[Array] Longest Subsequence With Limited Sum - Easy

The problem of finding the longest subsequence with a limited sum means we want to find the longest part of an array. This part should have a sum that does not go over a certain limit. To solve this, we usually sort the array first. Then we add elements to our subsequence one by one. We stop adding when the sum becomes too big. This way, we try to make the subsequence as long as we can while keeping the sum under control.

In this article, we will look closely at the longest subsequence with limited sum problem. We will understand what the problem means. We will also see the best way to solve it using sorting. We will share how to implement it in Java, Python, and C++. Additionally, we will talk about time and space complexity. We will also discuss common edge cases and answer some questions we often get about this topic.

[Array] Longest Subsequence With Limited Sum - Easy Overview
Understanding the Problem Statement for Longest Subsequence With Limited Sum
Optimal Approach for Longest Subsequence With Limited Sum Using Sorting
Java Implementation of Longest Subsequence With Limited Sum
Python Implementation of Longest Subsequence With Limited Sum
C++ Implementation of Longest Subsequence With Limited Sum
Time Complexity Analysis for Longest Subsequence With Limited Sum
Space Complexity Considerations for Longest Subsequence With Limited Sum
Common Edge Cases in Longest Subsequence With Limited Sum
Frequently Asked Questions

If you want to read more about related topics, we have some articles that might help. You can check out Array Two Sum - Easy, Array Maximum Subarray - Easy, and Array Rotate Array - Medium.

Understanding the Problem Statement for Longest Subsequence With Limited Sum

The problem is about finding the longest subsequence with a limited sum. We need to find the longest length of a subsequence from a given array. The sum of the elements in this subsequence should not go over a certain limit.

Problem Definition:

Input: An array of integers nums and an integer limit.
Output: The length of the longest subsequence where the sum of the subsequence elements is less than or equal to limit.

Requirements:

We must take the elements of the subsequence from the original array in the same order. They do not need to be next to each other.
Our goal is to have as many elements in the subsequence as possible while keeping the sum within the limit.

Example:

Input:
- nums = [3, 5, 2, 8, 1]
- limit = 10
Output: 4
Explanation: The subsequence [3, 2, 1] adds up to 6, which is okay. Its length is 3. If we add 5, the sum is 8. But if we add 8, it is too much. The longest valid subsequence is [3, 5, 1]. The sum is 9, so the length is 3.

We can solve this problem well by sorting the array. We can keep adding the smallest elements until we reach the limit. This way, we can make the subsequence as long as possible while keeping the sum in check.

Optimal Approach for Longest Subsequence With Limited Sum Using Sorting

To find the longest subsequence with a limited sum, we can use sorting. We will sort the array and go through it while keeping track of a total sum. This way, we start with the smallest numbers. It helps us to make the subsequence longer while staying within the sum limit.

Steps:

Sort the Array: First, we sort the input array in order from smallest to largest.
Initialize Variables: We create a variable for the total sum and another for counting the length of the subsequence.
Iterate Through the Array: We look at the sorted array and keep adding numbers to the total sum until adding the next number would go over the limit.
Count the Elements: Each time we add a number without going over the limit, we increase the counter.

Complexity:

Time Complexity: O(n log n) because of sorting, where n is how long the array is.
Space Complexity: O(1) if we sort it in place.

Example Code Implementations:

Java Implementation

import java.util.Arrays;

public class LongestSubsequence {
    public static int longestSubsequence(int[] arr, int limit) {
        Arrays.sort(arr);
        int sum = 0, count = 0;

        for (int num : arr) {
            if (sum + num <= limit) {
                sum += num;
                count++;
            } else {
                break;
            }
        }
        return count;
    }
}

Python Implementation

def longest_subsequence(arr, limit):
    arr.sort()
    total_sum = 0
    count = 0

    for num in arr:
        if total_sum + num <= limit:
            total_sum += num
            count += 1
        else:
            break
    return count

C++ Implementation

#include <vector>
#include <algorithm>

class Solution {
public:
    int longestSubsequence(std::vector<int>& arr, int limit) {
        std::sort(arr.begin(), arr.end());
        int total_sum = 0, count = 0;

        for (int num : arr) {
            if (total_sum + num <= limit) {
                total_sum += num;
                count++;
            } else {
                break;
            }
        }
        return count;
    }
};

This way is good for finding the longest subsequence that fits within the sum limit using sorting. It is simple and works well for this problem. If you want to see more related problems, you can check Array: Maximum Subarray. This may give more tips on how to work with arrays.

