Data Modelling & AI Data Structure & Algorithm

Longest subsequence with non negative prefix sum at each position

23 June 2025

0

Given an array arr[] consisting of N integers, the task is to find the longest subsequence such that the prefix sum at each position of the subsequence is non-negative.

Examples:

Input: arr[] = {4, -4, 1, -3, 1, -3}
Output: 5
Explanation:
Consider the subsequence as {4, 1, -3, 1, -3}. Now, the prefix sum of the chosen subsequence is {4, 5, 2, 3, 0}. Since, the prefix sum is non-negative at every possible index. Therefore, this subsequence is of maximum length having length 5.

Input: arr[] = {1, 3, 5, 7}
Output: 4

Naive Approach: The simplest approach to solve this problem is to generate all possible subsequences of the given array and print the length of that subsequence that has a non-negative prefix sum at each position and is of maximum length.

Time Complexity: O(2^N)
Auxiliary Space: O(2^N)

Efficient Approach: The above approach can also be optimized by using Dynamic Programming because it has Overlapping Subproblems property and Optimal Substructure property. Like other Dynamic Programming(DP) problems, recomputation of the same subproblems can be avoided by constructing a temporary array that stores the results of subproblems. Follow the steps below to solve the problem:

Initialize a matrix, say dp[][] where dp[i][j] stores the maximum sum possible if there are j valid elements till position i and initializing dp[][] array with -1.
Iterate over the range [0, N – 1] using the variable i and update dp[i][0] as 0.
If the value of arr[0] is at least 0, then update dp[0][1] as arr[0]. Otherwise, update it as -1.
Iterate over the range [1, N – 1] using the variable i:
- Iterate over the range [1, i + 1] using the variable j:
  - If current element is excluded i.e., if dp[i – 1][j] is not equal to -1, then update dp[i][j] as max of dp[i][j] and dp[i – 1][j].
  - If current element is included i.e., if dp[i – 1][j – 1] and dp[i – 1][j – 1] + arr[i] are greater than equal to 0, then update the value of dp[i][j] as the maximum of dp[i][j] and dp[i – 1][j – 1] + arr[i].
Initialize a variable say, ans as 0 to store the longest subsequence with non-negative prefix sum at each position.
Iterate in the range [0, N] using the variable j and if dp[N – 1][j] is greater than equal to 0, then update the value of ans as j.
After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of the
// longest subsequence with non negative
// prefix sum at each position
void longestSubsequence(int* arr, int N)
{
    // Stores the maximum sum possible
    // if we include j elements till
    // the position i
    int dp[N][N + 1];
 
    // Initialize dp array with -1
    memset(dp, -1, sizeof dp);
 
    // Maximum subsequence sum by
    // including no elements till
    // position 'i'
    for (int i = 0; i < N; ++i) {
        dp[i][0] = 0;
    }
 
    // Maximum subsequence sum by
    // including first element at
    // first position
    dp[0][1] = (arr[0] >= 0 ? arr[0] : -1);
 
    // Iterate over all the remaining
    // positions
    for (int i = 1; i < N; ++i) {
 
        for (int j = 1;
             j <= (i + 1); ++j) {
 
            // If the current element
            // is excluded
            if (dp[i - 1][j] != -1) {
                dp[i][j] = max(
                    dp[i][j], dp[i - 1][j]);
            }
 
            // If current element is
            // included and if total
            // sum is positive or not
            if (dp[i - 1][j - 1] >= 0
                && dp[i - 1][j - 1]
                           + arr[i]
                       >= 0) {
 
                dp[i][j] = max(
                    dp[i][j],
                    dp[i - 1][j - 1]
                        + arr[i]);
            }
        }
    }
 
    int ans = 0;
 
    // Select the maximum j by which
    // a non negative prefix sum
    // subsequence can be obtained
    for (int j = 0; j <= N; ++j) {
        if (dp[N - 1][j] >= 0) {
            ans = j;
        }
    }
 
    // Print the answer
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, -4, 1, -3, 1, -3 };
    int N = sizeof arr / sizeof arr[0];
    longestSubsequence(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the length of the
// longest subsequence with non negative
// prefix sum at each position
static void longestSubsequence(int[] arr, int N)
{
     
