Maximum Suffix length among all distinct elements

Given the array X[],   the task is to return the maximum suffix sub-array length of the same elements for each distinct element and the maximum suffix length among all by using the below operation:

• In each operation, you can select a sub-array, remove it, and concatenate the remaining part of X[].

Examples:

Input: N = 11, X[] = {2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2}
Output:

• Element = 2, Maximum suffix length = 3
• Element = 3, Maximum suffix length = 2
• Element = 5, Maximum suffix length = 3
• Maximum length among all = 3

Explanation: There are three distinct elements present in X[], Which are 2, 5 and 3. We will take a look on each of them:

• Element 2 = We can remove the sub-array = {X[ 3 ], X[ 4 ], . . . . ., X[ 10 ]} So the remaining X[] will be = {2, 2, 2}. Which gives the suffix length of element 2 in updated X[] as = 3.    
• Element 5 = We can remove the sub-array = {X[ 6 ], X[ 7 ], . . . . ., X[ 11 ]} So the remaining X[] will be = {2, 2, 5, 5, 5}. Which gives the suffix length of element 5 in updated X[] as = 3.
• Element 3 = We can remove the sub-array = { X[ 11 ], . . . X[ 11 ] } So the remaining X[] will be = {2, 2, 5, 5, 5, 3, 5, 5, 3, 3}. Which gives the suffix length of element 3 in updated X[] as = 2.   

So, the maximum suffix length for element 2, 5 and 3 using given operation is 3, 3, and 2 respectively. Maximum suffix length among all is 3. It can be verified that all the lengths are optimally longest possible suffix lengths. 

Input: N = 5, X[] = { 8, 3, 3, 3, 8 }
Output:

• Element = 8, Maximum suffix length = 2
• Element = 3, Maximum suffix length = 3
• Maximum length among all = 3

Explanation: For element 8, We can remove sub-array {3, 3, 3} so the remaining X[] will be {8, 8} Which has suffix length for element 8 as 2. While in case of 3 we can remove X[5], so the remaining X[] will be = {8, 3, 3, 3} and give suffix length for element 3 as 3. Maximum suffix length among all is 3.

Approach: Implement the idea below to solve the problem

The problem is based on observation and can be solved by using those observation. For understanding the concept see the Concept of approach section below. 

Concept of approach:

First reverse the array X[]. So that now the problem has changed, We have to find the maximum length prefix of same elements. This reversal on X[] is done because for solving the problem we have to traverse the array X[] number of times. So it will be convenient to traverse from left to right than right to left.

There can be two cases:

• If an element is present at first index of X[]: In this case the element at first index can be join with other consecutive frequencies for maximizing prefix length( reverse of suffix, Because X[] has reversed ). Let understand it with the example. X[] = {2, 2, 3, 3, 3, 2, 2, 2, 5, 5, 2, 2, 2, 2}. In the X[] there are three consecutive frequencies of element = 2 in X[], Which are 2, 3 and 4 ( mark bold in X[] ). Now we can remove sub-array { X[3] . . . X[5] } or { X[3] . . . . X[10] } for making X[] as: {2, 2, 2, 2, 2, 5, 5, 2, 2, 2, 2} or {2, 2, 2, 2, 2, 2}. Which gives maximum prefix length as 5 or 6. 6 is greater than 5, So 6 is the max possible prefix length for element 2. It should be noted that the element 2 is present at first index of X[] after reversing. Approach for such cases is as: 
  • Create a vector and initialize it with adjacent frequencies. For X[] = {2, 2, 3, 3, 3, 2, 2, 2, 5, 5, 2, 2, 2, 2}. Vector will contain values as: {2, 3, 4} mark bold in X[].
  • Then make the pair of first frequency(2 in list) with rest of the frequency (3 and 4 in list), With which sum will be max. Formally, max = max ( Vector[ 0 ] + Vector[ j ] ) for (2 <= j <= Vector.size()). The pair of first frequency with other frequency will be as follows: 
    • 2 + 3 = 5
    • 2 + 4 = 6
  • max(5, 6) =  6. Which is our max possible length for element 2 in X[].
• If an element is not present at first index: In such cases just output the maximum consecutive frequency from Vector. For example X[] = {2, 2, 5, 5, 5, 2,  5, 5, 5, 5, 3, 3, 3}. We can remove sub-array { X[1] . . .X[2] } or {X[1] . . . .X[6]} for making X[] as: {5, 5, 5, 2, 5, 5, 5, 5, 3, 3, 3} or {5, 5, 5, 5, 3, 3, 3} giving maximum prefix length for 5 as: 3 and 4. In which 4 is max possible prefix length for element 5. It should be noted that 5 was not initially present at first index of X[]. For such cases approach is as:
  • Just output the maximum frequency for such cases in frequency vector.
  • Create frequency vector. For element 5 in X[] = {2, 2, 5, 5, 5, 2, 5, 5, 5, 5, 3, 3, 3}. Frequency vector will be = {3, 4}, In which 4 is maximum.  