Java Implementation of Longest Subsequence With Limited Sum

To solve the Longest Subsequence With Limited Sum problem in Java, we can do these simple steps:

Sort the Input Array: First, we sort the array. This helps us to pick the smallest numbers first. This way, we can make the subsequence longer without going over the sum limit.
Iterate and Accumulate: Next, we set a counter for the length of the subsequence. We also need a variable to keep the current sum. Then we go through the sorted array. We add numbers to the sum until adding the next number will go over the limit.
Return the Length: Finally, when we finish checking all numbers, the counter tells us the length of the longest subsequence that does not go over the limit.

Here is the Java code that shows this logic:

import java.util.Arrays;

public class LongestSubsequenceWithLimitedSum {
    public static int maxLengthSubsequence(int[] nums, int limit) {
        Arrays.sort(nums);
        int currentSum = 0;
        int length = 0;

        for (int num : nums) {
            if (currentSum + num <= limit) {
                currentSum += num;
                length++;
            } else {
                break;
            }
        }

        return length;
    }

    public static void main(String[] args) {
        int[] nums = {4, 3, 2, 1, 5};
        int limit = 7;
        int result = maxLengthSubsequence(nums, limit);
        System.out.println("The length of the longest subsequence with limited sum is: " + result);
    }
}

Explanation of the Code:

Sorting: We use Arrays.sort(nums) to sort the input array.
Loop: We go through each number in the sorted array. If adding the current number to currentSum does not go over the limit, we update currentSum and increase the length.
Output: The program shows the length of the longest subsequence that we can create without going over the sum limit.

This implementation works well and helps us find the longest subsequence with a limited sum in Java. We can also use a similar idea in other programming languages. For more about working with arrays, we can read articles like Array Two Sum and Array Maximum Subarray.

Python Implementation of Longest Subsequence With Limited Sum

We want to find the longest subsequence with a limited sum using Python. We can do this easily by sorting the array and then going through it. The main idea is to sort the numbers and add them until we reach the limit.

Here is a simple way to do it:

def longest_subsequence_with_limited_sum(arr, limit):
    # Sort the array
    arr.sort()
    
    current_sum = 0
    count = 0
    
    for num in arr:
        if current_sum + num <= limit:
            current_sum += num
            count += 1
        else:
            break
    
    return count

# Example usage
arr = [3, 1, 5, 2, 8]
limit = 10
result = longest_subsequence_with_limited_sum(arr, limit)
print(f"The length of the longest subsequence is: {result}")

Explanation of the Code:

The function longest_subsequence_with_limited_sum takes an array arr and a number limit as inputs.
We sort the array in order from smallest to largest.
We set current_sum to keep track of the total sum and count to count how many numbers we can include.
We go through the sorted array. We add numbers to current_sum as long as they do not go over the limit. If adding a number goes over the limit, we stop.
Finally, the function returns how many numbers are in the longest subsequence.

This way runs in O(n log n) time because of sorting. Then, we have O(n) for going through the array. It is a good way for this problem.

If you want to learn more about similar problems, you can check these articles on Finding Maximum Subarray or Array Contains Duplicate.

C++ Implementation of Longest Subsequence With Limited Sum

We can solve the problem of finding the longest subsequence with a limited sum in C++. First, we sort the array. After sorting, we go through the elements. We keep track of the current sum and how many elements we add to the subsequence. We stop when we go over the limit.