    // Stores the maximum sum possible
    // if we include j elements till
    // the position i
    int dp[][] = new int[N][N + 1];
 
    // Initialize dp array with -1
    for(int i = 0; i < N; ++i) 
    {
        for(int j = 0; j < N + 1; ++j) 
        {
            dp[i][j] = -1;
        }
    }
     
    // Maximum subsequence sum by
    // including no elements till
    // position 'i'
    for(int i = 0; i < N; ++i)
    {
        dp[i][0] = 0;
    }
 
    // Maximum subsequence sum by
    // including first element at
    // first position
    dp[0][1] = (arr[0] >= 0 ? arr[0] : -1);
 
    // Iterate over all the remaining
    // positions
    for(int i = 1; i < N; ++i)
    {
        for(int j = 1; j <= (i + 1); ++j) 
        {
             
            // If the current element
            // is excluded
            if (dp[i - 1][j] != -1) 
            {
                dp[i][j] = Math.max(
                    dp[i][j], dp[i - 1][j]);
            }
 
            // If current element is
            // included and if total
            // sum is positive or not
            if (dp[i - 1][j - 1] >= 0 && 
                dp[i - 1][j - 1] + arr[i] >= 0)
            {
                dp[i][j] = Math.max(dp[i][j],
                                    dp[i - 1][j - 1] + 
                                   arr[i]);
            }
        }
    }
 
    int ans = 0;
 
    // Select the maximum j by which
    // a non negative prefix sum
    // subsequence can be obtained
    for(int j = 0; j <= N; ++j) 
    {
        if (dp[N - 1][j] >= 0) 
        {
            ans = j;
        }
    }
 
    // Print the answer
    System.out.println(ans);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, -4, 1, -3, 1, -3 };
    int N = arr.length;
     
    longestSubsequence(arr, N);
}
}
 
// This code is contributed by avijitmondal1998

Python3

# Python3 Program for the above approach
 
# Function to find the length of the
# longest subsequence with non negative
# prefix sum at each position
def longestSubsequence(arr, N):
 
    # Stores the maximum sum possible
    # if we include j elements till
    # the position i
 
    # Initialize dp array with -1
    dp = [[-1 for i in range(N + 1)] for i in range(N)]
 
    # Maximum subsequence sum by
    # including no elements till
    # position 'i'
    for i in range(N):
        dp[i][0] = 0
 
    # Maximum subsequence sum by
    # including first element at
    # first position
    dp[0][1] = arr[0] if arr[0] >= 0 else -1
 
    # Iterate over all the remaining
    # positions
    for i in range(1, N):
 
        for j in range(1, i + 2):
 
            # If the current element
            # is excluded
            if (dp[i - 1][j] != -1):
                dp[i][j] = max(dp[i][j], dp[i - 1][j])
 
            # If current element is
            # included and if total
            # sum is positive or not
            if (dp[i - 1][j - 1] >= 0 and dp[i - 1][j - 1] + arr[i] >= 0):
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + arr[i])
 
    ans = 0
 
    # Select the maximum j by which
    # a non negative prefix sum
    # subsequence can be obtained
    for j in range(N + 1):
        if (dp[N - 1][j] >= 0):
            ans = j
 
    # Print the answer
    print(ans)
 
# Driver Code
 
 
arr = [4, -4, 1, -3, 1, -3]
N = len(arr)
longestSubsequence(arr, N)
 
# This code is contributed by _saurabhy_jaiswal

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the length of the
// longest subsequence with non negative
// prefix sum at each position
static void longestSubsequence(int[] arr, int N)
{
     
    // Stores the maximum sum possible
    // if we include j elements till
    // the position i
    int[,] dp = new int[N, N + 1];
 
    // Initialize dp array with -1
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N + 1; ++j) 
        {
            dp[i, j] = -1;
        }
    }
 
    // Maximum subsequence sum by
    // including no elements till
    // position 'i'
    for(int i = 0; i < N; ++i)
    {
        dp[i, 0] = 0;
    }
 
    // Maximum subsequence sum by
    // including first element at
    // first position
    dp[0, 1] = (arr[0] >= 0 ? arr[0] : -1);
 
    // Iterate over all the remaining
    // positions
    for(int i = 1; i < N; ++i) 
    {
        for(int j = 1; j <= (i + 1); ++j)
        {
             