Steps were taken to solve the problem:

• At first, Reverse the X[].
• Create a Set and initialize it with distinct elements of X[].
• Create a variable let’s say max_all to hold the value of the longest suffix length among all distinct integers.
• Create an ArrayList<Integer> let’s say Adjacent_Frequency to hold the count of occurrence of an element consecutively. 
• Run a loop for the traversing set and follow the below-mentioned steps under the scope of the loop:
  • Traverse X[] and update consecutive frequencies in Adjacent_Frequency.
  • Traverse the Adjacent frequency list and get the max frequency in the variable let’s say max.
  • If the first element of X[] is the element for which we are finding the maximum suffix length, Then update max as max( Adjacent_frequency[ 0 ] + Adjacent_frequency[ j ] ), for (2 ? j ? N), else max = Adjacent_frequency[ 0 ].
  • Print the current element of the set and its max suffix length.
  • if ( max_all < max ), Then update max_all as max.
• Print the value of max_all.

Below is the Code to implement the approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// function for reversing array
vector<int> Reverse_Array(vector<int> X)
{
    int start = 0;
    int end = X.size() - 1;
    while (start < end) {
        int temp = X[start];
        X[start] = X[end];
        X[end] = temp;
        start++;
        end--;
    }
    return X;
}

// Method for getting distinct
// elements from X[]
vector<int> Set(int n, vector<int> X)
{
    vector<int> set;
    for (int i = 0; i < n; i++) {
        if (find(set.begin(), set.end(), X[i])
            != set.end()) {
            continue;
        }
        else {
            set.push_back(X[i]);
        }
    }
    return set;
}

// function for implementing approach
void approach(vector<int> set, int n, vector<int> X)
{

    // Variable initialized to hold
    // maximum suffix length among
    // distinct element
    int max_all = INT_MIN;

    // Loop for traversing over set
    // containing distinct elements
    for (int i = 0; i < set.size(); i++) {

        // Variable to hold element
        // from set
        int element = set[i];

        // Array initialized to
        // count the frequency
        // of same adjacent element
        vector<int> adjacent_frequency;

        int leftEnd = 0;

        // Loop for initializing
        // frequency in
        // adjacent_frequency list
        while (leftEnd < n) {
            if (X[leftEnd] != element) {
                leftEnd++;
            }
            else {
                int rightEnd = leftEnd;
                while (rightEnd < n
                       && X[rightEnd] == element)
                    rightEnd = rightEnd + 1;

                adjacent_frequency.push_back(rightEnd
                                             - leftEnd);
                leftEnd = rightEnd + 1;
            }
        }

        // Variable initialized to hold
        // maximum suffix length
        // for distinct element
        int max = INT_MIN;

        // Loop for traversing over
        // frequency Array
        for (int k = 0; k < adjacent_frequency.size();
             k++) {

            // Updating max with max
            // adjacent frequency
            max = adjacent_frequency[k] > max
                      ? adjacent_frequency[k]
                      : max;
        }
        bool flag = false;

        // Condition when an element is
        // not present at first
        // index of X[]
        if (X[0] != element) {
            cout << "Element : " << element
                 << " Max suffix length : " << max << endl;
            flag = true;
            max_all = max > max_all ? max : max_all;
        }

