Longest Zig-Zag Subsequence

The longest Zig-Zag subsequence problem is to find length of the longest subsequence of given sequence such that all elements of this are alternating.
If a sequence {x1, x2, .. xn} is alternating sequence then its element satisfy one of the following relation :

  x1 < x2 > x3 < x4 > x5 < …. xn or 
  x1 > x2 < x3 > x4 < x5 > …. xn 

Examples :

Input: arr[] = {1, 5, 4}
Output: 3
The whole arrays is of the form  x1 < x2 > x3 

Input: arr[] = {1, 4, 5}
Output: 2
All subsequences of length 2 are either of the form 
x1 < x2; or x1 > x2

Input: arr[] = {10, 22, 9, 33, 49, 50, 31, 60}
Output: 6
The subsequences {10, 22, 9, 33, 31, 60} or
{10, 22, 9, 49, 31, 60} or {10, 22, 9, 50, 31, 60}
are longest Zig-Zag of length 6.

This problem is an extension of longest increasing subsequence problem, but requires more thinking for finding optimal substructure property in this.

We will solve this problem by dynamic Programming method, Let A is given array of length n of integers. We define a 2D array Z[n][2] such that Z[i][0] contains longest Zig-Zag subsequence ending at index i and last element is greater than its previous element and Z[i][1] contains longest Zig-Zag subsequence ending at index i and last element is smaller than its previous element, then we have following recurrence relation between them,

Z[i][0] = Length of the longest Zig-Zag subsequence 
          ending at index i and last element is greater
          than its previous element
Z[i][1] = Length of the longest Zig-Zag subsequence 
          ending at index i and last element is smaller
          than its previous element

Recursive Formulation:
   Z[i][0] = max (Z[i][0], Z[j][1] + 1); 
             for all j < i and A[j] < A[i] 
   Z[i][1] = max (Z[i][1], Z[j][0] + 1); 
             for all j < i and A[j] > A[i]

The first recurrence relation is based on the fact that, If we are at position i and this element has to bigger than its previous element then for this sequence (upto i) to be bigger we will try to choose an element j ( < i) such that A[j] < A[i] i.e. A[j] can become A[i]’s previous element and Z[j][1] + 1 is bigger than Z[i][0] then we will update Z[i][0].
Remember we have chosen Z[j][1] + 1 not Z[j][0] + 1 to satisfy alternate property because in Z[j][0] last element is bigger than its previous one and A[i] is greater than A[j] which will break the alternating property if we update. So above fact derives first recurrence relation, similar argument can be made for second recurrence relation also.

C

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// C program to find longest Zig-Zag subsequence in
// an array
#include <stdio.h>
#include <stdlib.h>
  
// function to return max of two numbers
int max(int a, int b) {  return (a > b) ? a : b; }
  
// Function to return longest Zig-Zag subsequence length
int zzis(int arr[], int n)
{
    /*Z[i][0] = Length of the longest Zig-Zag subsequence
          ending at index i and last element is greater
          than its previous element
     Z[i][1] = Length of the longest Zig-Zag subsequence
          ending at index i and last element is smaller
          than its previous element   */
    int Z[n][2];
  
    /* Initialize all values from 1  */
    for (int i = 0; i < n; i++)
        Z[i][0] = Z[i][1] = 1;
  
    int res = 1; // Initialize result
  
    /* Compute values in bottom up manner */
    for (int i = 1; i < n; i++)
    {
        // Consider all elements as previous of arr[i]
        for (int j = 0; j < i; j++)
        {
            // If arr[i] is greater, then check with Z[j][1]
            if (arr[j] < arr[i] && Z[i][0] < Z[j][1] + 1)
                Z[i][0] = Z[j][1] + 1;
  
            // If arr[i] is smaller, then check with Z[j][0]
            if( arr[j] > arr[i] && Z[i][1] < Z[j][0] + 1)
                Z[i][1] = Z[j][0] + 1;
        }
  
        /* Pick maximum of both values at index i  */
        if (res < max(Z[i][0], Z[i][1]))
            res = max(Z[i][0], Z[i][1]);
    }
  
    return res;
}
  
/* Driver program */
int main()
{
    int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("Length of Longest Zig-Zag subsequence is %d\n",
            zzis(arr, n) );
    return 0;
}

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Java

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// Java program to find longest
// Zig-Zag subsequence in an array
import java.io.*;
  
class GFG {
  
// Function to return longest 
// Zig-Zag subsequence length
static int zzis(int arr[], int n)
{
    /*Z[i][0] = Length of the longest 
        Zig-Zag subsequence ending at
        index i and last element is 
        greater than its previous element
    Z[i][1] = Length of the longest 
        Zig-Zag subsequence ending at
        index i and last element is 
        smaller than its previous 
        element */
    int Z[][] = new int[n][2];
  