            // If the current element
            // is excluded
            if (dp[i - 1, j] != -1) 
            {
                dp[i, j] = Math.Max(dp[i, j], 
                                    dp[i - 1, j]);
            }
 
            // If current element is
            // included and if total
            // sum is positive or not
            if (dp[i - 1, j - 1] >= 0 && 
                dp[i - 1, j - 1] + arr[i] >= 0) 
            {
                dp[i, j] = Math.Max(dp[i, j],
                                    dp[i - 1, j - 1] + 
                                    arr[i]);
            }
        }
    }
 
    int ans = 0;
 
    // Select the maximum j by which
    // a non negative prefix sum
    // subsequence can be obtained
    for(int j = 0; j <= N; ++j) 
    {
        if (dp[N - 1, j] >= 0) 
        {
            ans = j;
        }
    }
 
    // Print the answer
    Console.Write(ans);
}
 
// Driver code
public static void Main()
{
    int[] arr = { 4, -4, 1, -3, 1, -3 };
    int N = arr.Length;
 
    longestSubsequence(arr, N);
}
}
 
// This code is contributed by ukasp

Javascript

<script>
       // JavaScript Program for the above approach
 
       // Function to find the length of the
       // longest subsequence with non negative
       // prefix sum at each position
       function longestSubsequence(arr, N)
       {
        
           // Stores the maximum sum possible
           // if we include j elements till
           // the position i
 
           // Initialize dp array with -1
           let dp = Array(N).fill().map(() => Array(N + 1).fill(-1));
 
 
           // Maximum subsequence sum by
           // including no elements till
           // position 'i'
           for (let i = 0; i < N; ++i) {
               dp[i][0] = 0;
           }
 
           // Maximum subsequence sum by
           // including first element at
           // first position
           dp[0][1] = (arr[0] >= 0 ? arr[0] : -1);
 
           // Iterate over all the remaining
           // positions
           for (let i = 1; i < N; ++i) {
 
               for (let j = 1;
                   j <= (i + 1); ++j) {
 
                   // If the current element
                   // is excluded
                   if (dp[i - 1][j] != -1) {
                       dp[i][j] = Math.max(
                           dp[i][j], dp[i - 1][j]);
                   }
 
                   // If current element is
                   // included and if total
                   // sum is positive or not
                   if (dp[i - 1][j - 1] >= 0
                       && dp[i - 1][j - 1]
                       + arr[i]
                       >= 0) {
 
                       dp[i][j] = Math.max(
                           dp[i][j],
                           dp[i - 1][j - 1]
                           + arr[i]);
                   }
               }
           }
 
           let ans = 0;
 
           // Select the maximum j by which
           // a non negative prefix sum
           // subsequence can be obtained
           for (let j = 0; j <= N; ++j) {
               if (dp[N - 1][j] >= 0) {
                   ans = j;
               }
           }
 
           // Print the answer
           document.write(ans);
       }
 
       // Driver Code
 
       let arr = [4, -4, 1, -3, 1, -3];
       let N = arr.length;
       longestSubsequence(arr, N);
 
   // This code is contributed by Potta Lokesh
 
   </script>

Output

Time Complexity: O(N²)
Auxiliary Space: O(N²)

Efficient approach : Space optimization

In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

Create a 1D vector dp of size N+1.
Set a base case by initializing the values of DP .
Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
Initialize a variable ans to store the final answer and update it by iterating through the Dp.
At last return and print the final answer stored in ans .

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
 
// This function finds the length of the longest subsequence
// with a non-negative prefix sum at each position
void longestSubsequence(int* arr, int N)
{
    // Initialize an array to store the maximum subsequence sum
    // with j elements till the position i
    int dp[N + 1];
    // Initialize the dp array with -1
    memset(dp, -1, sizeof dp);
     
    // Maximum subsequence sum with no elements till position i is 0
    dp[0] = 0;
     
    // Maximum subsequence sum with first element at first position
    dp[1] = (arr[0] >= 0 ? arr[0] : -1);
     
    // Iterate over all the remaining positions
    for (int i = 1; i < N; ++i) {
     
        // Iterate over all the possible j values
        // (from 1 to i+1 or N, whichever is smaller)
        for (int j = min(i + 1, N); j >= 1; --j) {
     