        // This will execute when an
        // element for which we are
        // finding max  suffix length,
        // is not present at first
        // index of X[]
        else if (flag != true) {
            for (int j = 1; j < adjacent_frequency.size();
                 j++) {
                max = (adjacent_frequency[0]
                       + adjacent_frequency[j])
                              > max
                          ? (adjacent_frequency[0]
                             + adjacent_frequency[j])
                          : max;
                max_all = max > max_all ? max : max_all;
            }

            // Printing the max length
            // for an element
            cout << "element : " << element
                 << " Max suffix length : " << max << endl;
        }
    }

    // Printing maximum length among
    // all distinct element
    cout << "Maximum length among all : " << max_all
         << endl;
}

// Driver Function
int main()
{

    // Input Array
    vector<int> X = { 2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2 };

    // Reversing and updating X[]
    X = Reverse_Array(X);

    // Variable to hold length of X[]
    int N = X.size();

    // List for storing distinct
    // elements present in X[]
    vector<int> set = ::Set(N, X);

    // Function call for applying
    // approach on X[]
    approach(set, N, X);

    return 0;
}

// This Code is Contributed by Prasad Kandekar(prasad264)

Java

// Java code to implement the approach

import java.util.*;
public class GFG {
    // Driver Function
    public static void main(String[] args)
    {

        // Input Array
        int[] X = { 2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2 };

        // Reversing and updating X[]
        X = Reverse_Array(X);

        // Variable to hold length of X[]
        int N = X.length;

        // List for storing distinct
        // elements present in X[]
        ArrayList<Integer> set = Set(N, X);

        // Function call for applying
        // approach on X[]
        Approach(set, N, X);
    }

    // Method for implementing approach
    static void Approach(ArrayList<Integer> set, int n,
                         int[] X)
    {

        // Variable initialized to hold
        // maximum suffix length among
        // distinct element
        int max_all = Integer.MIN_VALUE;

        // Loop for traversing over set
        // containing distinct elements
        for (int i = 0; i < set.size(); i++) {

            // Variable to hold element
            // from set
            int element = set.get(i);

            // Arraylist initialized to
            // count the frequency
            // of same adjacent element
            ArrayList<Integer> adjacent_frequency
                = new ArrayList<>();

            int leftEnd = 0;

            // Loop for initializing
            // frequency in
            // adjacent_frequency list
            while (leftEnd < n) {
                if (X[leftEnd] != element) {
                    leftEnd++;
                }
                else {
                    int rightEnd = leftEnd;
                    while (rightEnd < n
                           && X[rightEnd] == element)
                        rightEnd = rightEnd + 1;

                    adjacent_frequency.add(rightEnd
                                           - leftEnd);
                    leftEnd = rightEnd + 1;
                }
            }

            // Variable initialized to hold
            // maximum suffix length
            // for distinct element
            int max = Integer.MIN_VALUE;

            // Loop for traversing over
            // frequency ArrayList
            for (int k = 0; k < adjacent_frequency.size();
                 k++) {

                // Updating max with max
                // adjacent frequency
                max = adjacent_frequency.get(k) > max
                          ? adjacent_frequency.get(k)
                          : max;
            }

            boolean flag = false;

            // Condition when an element is
            // not present at first
            // index of X[]
            if (X[0] != element) {
                System.out.println(
                    "Element : " + element
                    + " Max suffix length :  " + max);
                flag = true;
                max_all = max > max_all ? max : max_all;
            }

            // This will execute when an
            // element for which we are
            // finding max  suffix length,
            // is not present at first
            // index of X[]
            else if (flag != true) {

                for (int j = 1;
                     j < adjacent_frequency.size(); j++) {
                    max = (adjacent_frequency.get(0)
                           + adjacent_frequency.get(j))
                                  > max
                              ? (adjacent_frequency.get(0)
                                 + adjacent_frequency.get(
                                       j))
                              : max;
                    max_all = max > max_all ? max : max_all;
                }

                // Printing the max length
                // for an element
                System.out.println(
                    "element : " + element
                    + " Max suffix length :  " + max);
            }
        }