Here is a simple implementation in C++:

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

int longestSubsequenceWithLimitedSum(std::vector<int>& nums, int limit) {
    // Sort the array
    std::sort(nums.begin(), nums.end());
    
    int count = 0;
    int currentSum = 0;

    // Iterate through the sorted array
    for (int num : nums) {
        // Check if adding the current number exceeds the limit
        if (currentSum + num <= limit) {
            currentSum += num;
            count++;
        } else {
            break; // Stop if limit is exceeded
        }
    }
    
    return count; // Return the length of the longest subsequence
}

int main() {
    std::vector<int> nums = {4, 5, 2, 1, 3};
    int limit = 5;
    int result = longestSubsequenceWithLimitedSum(nums, limit);
    std::cout << "Length of longest subsequence with limited sum: " << result << std::endl; // Output: 2
    return 0;
}

Explanation of the Code:

Sorting: We sort the array. This way, we can take the smallest elements first. This helps us get the longest subsequence without going over the limit.
Iterating: We go through the sorted array. We keep a running total called currentSum and count how many elements we add.
Condition Check: If adding the next number would go over the limit, we stop iterating.
Output: The function returns the count of the longest subsequence that fits the criteria.

This method finds the solution fast. The time complexity is (O(n n)) because of the sorting. Then we have an (O(n)) iteration. So it works well for arrays that are not too big.

For more details and similar problems, we can look at this article on maximum subarray.

Time Complexity Analysis for Longest Subsequence With Limited Sum

We look at the time complexity for finding the longest subsequence with a limited sum. This includes steps like sorting and going through the array.

Steps and Complexity Breakdown:

Sorting the Array:
- First, we need to sort the array. This takes time of (O(n n)). Here (n) is how many elements are in the array.
Iterating Through the Array:
- After we sort, we go through the array. We build the longest subsequence while checking the total sum. This needs one pass through the array. So, it adds (O(n)) to the total time.

Overall Time Complexity:

When we put these steps together, the overall time complexity is:

[ O(n n) + O(n) = O(n n) ]

The main part here is the sorting step. So, the time complexity for the longest subsequence with a limited sum is (O(n n)).

Example:

Let’s take the array [3, 1, 2, 5, 4] with a limit (L = 7): - Sorting: After sorting, the array looks like [1, 2, 3, 4, 5] in (O(n n)). - Iterating: We can add up the numbers and pick items until we reach the limit.

Summary:

To solve this problem well, we sort the array and then do a simple scan. This gives us a time complexity of (O(n n), which works well for bigger inputs.

Space Complexity Considerations for Longest Subsequence With Limited Sum

When we look at space complexity for finding the longest subsequence with a limited sum, we think about different things. We mainly focus on the data structures we use and the size of the input.

Input Space: We need O(n) space to store the input array. Here, n means the number of elements in the array.
Auxiliary Space: The extra space that the algorithm uses depends on our approach:
- Sorting Approach: If we sort the array as part of our solution, the space complexity can be O(n) in the worst case. This depends on the sorting method (like Timsort in Python or Merge Sort).
- Counting Variables: The extra variables we use for counting or storing things need O(1) space.
Overall Space Complexity: If we combine what we talked about:
- If our algorithm sorts the input, the total space complexity is O(n) because of input storage and sorting.
- If we do not sort, the space complexity stays O(1) for extra storage, but it is still O(n) for the input.
Final Consideration: In real life, space complexity can change based on how we implement and optimize our code. But usually, we can expect O(n) for input storage and O(1) or O(n) for extra storage depending on sorting.

Here is a simple example showing the space complexity in Java:

public int longestSubsequence(int[] arr, int limit) {
    Arrays.sort(arr); // O(n log n)
    int sum = 0, count = 0; // O(1)
    for (int num : arr) {
        if (sum + num <= limit) {
            sum += num;
            count++;
        }
    }
    return count;
}

In this example, the main space usage comes from the input array and any temporary variables we use. This gives an overall complexity of O(n).

For more info on similar problems, we can check out Array Two Sum. It provides insights on space management and optimization strategies.

