Longest ZigZag Subsequence (original) (raw)

Longest Zig-Zag Subsequence

Last Updated : 23 Jul, 2025

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.

Try It Yourself

C++ `

// C++ program to find longest Zig-Zag subsequence in // an array #include <bits/stdc++.h> using namespace std;

// 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]); cout<<"Length of Longest Zig-Zag subsequence is "<<zzis(arr, n)<<endl; return 0; }

// This code is contributed by noob2000.

C

// 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; }

Java

// 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

Python3

Python3 program to find longest

Zig-Zag subsequence in an array

Function to return max of two numbers

Function to return longest

Zig-Zag subsequence length

def 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 = [[1 for i in range(2)] for i in range(n)]


res = 1 # Initialize result

# Compute values in bottom up manner '''
for i in range(1, n):
    
    # Consider all elements as previous of arr[i]
    for j in range(i):

        # If arr[i] is greater, then check with Z[j][1]
        if (arr[j] < arr[i] and 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] and 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 Code

arr = [10, 22, 9, 33, 49, 50, 31, 60] n = len(arr) print("Length of Longest Zig-Zag subsequence is", zzis(arr, n))

This code is contributed by Mohit Kumar

C#

// 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

PHP

JavaScript

`

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++ `

/CPP program to find the maximum length of zig-zag sub-sequence in given sequence/ #include <bits/stdc++.h> #include using namespace std;

// Function prototype. int signum(int n);

/* 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]);

    // It qualifies
    if (Sign != lastSign && Sign != 0) 
    {
      
        // Updating lastSign
        lastSign = Sign; 
        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;

}

Java

// Java code to find out maximum length of zig-zag // sub-sequence in given sequence import java.util.; import java.io.;

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;
    }
}

}

Python3

Python3 program to find the maximum

length of zig-zag sub-sequence in

given sequence

Function to calculate maximum length

of zig-zag sub-sequence in given sequence.

def maxZigZag(seq, n):

if (n == 0):
    return 0

lastSign = 0

# Length is initialized to 1 as that is
# minimum value for arbitrary sequence
length = 1

for i in range(1, n):
    Sign = signum(seq[i] - seq[i - 1])

    # It qualifies
    if (Sign != lastSign and Sign != 0):

        # Updating lastSign
        lastSign = Sign
        length += 1

return length

Signum function :

Returns 1 when passed a positive integer

Returns -1 when passed a negative integer

Returns 0 when passed 0.

def signum(n):

if (n != 0):
    return 1 if n > 0 else -1
else:
    return 0

Driver code

if name == 'main':

sequence1 = [1, 3, 6, 2]
sequence2 = [5, 0, 3, 1, 0]

n1 = len(sequence1)
n2 = len(sequence2)

# Function call
maxLength1 = maxZigZag(sequence1, n1)
maxLength2 = maxZigZag(sequence2, n2)

print("The maximum length of zig-zag sub-sequence "
      "in first sequence is:", maxLength1)
print("The maximum length of zig-zag sub-sequence "
      "in second sequence is:", maxLength2)

This code is contributed by himanshu77

C#

// C# code to find out maximum length of // zig-zag sub-sequence in given sequence using System;

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

    Console.WriteLine(
        "The maximum length of zig-zag sub-sequence"
        + " in first sequence is: " + maxLength1);
    Console.WriteLine(
        "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;
    }

    // length is initialized to 1 as that is minimum
    // value for arbitrary sequence.
    int lastSign = 0, length = 1;

    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;
    }
}

}

// This code is contributed by Rajput-Ji

JavaScript

`

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(1)