Maximize length of Non-Decreasing Subsequence by reversing at most one Subarray

Given a binary array arr[], the task is to find the maximum possible length of non-decreasing subsequence that can be generated by reversing a subarray at most once.

Examples:

Input: arr[] = {0, 1, 0, 1} 
Output:
Explanation: 
After reversing the subarray from index [2, 3], the array modifies to {0, 0, 1, 1}. 
Hence, the longest non-decreasing subsequence is {0, 0, 1, 1}.

Input: arr[] = {0, 1, 1, 1, 0, 0, 1, 1, 0} 
Output:
Explanation: 
After reversing the subarray from index [2, 6], the array modifies to {0, 0, 0, 1, 1, 1, 1, 1, 0}. 
Hence, the longest non-decreasing subsequence is {0, 0, 0, 1, 1, 1, 1, 1}.

Naive Approach: The simplest approach to solve the problem is to reverse each possible subarray in the given array, and find the longest non-decreasing subsequence possible from the array after reversing the subarray.



Time Complexity: O(N3) 
Auxiliary Space: O(N)

Efficient Approach: The idea is to use Dynamic Programming to solve the problem. Follow the steps below: 

  • Since the array is a binary array the idea is to find the longest subsequence among the subsequences of the forms {0….0}, {0…1…}, {0..1..0…}, 0..1..0..1.
  • Initialize a dynamic programming table as dp[][] which stores the following:

    dp[i][0] : Stores the length of the longest subsequence (0..) from a[0 to i]. 
    dp[i][1] : Stores the length of the longest subsequence (0..1..) from a[0 to i]. 
    dp[i][2] : Stores the length of the longest subsequence (0..1..0..) from a[0 to i]. 
    dp[i][3] : Stores the length of the longest subsequence (0..1..0..1..) from a[0 to i].

  • Therefore, the answer is the longest subsequence or the maximum of all the 4 given possibilities ( dp[n-1][0], d[n-1][1], dp[n-1][2], dp[n-1][3] ).

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
  
// Function to find the maximum length
// non decreasing subarray by reversing
// at most one subarray
void main_fun(int arr[], int n)
{
  
    // dp[i][j] be the longest
    // subsequence of a[0...i]
    // with first j parts
    int dp[4][n];
    memset(dp, 0, sizeof(dp[0][0] * 4 * n));
  
    if (arr[0] == 0)
        dp[0][0] = 1;
    else
        dp[1][0] = 1;
  
    // Maximum length sub-sequence
    // of (0..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 0)
            dp[0][i] = dp[0][i - 1] + 1;
        else
            dp[0][i] = dp[0][i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 1)
            dp[1][i] = max(dp[1][i - 1] + 1, 
                           dp[0][i - 1] + 1);
        else
            dp[1][i] = dp[1][i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..0..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 0)
        {
            dp[2][i] = max(dp[2][i - 1] + 1,
                           max(dp[1][i - 1] + 1, 
                               dp[0][i - 1] + 1));
        }
        else
            dp[2][i] = dp[2][i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..0..1..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 1)
        {
            dp[3][i] = max(dp[3][i - 1] + 1,
                            max(dp[2][i - 1] + 1, 
                                max(dp[1][i - 1] + 1,
                                    dp[0][i - 1] + 1)));
        }
        else
            dp[3][i] = dp[3][i - 1];
    }
  
    // Find the max length subsequence
    int ans = max(dp[2][n - 1], max(dp[1][n - 1],
              max(dp[0][n - 1], dp[3][n - 1])));
  
    // Print the answer
    cout << (ans);
}
  
// Driver Code
int main()
{
    int n = 4;
    int arr[] = {0, 1, 0, 1};
      
    main_fun(arr, n);
    return 0;
}
  
// This code is contributed by chitranayal

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Java

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// Java program to implement 
// the above approach 
import java.util.*;
  
class GFG{
  
// Function to find the maximum length
// non decreasing subarray by reversing
// at most one subarray
static void main_fun(int arr[], int n)
{
      
    // dp[i][j] be the longest
    // subsequence of a[0...i]
    // with first j parts
    int[][] dp = new int[4][n];
  
    if (arr[0] == 0)
        dp[0][0] = 1;
    else
        dp[1][0] = 1;
  
    // Maximum length sub-sequence
    // of (0..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 0)
            dp[0][i] = dp[0][i - 1] + 1;
        else
            dp[0][i] = dp[0][i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 1)
            dp[1][i] = Math.max(dp[1][i - 1] + 1
                                dp[0][i - 1] + 1);
        else
            dp[1][i] = dp[1][i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..0..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 0)
        {
            dp[2][i] = Math.max(dp[2][i - 1] + 1,
                       Math.max(dp[1][i - 1] + 1
                                dp[0][i - 1] + 1));
        }
        else
            dp[2][i] = dp[2][i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..0..1..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 1)
        {
            dp[3][i] = Math.max(dp[3][i - 1] + 1,
                       Math.max(dp[2][i - 1] + 1
                       Math.max(dp[1][i - 1] + 1,
                                dp[0][i - 1] + 1)));
        }
        else
            dp[3][i] = dp[3][i - 1];
    }
  