            // If the current element is excluded
            if (dp[j] != -1) {
                dp[j] = max(dp[j], dp[j - 1] + arr[i]);
            }
     
            // If the current element is included and the total
            // sum is positive or zero
            if (dp[j - 1] >= 0 && dp[j - 1] + arr[i] >= 0) {
                dp[j] = max(dp[j], dp[j - 1] + arr[i]);
            }
        }
    }
     
    // Find the maximum j value for which a non-negative prefix
    // sum subsequence can be obtained
    int ans = 0;
    for (int j = 0; j <= N; ++j) {
        if (dp[j] >= 0) {
            ans = j;
        }
    }
     
    // Print the answer
    cout << ans << endl;
}
 
// Driver code to test the function
int main()
{
    int arr[] = { 4, -4, 1, -3, 1, -3 };
    int N = sizeof arr / sizeof arr[0];
    longestSubsequence(arr, N);
    return 0;
}

Java

import java.util.Arrays;
 
public class Main 
{
   
  // This function finds the length of the longest subsequence
  // with a non-negative prefix sum at each position
  public static void longestSubsequence(int[] arr, int N)
  {
     
    // Initialize an array to store the maximum subsequence sum
    // with j elements till the position i
    int[] dp = new int[N + 1];
     
    // Initialize the dp array with -1
    Arrays.fill(dp, -1);
 
    // Maximum subsequence sum with no elements till position i is 0
    dp[0] = 0;
 
    // Maximum subsequence sum with first element at first position
    dp[1] = (arr[0] >= 0 ? arr[0] : -1);
 
    // Iterate over all the remaining positions
    for (int i = 1; i < N; ++i) {
 
      // Iterate over all the possible j values
      // (from 1 to i+1 or N, whichever is smaller)
      for (int j = Math.min(i + 1, N); j >= 1; --j) {
 
        // If the current element is excluded
        if (dp[j] != -1) {
          dp[j] = Math.max(dp[j], dp[j - 1] + arr[i]);
        }
 
        // If the current element is included and the total
        // sum is positive or zero
        if (dp[j - 1] >= 0 && dp[j - 1] + arr[i] >= 0) {
          dp[j] = Math.max(dp[j], dp[j - 1] + arr[i]);
        }
      }
    }
 
    // Find the maximum j value for which a non-negative prefix
    // sum subsequence can be obtained
    int ans = 0;
    for (int j = 0; j <= N; ++j) {
      if (dp[j] >= 0) {
        ans = j;
      }
    }
 
    // Print the answer
    System.out.println(ans);
  }
 
  // Driver code to test the function
  public static void main(String[] args) {
    int[] arr = {4, -4, 1, -3, 1, -3};
    int N = arr.length;
    longestSubsequence(arr, N);
  }
}

Python

def longestSubsequence(arr, N):
    # Initialize an array to store the maximum subsequence sum
    # with j elements till the position i
    dp = [-1] * (N + 1)
 
    # Maximum subsequence sum with no elements till position i is 0
    dp[0] = 0
 
    # Maximum subsequence sum with first element at first position
    dp[1] = arr[0] if arr[0] >= 0 else -1
 
    # Iterate over all the remaining positions
    for i in range(1, N):
        # Iterate over all the possible j values
        # (from 1 to i+1 or N, whichever is smaller)
        for j in range(min(i + 1, N), 0, -1):
            # If the current element is excluded
            if dp[j] != -1:
                dp[j] = max(dp[j], dp[j - 1] + arr[i])
 
            # If the current element is included and the total
            # sum is positive or zero
            if dp[j - 1] >= 0 and dp[j - 1] + arr[i] >= 0:
                dp[j] = max(dp[j], dp[j - 1] + arr[i])
 
    # Find the maximum j value for which a non-negative prefix
    # sum subsequence can be obtained
    ans = 0
    for j in range(N + 1):
        if dp[j] >= 0:
            ans = j
 
    # Print the answer
    print(ans)
 
# Driver code to test the function
if __name__ == '__main__':
    arr = [4, -4, 1, -3, 1, -3]
    N = len(arr)
    longestSubsequence(arr, N)