    /* Initialize all values from 1 */
    for (int i = 0; i < n; i++)
        Z[i][0] = Z[i][1] = 1;
  
    int res = 1; // Initialize result
  
    /* Compute values in bottom up manner */
    for (int i = 1; i < n; i++)
    {
        // Consider all elements as 
        // previous of arr[i]
        for (int j = 0; j < i; j++)
        {
            // If arr[i] is greater, then 
            // check with Z[j][1]
            if (arr[j] < arr[i] && 
                Z[i][0] < Z[j][1] + 1)
                Z[i][0] = Z[j][1] + 1;
  
            // If arr[i] is smaller, then
            // check with Z[j][0]
            if( arr[j] > arr[i] &&
              Z[i][1] < Z[j][0] + 1)
                Z[i][1] = Z[j][0] + 1;
        }
  
        /* Pick maximum of both values at
        index i */
        if (res < Math.max(Z[i][0], Z[i][1]))
            res = Math.max(Z[i][0], Z[i][1]);
    }
  
    return res;
}
  
/* Driver program */
public static void main(String[] args)
{
    int arr[] = { 10, 22, 9, 33, 49
                  50, 31, 60 };
    int n = arr.length;
    System.out.println("Length of Longest "+
                    "Zig-Zag subsequence is " +
                    zzis(arr, n));
}
}
// This code is contributed by Prerna Saini

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C#

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// C# program to find longest
// Zig-Zag subsequence in an array
using System;
  
class GFG
{
      
// Function to return longest 
// Zig-Zag subsequence length
static int zzis(int []arr, int n)
{
    /*Z[i][0] = Length of the longest 
        Zig-Zag subsequence ending at
        index i and last element is 
        greater than its previous element
    Z[i][1] = Length of the longest 
        Zig-Zag subsequence ending at
        index i and last element is 
        smaller than its previous 
        element */
    int [,]Z = new int[n, 2];
  
    /* Initialize all values from 1 */
    for (int i = 0; i < n; i++)
        Z[i, 0] = Z[i, 1] = 1;
  
    // Initialize result
    int res = 1; 
  
    /* Compute values in
    bottom up manner */
    for (int i = 1; i < n; i++)
    {
        // Consider all elements as 
        // previous of arr[i]
        for (int j = 0; j < i; j++)
        {
            // If arr[i] is greater, then 
            // check with Z[j][1]
            if (arr[j] < arr[i] && 
                Z[i, 0] < Z[j, 1] + 1)
                Z[i, 0] = Z[j, 1] + 1;
                  
            // If arr[i] is smaller, then
            // check with Z[j][0]
            if( arr[j] > arr[i] &&
            Z[i, 1] < Z[j, 0] + 1)
                Z[i, 1] = Z[j, 0] + 1;
        }
  
        /* Pick maximum of both values at
        index i */
        if (res < Math.Max(Z[i, 0], Z[i, 1]))
            res = Math.Max(Z[i, 0], Z[i, 1]);
    }
  
    return res;
}
  
// Driver Code 
static public void Main ()
{
    int []arr = {10, 22, 9, 33, 
                 49, 50, 31, 60};
    int n = arr.Length;
    Console.WriteLine("Length of Longest "+
                      "Zig-Zag subsequence is " +
                                   zzis(arr, n));
    }
}
  
// This code is contributed by ajit

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PHP

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<?php
//PHP program to find longest Zig-Zag 
//subsequence in  an array
  
// function to return max of two numbers
function  maxD($a, $b) {
    return ($a > $b) ? $a : $b;
    }
  
// Function to return longest Zig-Zag subsequence length
function  zzis($arr, $n)
{
    /*Z[i][0] = Length of the longest Zig-Zag subsequence
        ending at index i and last element is greater
        than its previous element
    Z[i][1] = Length of the longest Zig-Zag subsequence
        ending at index i and last element is smaller
        than its previous element */
     //$Z[$n][2];
  
    /* Initialize all values from 1 */
    for ($i = 0; $i < $n; $i++)
        $Z[$i][0] = $Z[$i][1] = 1;
  
     $res = 1; // Initialize result
  
    /* Compute values in bottom up manner */
    for ($i = 1; $i < $n; $i++)
    {
        // Consider all elements as previous of arr[i]
        for ($j = 0; $j < $i; $j++)
        {
            // If arr[i] is greater, then check with Z[j][1]
            if ($arr[$j] < $arr[$i] && $Z[$i][0] < $Z[$j][1] + 1)
                $Z[$i][0] = $Z[$j][1] + 1;
  
            // If arr[i] is smaller, then check with Z[j][0]
            if( $arr[$j] > $arr[$i] && $Z[$i][1] < $Z[$j][0] + 1)
                $Z[$i][1] = $Z[$j][0] + 1;
        }
  
        /* Pick maximum of both values at index i */
        if ($res < max($Z[$i][0], $Z[$i][1]))
            $res = max($Z[$i][0], $Z[$i][1]);
    }
  
    return $res;
}
  
/* Driver program */
  
     $arr = array( 10, 22, 9, 33, 49, 50, 31, 60 );
    $n = sizeof($arr);
    echo "Length of Longest Zig-Zag subsequence is ",
            zzis($arr, $n) ;
    echo "\n";
  