        // Printing maximum length among
        // all distinct element
        System.out.println("Maximum length among all : "
                           + max_all);
    }

    // Method for reversing array
    static int[] Reverse_Array(int[] X)
    {
        int start = 0;
        int end = X.length - 1;
        while (start < end) {
            int temp = X[start];
            X[start] = X[end];
            X[end] = temp;

            start++;
            end--;
        }
        return X;
    }

    // Method for getting distinct
    // elements from X[]
    static ArrayList<Integer> Set(int n, int[] X)
    {
        ArrayList<Integer> set = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            if (set.contains(X[i])) {
                continue;
            }
            else {
                set.add(X[i]);
            }
        }
        return (set);
    }
}

Python3

# Python3 code to implement the approach

# function for reversing array
def reverse_array(arr):
    """
    Function to reverse an array
    """
    start = 0
    end = len(arr) - 1
    while start < end:
        arr[start], arr[end] = arr[end], arr[start]
        start += 1
        end -= 1
    return arr

# Method for getting distinct
# elements from X[]
def get_distinct_elements(arr):
    """
    Method to get distinct elements from an array
    """
    distinct_set = []
    for i in range(len(arr)):
        if arr[i] not in distinct_set:
            distinct_set.append(arr[i])
    return distinct_set

# function for implementing approach
def approach(distinct_set, arr):
    """
    Function to implement the approach
    """
    max_all = float('-inf') # Variable to hold maximum suffix length among distinct elements

    # Loop for traversing over set
    # containing distinct elements
    for element in distinct_set:
        adjacent_frequency = [] # List initialized to count the frequency of same adjacent element

        left_end = 0
        # Loop for initializing
        # frequency in
        # adjacent_frequency list
        while left_end < len(arr):
            if arr[left_end] != element:
                left_end += 1
            else:
                right_end = left_end
                while right_end < len(arr) and arr[right_end] == element:
                    right_end += 1

                adjacent_frequency.append(right_end - left_end)
                left_end = right_end + 1

        maxm = float('-inf') # Variable to hold maximum suffix length for distinct element
        for k in range(len(adjacent_frequency)):
            maxm = max(adjacent_frequency[k], maxm) # Updating max with max adjacent frequency

        flag = False
        if arr[0] != element:
            print(f"Element: {element} Max suffix length: {maxm}")
            flag = True
            max_all = maxm if maxm > max_all else max_all
        elif not flag:
            for j in range(1, len(adjacent_frequency)):
                maxm = max(adjacent_frequency[0] + adjacent_frequency[j], maxm)
                max_all = maxm if maxm > max_all else max_all

            print(f"Element: {element} Max suffix length: {maxm}")

    print(f"Maximum length among all: {max_all}")

# Driver code
X = [2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2]
X = reverse_array(X)
distinct_set = get_distinct_elements(X)
approach(distinct_set, X)

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
using System.Linq;

public class GFG {

    // function for reversing array
    static List<int> Reverse_Array(List<int> X)
    {
        int start = 0;
        int end = X.Count - 1;
        while (start < end) {
            int temp = X[start];
            X[start] = X[end];
            X[end] = temp;
            start++;
            end--;
        }
        return X;
    }

    // Method for getting distinct
    // elements from X[]
    static List<int> Set(int n, List<int> X)
    {
        List<int> set = new List<int>();
        for (int i = 0; i < n; i++) {
            if (set.Contains(X[i])) {
                continue;
            }
            else {
                set.Add(X[i]);
            }
        }
        return set;
    }

    // function for implementing approach
    static void approach(List<int> set, int n, List<int> X)
    {

        // Variable initialized to hold
        // maximum suffix length among
        // distinct element
        int max_all = int.MinValue;

        // Loop for traversing over set
        // containing distinct elements
        for (int i = 0; i < set.Count; i++) {

            // Variable to hold element
            // from set
            int element = set[i];

            // Array initialized to
            // count the frequency
            // of same adjacent element
            List<int> adjacent_frequency = new List<int>();

            int leftEnd = 0;