    // Find the max length subsequence
    int ans = Math.max(dp[2][n - 1], 
              Math.max(dp[1][n - 1],
              Math.max(dp[0][n - 1],
                       dp[3][n - 1])));
  
    // Print the answer
    System.out.print(ans);
  
// Driver code
public static void main (String[] args)
{
    int n = 4;
    int arr[] = { 0, 1, 0, 1 };
      
    main_fun(arr, n);
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program to implement 
# the above approach 
import sys 
  
# Function to find the maximum length 
# non decreasing subarray by reversing 
# at most one subarray 
def main(arr, n): 
  
    # dp[i][j] be the longest 
    # subsequence of a[0...i] 
    # with first j parts 
    dp = [[0 for x in range(n)] for y in range(4)] 
  
    if arr[0] == 0
        dp[0][0] = 1
    else
        dp[1][0] = 1
  
    # Maximum length sub-sequence 
    # of (0..) 
    for i in range(1, n): 
        if arr[i] == 0
            dp[0][i] = dp[0][i-1] + 1
        else
            dp[0][i] = dp[0][i-1
  
    # Maximum length sub-sequence 
    # of (0..1..) 
    for i in range(1, n): 
        if arr[i] == 1
            dp[1][i] = max(dp[1][i-1] + 1, dp[0][i-1] + 1
        else
            dp[1][i] = dp[1][i-1
  
    # Maximum length sub-sequence 
    # of (0..1..0..) 
    for i in range(1, n): 
        if arr[i] == 0
            dp[2][i] = max([dp[2][i-1] + 1
                            dp[1][i-1] + 1
                            dp[0][i-1] + 1]) 
        else
            dp[2][i] = dp[2][i-1
  
    # Maximum length sub-sequence 
    # of (0..1..0..1..) 
    for i in range(1, n): 
        if arr[i] == 1
            dp[3][i] = max([dp[3][i-1] + 1
                            dp[2][i-1] + 1
                            dp[1][i-1] + 1
                            dp[0][i-1] + 1]) 
        else
            dp[3][i] = dp[3][i-1
  
    # Find the max length subsequence 
    ans = max([dp[2][n-1], dp[1][n-1], 
            dp[0][n-1], dp[3][n-1]]) 
  
    # Print the answer 
    print(ans) 
  
  
# Driver Code 
if __name__ == "__main__"
    n = 4
    arr = [0, 1, 0, 1
    main(arr, n) 

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C#

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// C# program to implement 
// the above approach 
using System;
  
class GFG{
  
// Function to find the maximum length
// non decreasing subarray by reversing
// at most one subarray
static void main_fun(int []arr, int n)
{
      
    // dp[i,j] be the longest
    // subsequence of a[0...i]
    // with first j parts
    int[,] dp = new int[4, n];
  
    if (arr[0] == 0)
        dp[0, 0] = 1;
    else
        dp[1, 0] = 1;
  
    // Maximum length sub-sequence
    // of (0..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 0)
            dp[0, i] = dp[0, i - 1] + 1;
        else
            dp[0, i] = dp[0, i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 1)
            dp[1, i] = Math.Max(dp[1, i - 1] + 1, 
                                dp[0, i - 1] + 1);
        else
            dp[1, i] = dp[1, i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..0..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 0)
        {
            dp[2, i] = Math.Max(dp[2, i - 1] + 1,
                       Math.Max(dp[1, i - 1] + 1, 
                                dp[0, i - 1] + 1));
        }
        else
            dp[2, i] = dp[2, i - 1];
    }
  
    // Maximum length sub-sequence
    // of (0..1..0..1..)
    for(int i = 1; i < n; i++)
    {
        if (arr[i] == 1)
        {
            dp[3, i] = Math.Max(dp[3, i - 1] + 1,
                       Math.Max(dp[2, i - 1] + 1, 
                       Math.Max(dp[1, i - 1] + 1,
                                dp[0, i - 1] + 1)));
        }
        else
            dp[3, i] = dp[3, i - 1];
    }
  
    // Find the max length subsequence
    int ans = Math.Max(dp[2, n - 1], 
              Math.Max(dp[1, n - 1],
              Math.Max(dp[0, n - 1],
                       dp[3, n - 1])));
  
    // Print the answer
    Console.Write(ans);
  
// Driver code
public static void Main(String[] args)
{
    int n = 4;
    int []arr = { 0, 1, 0, 1 };
      
    main_fun(arr, n);
}
}
  
// This code is contributed by Amit Katiyar  

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Output: 

4

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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