  
  
#This code is contributed by aj_36
?>

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Output :

Length of Longest Zig-Zag subsequence is 6

Time Complexity : O(n2)
Auxiliary Space : O(n)

A better approach with time complexity O(n) is explained below:
Let the sequence be stored in an unsorted integer array arr[N].
We shall proceed by comparing the mathematical signs(negative or positive) of the difference of two consecutive elements of arr. To achieve this, we shall store the sign of (arr[i] – arr[i-1]) in a variable, subsequently comparing it with that of (arr[i+1] – arr[i]). If it is different, we shall increment our result. For checking the sign, we shall use a simple Signum Function, which shall determine the sign of a number passed to it. That is,
 signum(n) =       \begin{cases}        1 &\quad\text{if }n > 0\\        -1 &\quad\text{if }n < 0\\        0 &\quad\text{if }n = 0\\       \end{cases}

Considering the fact that we traverse the sequence only once, this becomes an O(n) solution.

The algorithm for the approach discussed above is :

Input integer array seq[N].
Initialize integer lastSign to 0. 
FOR i in range 1 to N - 1
    integer sign = signum(seq[i] - seq[i-1])
    IF sign != lastSign AND IF sign != 0
        increment length by 1. lastSign = sign.
    END IF
END FOR
return length.

Following is the implementation of the above approach:

C++

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#include <iostream>
#include<bits/stdc++.h>
using namespace std;
/*CPP program to find the maximum length of zig-zag sub-sequence
in given sequence*/
int signum(int n); //function prototype.
  
/* Function to calculate maximum length of zig-zag
sub-sequence in given sequence.
*/
int maxZigZag(int seq[], int n)
{
    if (n == 0)
    {
        return 0;
    }
      
    int lastSign = 0, length = 1;
                    // length is initialized to 1 as that is minimum value
                    // for arbitrary sequence.
    for (int i = 1; i < n; ++i)
    {
        int Sign = signum(seq[i] - seq[i-1]);
          
        if (Sign != lastSign && Sign != 0) // it qualifies
        {
            lastSign = Sign;   //updating lastSign
            length++;
        }
    }
return length;    
}
  
/* Signum function :
Returns 1 when passed a positive integer
Returns -1 when passed a negative integer
Returns 0 when passed 0. */
int signum(int n)
{
    if (n != 0)
    {
        return n > 0 ? 1 : -1;
    }
          
    else
    {
        return 0;
    
}
//Driver method
int main()
{
    int sequence1[4] = {1, 3 ,6, 2};
    int sequence2[5] = {5, 0, 3, 1, 0};
          
    int n1 = sizeof(sequence1) / sizeof(*sequence1);  //size of sequences
    int n2 = sizeof(sequence2) / sizeof(*sequence2);
          
    int maxLength1 = maxZigZag(sequence1,n1);
    int maxLength2 = maxZigZag(sequence2,n2); //function call
          
     cout << "The maximum length of zig-zag sub-sequence in first sequence is: " << maxLength1;
     cout << endl;
    cout << "The maximum length of zig-zag sub-sequence in second sequence is: " << maxLength2;    
}

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Java

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// Java code to find out maximum length of zig-zag sub-sequence
// in given sequence
class zigZagMaxLength
{
    //Driver method
    public static void main(String[] args)
    {
        int[] sequence1 = {1, 3 ,6, 2};
        int[] sequence2 = {5, 0, 3, 1, 0};
          
        int n1 = sequence1.length;  //size of sequences
        int n2 = sequence2.length;
          
        int maxLength1 = maxZigZag(sequence1,n1);
        int maxLength2 = maxZigZag(sequence2,n2); //function call
          
        System.out.println("The maximum length of zig-zag sub-sequence in first sequence is: " + maxLength1);
       System.out.println("The maximum length of zig-zag sub-sequence in second sequence is: " + maxLength2);
    }
      
    /* Function to calculate maximum length of zig-zag
    sub-sequence in given sequence.
    */
      
    static int maxZigZag(int[] seq, int n)
    {
        if (n == 0
        {
            return 0;
        
          
        int lastSign = 0 , length = 1
                    // length is initialized to 1 as that is minimum value
                    // for arbitrary sequence.                                               
                                                                      
          
        for (int i = 1; i < n; ++i)
        {        
            int Sign = signum(seq[i] - seq[i-1]);
              
            if (Sign != 0 && Sign!= lastSign) //it qualifies
            {
                lastSign = Sign;         //updating lastSign
                length++;                    
            }        
        }
          
        return length;
    }
      
      
    /* Signum function :
    Returns 1 when passed a positive integer
    Returns -1 when passed a negative integer
    Returns 0 when passed 0. */
    static int signum(int n)
    {
        if (n != 0)
        {
            return n > 0 ? 1 : -1;
        }
          
        else
        {
            return 0;
        
    }
      
}

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Output :

The maximum length of zig-zag sub-sequence in first sequence is: 3
The maximum length of zig-zag sub-sequence in second sequence is: 4

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

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : jit_t, santhanambr2002