            // Loop for initializing
            // frequency in
            // adjacent_frequency list
            while (leftEnd < n) {
                if (X[leftEnd] != element) {
                    leftEnd++;
                }
                else {
                    int rightEnd = leftEnd;
                    while (rightEnd < n
                           && X[rightEnd] == element) {
                        rightEnd = rightEnd + 1;
                    }

                    adjacent_frequency.Add(rightEnd
                                           - leftEnd);
                    leftEnd = rightEnd + 1;
                }
            }

            // Variable initialized to hold
            // maximum suffix length
            // for distinct element
            int max = int.MinValue;

            // Loop for traversing over
            // frequency Array
            for (int k = 0; k < adjacent_frequency.Count;
                 k++) {

                // Updating max with max
                // adjacent frequency
                max = adjacent_frequency[k] > max
                          ? adjacent_frequency[k]
                          : max;
            }
            bool flag = false;

            // Condition when an element is
            // not present at first
            // index of X[]
            if (X[0] != element) {
                Console.WriteLine("Element : " + element
                                  + " Max suffix length : "
                                  + max);
                flag = true;
                max_all = max > max_all ? max : max_all;
            }

            // This will execute when an
            // element for which we are
            // finding max  suffix length,
            // is not present at first
            // index of X[]
            else if (flag != true) {
                for (int j = 1;
                     j < adjacent_frequency.Count; j++) {
                    max = (adjacent_frequency[0]
                           + adjacent_frequency[j])
                                  > max
                              ? (adjacent_frequency[0]
                                 + adjacent_frequency[j])
                              : max;
                    max_all = max > max_all ? max : max_all;
                }

                // Printing the max length
                // for an element
                Console.WriteLine("element : " + element
                                  + " Max suffix length : "
                                  + max);
            }
        }

        // Printing maximum length among
        // all distinct element
        Console.WriteLine("Maximum length among all : "
                          + max_all);
    }

    // Driver Function
    static public void Main()
    {

        // Input Array
        List<int> X = new List<int>{ 2, 2, 5, 5, 5, 3,
                                     5, 5, 3, 3, 2 };

        // Reversing and updating X[]
        X = Reverse_Array(X);

        // Variable to hold length of X[]
        int N = X.Count;

        // List for storing distinct
        // elements present in X[]
        List<int> set = Set(N, X);

        // Function call for applying
        // approach on X[]
        approach(set, N, X);
    }
}
// This code is contributed by Prasad Kandekar(prasad264)

Javascript

// JavaScript code to implement the approach

// function for reversing array
function reverseArray(X){
    let start = 0;
    let end = X.length - 1;
    while (start < end) {
        let temp = X[start];
        X[start] = X[end];
        X[end] = temp;
        start++;
        end--;
    }
    return X;
}

// Method for getting distinct
// elements from X[]
function getDistinctElements(X) { return [... new Set(X)]; }

// function for implementing approach
function approach(set, X){
    let n = X.length;
    // Variable initialized to hold
    // maximum suffix length among
    // distinct element
    let max_all = Number.MIN_SAFE_INTEGER;

    // Loop for traversing over set
    // containing distinct elements
    for (let i = 0; i < set.length; i++) {
        // Variable to hold element
        // from set
        let element = set[i];
        // Array initialized to
        // count the frequency
        // of same adjacent element
        let adjacentFrequency = [];
        let leftEnd = 0;

        // Loop for initializing
        // frequency in
        // adjacent_frequency list
        while (leftEnd < n) {
            if (X[leftEnd] !== element) {
                leftEnd++;
            }
            else {
                let rightEnd = leftEnd;
                while (rightEnd < n
                           && X[rightEnd] === element) {
                    rightEnd++;
                }
                adjacentFrequency.push(rightEnd - leftEnd);
                leftEnd = rightEnd + 1;
            }
        }

        // Variable initialized to hold
        // maximum suffix length
        // for distinct element
        let max = Number.MIN_SAFE_INTEGER;

        // Loop for traversing over
        // frequency Array
        for (let k = 0; k < adjacentFrequency.length; k++) {
            // Updating max with max
            // adjacent frequency
            max = adjacentFrequency[k] > max
                      ? adjacentFrequency[k]
                      : max;
        }
        let flag = false;

