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

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  • Difficulty Level : Hard
  • Last Updated : 01 Jun, 2021
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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++




// 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

Java




// 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

Python3




# 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)

C#




// 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 

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find the maximum length
// non decreasing subarray by reversing
// at most one subarray
function main_fun(arr, n)
{
 
    // dp[i][j] be the longest
    // subsequence of a[0...i]
    // with first j parts
    var dp = Array.from(Array(4), ()=>Array(n).fill(0));
 
    if (arr[0] == 0)
        dp[0][0] = 1;
    else
        dp[1][0] = 1;
 
    // Maximum length sub-sequence
    // of (0..)
    for(var 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(var 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(var 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(var 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
    var 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
    document.write(ans);
}
 
// Driver Code
var n = 4;
var arr = [0, 1, 0, 1];
 
main_fun(arr, n);
 
</script>

Output: 

4

 

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


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