Common Edge Cases in Longest Subsequence With Limited Sum

When we solve the problem to find the longest subsequence with a limited sum, we need to think about some edge cases. These cases can change the result. Here are some common situations to keep in mind:

Empty Array: If the input array is empty, then the longest subsequence is also empty. So, the result should be 0.
```
int[] nums = {};
int limit = 10;
int result = longestSubsequence(nums, limit); // result = 0
```
Single Element Array: If there is only one element in the array, we check if it is less than or equal to the limit. If it is, the longest subsequence is 1. If not, it is 0.
```
int[] nums = {5};
int limit = 5; // result = 1
```
All Elements Exceed Limit: If all elements in the array are bigger than the limit, then the longest subsequence is 0. No element can be included.
```
int[] nums = {10, 15, 20};
int limit = 5; // result = 0
```
Elements Adding Up to Limit: If we can add elements in the array to exactly meet the limit, we should find the combination that gives us the longest subsequence.
```
int[] nums = {1, 2, 3, 4};
int limit = 6; // result = 3 (subsequence: 1, 2, 3)
```
Negative Numbers: If the array has negative numbers, we can use these to make the subsequence longer without going over the limit.
```
int[] nums = {-1, 2, 3, 4};
int limit = 5; // result = 3 (subsequence: -1, 2, 3)
```
Duplicates: If the array has duplicates, we must make sure the algorithm counts them correctly in the subsequence.
```
int[] nums = {1, 1, 1, 2};
int limit = 3; // result = 3 (subsequence: 1, 1, 1)
```
Limit Exactly Matches the Sum of Multiple Elements: When we can combine several elements to match the limit exactly, we should maximize the count of elements.
```
int[] nums = {2, 3, 4, 5};
int limit = 7; // result = 2 (subsequence: 2, 5 or 3, 4)
```
Large Numbers: We should be careful with large numbers that can cause overflow. We need to handle the sum checks and comparisons right.
```
int[] nums = {Integer.MAX_VALUE, Integer.MAX_VALUE};
int limit = Integer.MAX_VALUE; // result = 0
```
All Elements are Zero: If the array has only zeros, the length of the longest subsequence can be as many as the number of zeros. This is true if the limit is also non-negative.
```
int[] nums = {0, 0, 0};
int limit = 0; // result = 3
```

By thinking about these edge cases, we can make the implementation of the longest subsequence with a limited sum stronger and more reliable.

Frequently Asked Questions

1. What is the problem statement for the Longest Subsequence With Limited Sum?

The problem statement for “Longest Subsequence With Limited Sum” is about finding the longest subsequence in an array. We need to make sure the sum of this subsequence does not go over a certain limit. We choose elements from the array to get the highest count of selected elements. At the same time, we keep their total sum within the limit.

2. How can sorting help in finding the longest subsequence with a limited sum?

Sorting the array is very important to solve the “Longest Subsequence With Limited Sum” problem better. When we sort the elements in increasing order, we can pick the smallest values first. This way, we include more elements in the subsequence without going over the limit. This greedy method makes it easier to add elements until we reach the sum limit.

3. What are the time and space complexities of the optimal solution?

The best way to solve “Longest Subsequence With Limited Sum” using sorting has a time complexity of O(n log n). This is because of the sorting part, where n is the number of elements in the array. The space complexity can be O(1) if we only need a few extra variables. It can also be O(n) if we save the sorted array. Knowing these complexities is important to check how good the solution is.

4. Can you provide examples of common edge cases for this problem?

Some common edge cases in “Longest Subsequence With Limited Sum” are when all elements are bigger than the limit. In this case, we get zero valid subsequences. We should also think about arrays with negative numbers, zeros, or duplicates. These can change the total sum and the number of elements we can use in a valid subsequence. Testing these cases helps make our solution strong.

5. What programming languages can I use to implement the solution?

We can use many programming languages to solve “Longest Subsequence With Limited Sum”. Some of them are Java, Python, and C++. Each language has its own features and libraries that can help us code easier. For example, Python lets us do list operations quickly. Java gives us strong type safety. We can look at detailed examples in Java, Python, and C++ to see how each language works best for this problem.