        // Condition when an element is
        // not present at first
        // index of X[]
        if (X[0] !== element) {
            console.log("Element: " + element
                        + " Max suffix length: " + max);
            flag = true;
            max_all = max > max_all ? max : max_all;
        }
        // This will execute when an
        // element for which we are
        // finding max  suffix length,
        // is not present at first
        // index of X[]
        else if (flag !== true) {
            for (let j = 1; j < adjacentFrequency.length;
                 j++) {
                max = adjacentFrequency[0]
                                  + adjacentFrequency[j]
                              > max
                          ? adjacentFrequency[0]
                                + adjacentFrequency[j]
                          : max;
                max_all = max > max_all ? max : max_all;
            }
            // Printing the max length
            // for an element
            console.log("Element: " + element
                        + " Max suffix length: " + max);
        }
    }

    // Printing maximum length among
    // all distinct element
    console.log("Maximum length among all: " + max_all);
}
// Driver Code

// Input Array
let X = [ 2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2 ];

// Reversing and updating X[]
X = reverseArray(X);

// List for storing distinct
// elements present in X[]
let set = getDistinctElements(X);

// Function call for applying
// approach on X[]
approach(set, X);

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// function for reversing array
vector<int> Reverse_Array(vector<int> X)
{
    int start = 0;
    int end = X.size() - 1;
    while (start < end) {
        int temp = X[start];
        X[start] = X[end];
        X[end] = temp;
        start++;
        end--;
    }
    return X;
}

// Method for getting distinct
// elements from X[]
vector<int> Set(int n, vector<int> X)
{
    vector<int> set;
    for (int i = 0; i < n; i++) {
        if (find(set.begin(), set.end(), X[i])
            != set.end()) {
            continue;
        }
        else {
            set.push_back(X[i]);
        }
    }
    return set;
}

// function for implementing approach
void approach(vector<int> set, int n, vector<int> X)
{

    // Variable initialized to hold
    // maximum suffix length among
    // distinct element
    int max_all = INT_MIN;

    // Loop for traversing over set
    // containing distinct elements
    for (int i = 0; i < set.size(); i++) {

        // Variable to hold element
        // from set
        int element = set[i];

        // Array initialized to
        // count the frequency
        // of same adjacent element
        vector<int> adjacent_frequency;

        int leftEnd = 0;

        // Loop for initializing
        // frequency in
        // adjacent_frequency list
        while (leftEnd < n) {
            if (X[leftEnd] != element) {
                leftEnd++;
            }
            else {
                int rightEnd = leftEnd;
                while (rightEnd < n
                       && X[rightEnd] == element)
                    rightEnd = rightEnd + 1;

                adjacent_frequency.push_back(rightEnd
                                             - leftEnd);
                leftEnd = rightEnd + 1;
            }
        }

        // Variable initialized to hold
        // maximum suffix length
        // for distinct element
        int max = INT_MIN;

        // Loop for traversing over
        // frequency Array
        for (int k = 0; k < adjacent_frequency.size();
             k++) {

            // Updating max with max
            // adjacent frequency
            max = adjacent_frequency[k] > max
                      ? adjacent_frequency[k]
                      : max;
        }
        bool flag = false;

        // Condition when an element is
        // not present at first
        // index of X[]
        if (X[0] != element) {
            cout << "Element : " << element
                 << " Max suffix length : " << max << endl;
            flag = true;
            max_all = max > max_all ? max : max_all;
        }

        // This will execute when an
        // element for which we are
        // finding max  suffix length,
        // is not present at first
        // index of X[]
        else if (flag != true) {
            for (int j = 1; j < adjacent_frequency.size();
                 j++) {
                max = (adjacent_frequency[0]
                       + adjacent_frequency[j])
                              > max
                          ? (adjacent_frequency[0]
                             + adjacent_frequency[j])
                          : max;
                max_all = max > max_all ? max : max_all;
            }

            // Printing the max length
            // for an element
            cout << "element : " << element
                 << " Max suffix length : " << max << endl;
        }
    }