C#

using System;
 
class GFG
{
    // This function finds the length of the longest subsequence
    // with a non-negative prefix sum at each position
    static void longestSubsequence(int[] arr, int N)
    {
        // Initialize an array to store the maximum subsequence sum
        // with j elements till the position i
        int[] dp = new int[N + 1];
         
         
        // Initialize the dp array with -1
        Array.Fill(dp, -1);
         
        // Maximum subsequence sum with no elements till position i is 0
        dp[0] = 0;
         
        // Maximum subsequence sum with first element at first position
        dp[1] = (arr[0] >= 0 ? arr[0] : -1);
         
        // Iterate over all the remaining positions
        for (int i = 1; i < N; ++i)
        {
            // Iterate over all the possible j values
            // (from 1 to i+1 or N, whichever is smaller)
            for (int j = Math.Min(i + 1, N); j >= 1; --j)
            {
                // If the current element is excluded
                if (dp[j] != -1)
                {
                    dp[j] = Math.Max(dp[j], dp[j - 1] + arr[i]);
                }
                 
                // If the current element is included and the total
                // sum is positive or zero
                if (dp[j - 1] >= 0 && dp[j - 1] + arr[i] >= 0)
                {
                    dp[j] = Math.Max(dp[j], dp[j - 1] + arr[i]);
                }
            }
        }
         
        // Find the maximum j value for which a non-negative prefix
        // sum subsequence can be obtained
        int ans = 0;
        for (int j = 0; j <= N; ++j)
        {
            if (dp[j] >= 0)
            {
                ans = j;
            }
        }
         
        // Print the answer
        Console.WriteLine(ans);
    }
     
    // Driver code to test the function
    static void Main(string[] args)
    {
        int[] arr = { 4, -4, 1, -3, 1, -3 };
        int N = arr.Length;
        longestSubsequence(arr, N);
    }
}

Javascript

// This function finds the length of the longest subsequence
// with a non-negative prefix sum at each position
function longestSubsequence(arr, N) {
 
  // Initialize an array to store the maximum subsequence sum
  // with j elements till the position i
  // Initialize the dp array with -1
  let dp = new Array(N + 1).fill(-1);
       
  // Maximum subsequence sum with no elements till position i is 0
  dp[0] = 0;
   
       
  // Maximum subsequence sum with first element at first position
  dp[1] = arr[0] >= 0 ? arr[0] : -1;
 
     
  // Iterate over all the remaining positions
  for (let i = 1; i < N; ++i) {
     
    // Iterate over all the possible j values
    // (from 1 to i+1 or N, whichever is smaller)
    for (let j = Math.min(i + 1, N); j >= 1; --j) {
     
       // If the current element is excluded
       if (dp[j] !== -1) {
        dp[j] = Math.max(dp[j], dp[j - 1] + arr[i]);
      }
 
 
      // If the current element is included and the total
      // sum is positive or zero
      if (dp[j - 1] >= 0 && dp[j - 1] + arr[i] >= 0) {
        dp[j] = Math.max(dp[j], dp[j - 1] + arr[i]);
      }
    }
  }
 
  // Find the maximum j value for which a non-negative prefix
 // sum subsequence can be obtained
  let ans = 0;
  for (let j = 0; j <= N; ++j) {
    if (dp[j] >= 0) {
      ans = j;
    }
  }
 
     
  // Print the answer
  console.log(ans);
}
 
// Test case
let arr = [4, -4, 1, -3, 1, -3];
let N = arr.length;
longestSubsequence(arr, N);

Output:

Time Complexity: O(N^2)
Auxiliary Space: O(N)

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Longest subsequence with non negative prefix sum at each position

C++

Java

Python3

C#

Javascript

C++

Java

Python

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Javascript

Introducing the Milvus Ngram Index: Faster Keyword Matching and LIKE Queries for Agent Workloads

LangChain 1.0 and Milvus: How to Build Production-Ready Agents with Real Long-Term Memory

Building AI Search Pipelines That Understand What Users Want with Milvus

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YouTube Music’s long-awaited Now Playing redesign is nearly here

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Introducing the Milvus Ngram Index: Faster Keyword Matching and LIKE Queries for Agent Workloads

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YouTube Music’s long-awaited Now Playing redesign is nearly here

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Vale Current Release Athlete Approved by

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YouTube Music’s long-awaited Now Playing redesign is nearly here

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Vale Current Release Athlete Approved by

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