve the problem of finding the maximum sum of a zigzag subsequence in Java, we use dynamic programming. A zigzag subsequence is a sequence where the differences between the numbers change signs. If one number is bigger than the next, the next number must be smaller than the one after it, and the opposite is true too.

Here is the Java code:

public class ZigzagSubsequence {
    public static int maxZigzagSum(int[] nums) {
        if (nums == null || nums.length == 0) return 0;

        int n = nums.length;
        int[] up = new int[n]; // Store max sum ending with a peak
        int[] down = new int[n]; // Store max sum ending with a valley

        up[0] = nums[0]; // Start with the first number
        down[0] = 0; // No valley at the start

        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) {
                up[i] = down[i - 1] + nums[i]; // If current is a peak
                down[i] = down[i - 1]; // Keep previous valley
            } else if (nums[i] < nums[i - 1]) {
                down[i] = up[i - 1] + nums[i]; // If current is a valley
                up[i] = up[i - 1]; // Keep previous peak
            } else {
                up[i] = up[i - 1]; // No change
                down[i] = down[i - 1]; // No change
            }
        }

        return Math.max(up[n - 1], down[n - 1]); // Maximum of last peak or valley
    }

    public static void main(String[] args) {
        int[] nums = {1, 7, 4, 9, 2, 5};
        System.out.println("Maximum Sum of Zigzag Subsequence: " + maxZigzagSum(nums));
    }
}

Explanation of the Code:

  • We have two arrays named up and down.
    • up[i] shows the maximum sum of a zigzag subsequence that ends at index i with a peak.
    • down[i] shows the maximum sum of a zigzag subsequence that ends at index i with a valley.
  • We set the first element of up to the first number in the array and down to 0.
  • As we go through the array, we update the up and down arrays by comparing the numbers next to each other.
  • Finally, we return the biggest value from the last elements of up and down.

This Java code finds the maximum sum of a zigzag subsequence using dynamic programming. It runs in O(n) time and uses O(n) space. If we want to learn more about dynamic programming, we can check articles like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Python Code Example for Maximum Sum of a Zigzag Subsequence

To find the maximum sum of a zigzag subsequence, we can use dynamic programming. A zigzag subsequence is a sequence of numbers that goes up and down. We will use two lists to keep track of the maximum sums of zigzag subsequences. One list will track sums for upward movements and the other for downward movements.

Here is a simple code example in Python:

def maxZigzagSum(nums):
    if not nums:
        return 0
    
    n = len(nums)
    up = [0] * n  # Maximum sum of zigzag subsequence ending with upward movement
    down = [0] * n  # Maximum sum of zigzag subsequence ending with downward movement
    
    up[0] = nums[0]
    down[0] = nums[0]
    
    for i in range(1, n):
        up[i] = down[i-1] + nums[i]  # If we are moving up
        down[i] = up[i-1] + nums[i]  # If we are moving down
        
        # Make sure the sums don't go negative
        up[i] = max(up[i], nums[i])
        down[i] = max(down[i], nums[i])
    
    return max(max(up), max(down))

# Example usage
nums = [1, 2, 3, 4, 3, 2, 5]
result = maxZigzagSum(nums)
print(f"Maximum Sum of Zigzag Subsequence: {result}")

Explanation:

  • Starting Point: We begin by creating two lists called up and down. We set the first value of each list to the first number in nums.
  • Dynamic Programming Steps: For each number after the first, we find the maximum sums for both upward and downward zigzag sequences using the previous values.
  • Final Result: At the end, we get the maximum sum by looking at the highest values from both lists.

This Python code shows how to use dynamic programming to find the maximum sum of a zigzag subsequence. It makes sure that the zigzag rules are followed while trying to get the highest sum. If you want to learn more about dynamic programming, you can check Dynamic Programming Fibonacci Number.

C++ Solution for Maximum Sum of a Zigzag Subsequence

We can solve the problem to find the maximum sum of a zigzag subsequence in an array using C++. We will use a simple method called dynamic programming. A zigzag subsequence means the numbers go up and down alternately.