    // Printing maximum length among
    // all distinct element
    cout << "Maximum length among all : " << max_all
         << endl;
}

// Driver Function
int main()
{

    // Input Array
    vector<int> X = { 2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2 };

    // Reversing and updating X[]
    X = Reverse_Array(X);

    // Variable to hold length of X[]
    int N = X.size();

    // List for storing distinct
    // elements present in X[]
    vector<int> set = ::Set(N, X);

    // Function call for applying
    // approach on X[]
    approach(set, N, X);

    return 0;
}

// This Code is Contributed by Prasad Kandekar(prasad264)

Java

// Java code to implement the approach

import java.util.*;
public class GFG {
    // Driver Function
    public static void main(String[] args)
    {

        // Input Array
        int[] X = { 2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2 };

        // Reversing and updating X[]
        X = Reverse_Array(X);

        // Variable to hold length of X[]
        int N = X.length;

        // List for storing distinct
        // elements present in X[]
        ArrayList<Integer> set = Set(N, X);

        // Function call for applying
        // approach on X[]
        Approach(set, N, X);
    }

    // Method for implementing approach
    static void Approach(ArrayList<Integer> set, int n,
                         int[] X)
    {

        // Variable initialized to hold
        // maximum suffix length among
        // distinct element
        int max_all = Integer.MIN_VALUE;

        // Loop for traversing over set
        // containing distinct elements
        for (int i = 0; i < set.size(); i++) {

            // Variable to hold element
            // from set
            int element = set.get(i);

            // Arraylist initialized to
            // count the frequency
            // of same adjacent element
            ArrayList<Integer> adjacent_frequency
                = new ArrayList<>();

            int leftEnd = 0;

            // Loop for initializing
            // frequency in
            // adjacent_frequency list
            while (leftEnd < n) {
                if (X[leftEnd] != element) {
                    leftEnd++;
                }
                else {
                    int rightEnd = leftEnd;
                    while (rightEnd < n
                           && X[rightEnd] == element)
                        rightEnd = rightEnd + 1;

                    adjacent_frequency.add(rightEnd
                                           - leftEnd);
                    leftEnd = rightEnd + 1;
                }
            }

            // Variable initialized to hold
            // maximum suffix length
            // for distinct element
            int max = Integer.MIN_VALUE;

            // Loop for traversing over
            // frequency ArrayList
            for (int k = 0; k < adjacent_frequency.size();
                 k++) {

                // Updating max with max
                // adjacent frequency
                max = adjacent_frequency.get(k) > max
                          ? adjacent_frequency.get(k)
                          : max;
            }

            boolean flag = false;

            // Condition when an element is
            // not present at first
            // index of X[]
            if (X[0] != element) {
                System.out.println(
                    "Element : " + element
                    + " Max suffix length :  " + max);
                flag = true;
                max_all = max > max_all ? max : max_all;
            }

            // This will execute when an
            // element for which we are
            // finding max  suffix length,
            // is not present at first
            // index of X[]
            else if (flag != true) {

                for (int j = 1;
                     j < adjacent_frequency.size(); j++) {
                    max = (adjacent_frequency.get(0)
                           + adjacent_frequency.get(j))
                                  > max
                              ? (adjacent_frequency.get(0)
                                 + adjacent_frequency.get(
                                       j))
                              : max;
                    max_all = max > max_all ? max : max_all;
                }

                // Printing the max length
                // for an element
                System.out.println(
                    "element : " + element
                    + " Max suffix length :  " + max);
            }
        }

        // Printing maximum length among
        // all distinct element
        System.out.println("Maximum length among all : "
                           + max_all);
    }

    // Method for reversing array
    static int[] Reverse_Array(int[] X)
    {
        int start = 0;
        int end = X.length - 1;
        while (start < end) {
            int temp = X[start];
            X[start] = X[end];
            X[end] = temp;

            start++;
            end--;
        }
        return X;
    }

    // Method for getting distinct
    // elements from X[]
    static ArrayList<Integer> Set(int n, int[] X)
    {
        ArrayList<Integer> set = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            if (set.contains(X[i])) {
                continue;
            }
            else {
                set.add(X[i]);
            }
        }
        return (set);
    }
}