C++ Implementation

Here is a simple code to solve the problem in C++:

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

using namespace std;

int maxZigzagSum(vector<int>& nums) {
    int n = nums.size();
    if (n == 0) return 0;

    vector<int> up(n, 0), down(n, 0);
    up[0] = nums[0];
    down[0] = 0;

    for (int i = 1; i < n; ++i) {
        up[i] = down[i - 1] + nums[i];
        down[i] = max(down[i - 1], up[i - 1]);
    }

    return max(up[n - 1], down[n - 1]);
}

int main() {
    vector<int> nums = {1, 2, 3, 4, 3, 2, 5};
    cout << "Maximum Sum of Zigzag Subsequence: " << maxZigzagSum(nums) << endl;
    return 0;
}

Explanation of the Code

  • Initialization: We make two lists, up and down. The up[i] shows the maximum sum of a zigzag sequence ending at index i with an upward move. The down[i] shows the maximum sum with a downward move.
  • Base Case: For the first number, the maximum sum when it is an upward zigzag is its value. The downward sum is 0.
  • Dynamic Programming Transition: For each next number:
    • The maximum sum for an upward move at index i comes from the maximum downward sum of the last number plus the current number.
    • The maximum downward sum is the higher value between the last downward sum and the last upward sum.
  • Result: At the end, we get the maximum value from the last elements of both up and down.

Performance

  • Time Complexity: O(n), where n is the number of numbers in the input array.
  • Space Complexity: O(n) because we need extra space for the up and down lists. We can make it O(1) if we only keep the last values.

This C++ solution works well to find the maximum sum of a zigzag subsequence. It is good for competitive programming and job interviews.

Optimizing the Dynamic Programming Solution for Zigzag Subsequences

We can make the dynamic programming solution for the Maximum Sum of a Zigzag Subsequence better. By using some simple ideas, we can lower the space needed without losing speed.

Key Observations

  1. State Representation: The zigzag rule means the subsequence must switch between going up and down. So, we keep track of two states for each item:
    • up[i]: This is the maximum sum of a zigzag subsequence that ends at index i and goes up.
    • down[i]: This is the maximum sum of a zigzag subsequence that ends at index i and goes down.
  2. Transitions:

    • If the current item is bigger than the last one, it can continue a downward subsequence:

      up[i] = down[j] + nums[i]  (for all j < i where nums[j] < nums[i])

    • If it’s smaller, it can continue an upward subsequence:

      down[i] = up[j] + nums[i]  (for all j < i where nums[j] > nums[i])

Space Optimization

Instead of using arrays for up and down, we just need two variables to keep the maximum sums. This is because the state only relies on the last states. This change cuts down the space needed from O(n) to O(1).

Optimized Algorithm

Here is how we can write the optimized dynamic programming solution in Java:

public class ZigzagSubsequence {
    public static int maxZigzagSum(int[] nums) {
        if (nums.length == 0) return 0;

        int up = nums[0];   // This is the max sum of a zigzag subsequence going up
        int down = 0;       // This is the max sum of a zigzag subsequence going down

        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > nums[i - 1]) {
                up = Math.max(up, down + nums[i]); // Move from down to up
            } else if (nums[i] < nums[i - 1]) {
                down = Math.max(down, up + nums[i]); // Move from up to down
            }
        }

        return Math.max(up, down); // We return the bigger of both states
    }
}

Python Implementation

Now here is the same optimized solution in Python:

def max_zigzag_sum(nums):
    if not nums:
        return 0

    up = nums[0]
    down = 0

    for i in range(1, len(nums)):
        if nums[i] > nums[i - 1]:
            up = max(up, down + nums[i])
        elif nums[i] < nums[i - 1]:
            down = max(down, up + nums[i])

    return max(up, down)

C++ Implementation

For C++, we can write the optimized solution like this:

class Solution {
public:
    int maxZigzagSum(vector<int>& nums) {
        if (nums.empty()) return 0;

        int up = nums[0];
        int down = 0;

        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] > nums[i - 1]) {
                up = max(up, down + nums[i]);
            } else if (nums[i] < nums[i - 1]) {
                down = max(down, up + nums[i]);
            }
        }

        return max(up, down);
    }
};

Complexity Analysis

  • Time Complexity: O(n). This is for n items in the input array.
  • Space Complexity: O(1). We only use a fixed amount of space.

This better way calculates the maximum sum of a zigzag subsequence while using less memory. This is good for bigger inputs. For more on dynamic programming, see articles like Dynamic Programming - Fibonacci Number.

Common Mistakes to Avoid in Maximum Sum of a Zigzag Subsequence

When we try to find the maximum sum of a zigzag subsequence, we need to avoid some common mistakes. These mistakes can lead us to wrong answers. Here are some key points to keep in mind:

  1. Ignoring Zigzag Conditions:
    • We must make sure that the subsequence goes up and down. A mistake is including numbers that do not follow this rule. This will make our zigzag pattern incorrect.
  2. Incorrect Initialization of DP Arrays:
    • When we use dynamic programming, we need to set up our arrays right. For example, if we have two arrays for increasing and decreasing patterns, we should start them with the right values based on the first number in the input array.
  3. Improper State Transition:
    • We should pay attention to how we change states. For each number, we must decide if we want to continue the current zigzag subsequence or start a new one. We need to make sure our transitions show both increasing and decreasing patterns correctly.
  4. Neglecting Edge Cases:
    • We must be careful with edge cases like arrays with one number or arrays where all numbers are the same. If we do not handle these cases, we can get unexpected results.
  5. Not Considering All Possible Subsequences:
    • Sometimes, we might miss some valid subsequences. We need to look at all possible subsequences to make sure we find the maximum sum.
  6. Inadequate Testing:
    • We should always test our solution with different cases. This includes big arrays, those with negative numbers, or those with many repeating values. This helps us find problems in our logic or code.
  7. Overcomplicating the Algorithm:
    • It is good to make sure our solution is correct, but we should not make it too complex. Simple dynamic programming methods can often get the result we want more quickly.

By knowing these common mistakes, we can improve how we solve the Maximum Sum of a Zigzag Subsequence problem. For more information about dynamic programming, we can check out articles like Dynamic Programming Fibonacci Number or Dynamic Programming Maximum Subarray (Kadane’s Algorithm).

Performance Analysis of Zigzag Subsequence Solutions

We can look at how well the solutions work for the Maximum Sum of a Zigzag Subsequence problem by checking time and space complexity. We often use a dynamic programming method for this problem. It helps us find the best solution quickly.

Time Complexity

  • The time complexity of our dynamic programming solution for the Maximum Sum of a Zigzag Subsequence is O(n). Here n means the length of the input array. We only need to go through the array once to fill the dynamic programming table.

Space Complexity

  • We can lower the space complexity to O(1) if we only keep track of the last two states. This way, we do not need an extra array. But, if we want a simpler solution, it might use O(n) space since we will store results for all indices.

Example

If we have an input array, we can find the maximum sum of a zigzag subsequence like this:

public int maxZigzagSum(int[] nums) {
    int n = nums.length;
    if (n == 0) return 0;

    int[] up = new int[n];
    int[] down = new int[n];

    up[0] = nums[0];
    down[0] = 0;

    for (int i = 1; i < n; i++) {
        up[i] = down[i - 1] + nums[i];
        down[i] = Math.max(down[i - 1], up[i - 1]);
    }

    return Math.max(up[n - 1], down[n - 1]);
}

Performance Metrics

  • Input Size: If we have larger inputs, the execution time will increase. But the linear complexity keeps it manageable for usual limits.
  • Space Usage: How we choose the space complexity can really change memory use, especially for big sequences. For large sequences, using O(1) space is best to save memory.

Summary of Performance

Our dynamic programming method for the Maximum Sum of a Zigzag Subsequence gives good performance. It has linear time complexity and flexible space complexity. This makes it a good choice for competitive programming and real-world tasks where being efficient is important. We can also add more optimizations based on specific needs or changes in the problem. If you want to learn more about dynamic programming techniques, you can check articles like Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Frequently Asked Questions

1. What is a zigzag subsequence in dynamic programming?

A zigzag subsequence is a series of numbers. The differences between the numbers go up and down. In dynamic programming, we look for the zigzag subsequence in an array that has the highest sum. We can solve this problem well using dynamic programming methods. We store results that we find along the way.

2. How can I implement the maximum sum of a zigzag subsequence in Python?

To do this in Python, we can use dynamic programming. We need two lists. One list is for the maximum sum when we go up. The other list is for when we go down. We go through the array and update these two lists based on zigzag rules. This way, we can find the maximum sum quickly. For code examples, please see the Python part in this article.

3. What is the time complexity of the maximum sum of a zigzag subsequence algorithm?

The time complexity for this algorithm is O(n). Here n is the number of items in the input array. We achieve this speed by using dynamic programming. We only need to go through the array one time to find maximum sums. We use results we found before.

4. Are there any common pitfalls to avoid when solving the maximum sum of a zigzag subsequence?

Yes, there are some common mistakes. One mistake is not following zigzag rules. This means not switching between going up and down. Another mistake is forgetting to update the lists correctly. This can make our results wrong. We need to keep track of the previous numbers and how they help the current maximum sum.

5. Can you provide a C++ solution for the maximum sum of a zigzag subsequence?

Sure! The C++ solution uses a similar dynamic programming method like other solutions. We make two variables to keep track of the maximum sums for going up and down. We loop through the array and update these sums. For full C++ code, please look in the C++ solutions part of this article.

For more reading on similar dynamic programming problems, we can check articles like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) or Dynamic Programming: Longest Increasing Subsequence.