Python3

# Python3 code to implement the approach
import sys

# function for reversing array
def reverse_array(X):
    start = 0
    end = len(X) - 1
    while start < end:
        temp = X[start]
        X[start] = X[end]
        X[end] = temp
        start += 1
        end -= 1
    return X

# Method for getting distinct
# elements from X[]
def set_fn(n, X):
    s = []
    for i in range(n):
        if X[i] in s:
            continue
        else:
            s.append(X[i])
    return s

# function for implementing approach
def approach(s, n, X):

    # Variable initialized to hold
    # maximum suffix length among
    # distinct element
    max_all = -sys.maxsize - 1

    # Loop for traversing over set
    # containing distinct elements
    for element in s:

        # Array initialized to
        # count the frequency
        # of same adjacent element
        adjacent_frequency = []

        leftEnd = 0

        # Loop for initializing
        # frequency in
        # adjacent_frequency list
        while leftEnd < n:
            if X[leftEnd] != element:
                leftEnd += 1
            else:
                rightEnd = leftEnd
                while rightEnd < n and X[rightEnd] == element:
                    rightEnd += 1

                adjacent_frequency.append(rightEnd - leftEnd)
                leftEnd = rightEnd + 1

        # maximum suffix length
        # for distinct element
        max_val = -sys.maxsize - 1

        # Loop for traversing over
        # frequency Array
        for k in range(len(adjacent_frequency)):

            # Updating max with max
            # adjacent frequency
            max_val = max(adjacent_frequency[k], max_val)

        flag = False

        # Condition when an element is
        # not present at first
        # index of X[]
        if X[0] != element:
            print("Element :", element, "Max suffix length :", max_val)
            flag = True
            max_all = max(max_val, max_all)

        # This will execute when an
        # element for which we are
        # finding max  suffix length,
        # is not present at first
        # index of X[]
        elif not flag:
            for j in range(1, len(adjacent_frequency)):
                max_val = max(adjacent_frequency[0] + adjacent_frequency[j], max_val)
                max_all = max(max_val, max_all)

            # Printing the max length
            # for an elemen
            print("Element :", element, "Max suffix length :", max_val)

    # Printing maximum length among
    # all distinct element
    print("Maximum length among all :", max_all)

# Driver code

# Input Array
X = [2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2]

# Reversing and updating X[]
X = reverse_array(X)

# Variable to hold length of X[]
N = len(X)

# List for storing distinct
# elements present in X[]
s = set_fn(N, X)

# Function call for applying
# approach on X[]
approach(s, N, X)

# This Code is Contributed by Prajwal Kandekar

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
using System.Linq;

public class GFG {

    // function for reversing array
    static List<int> Reverse_Array(List<int> X)
    {
        int start = 0;
        int end = X.Count - 1;
        while (start < end) {
            int temp = X[start];
            X[start] = X[end];
            X[end] = temp;
            start++;
            end--;
        }
        return X;
    }

    // Method for getting distinct
    // elements from X[]
    static List<int> Set(int n, List<int> X)
    {
        List<int> set = new List<int>();
        for (int i = 0; i < n; i++) {
            if (set.Contains(X[i])) {
                continue;
            }
            else {
                set.Add(X[i]);
            }
        }
        return set;
    }

    // function for implementing approach
    static void approach(List<int> set, int n, List<int> X)
    {

        // Variable initialized to hold
        // maximum suffix length among
        // distinct element
        int max_all = int.MinValue;

        // Loop for traversing over set
        // containing distinct elements
        for (int i = 0; i < set.Count; i++) {

            // Variable to hold element
            // from set
            int element = set[i];

            // Array initialized to
            // count the frequency
            // of same adjacent element
            List<int> adjacent_frequency = new List<int>();

            int leftEnd = 0;

            // Loop for initializing
            // frequency in
            // adjacent_frequency list
            while (leftEnd < n) {
                if (X[leftEnd] != element) {
                    leftEnd++;
                }
                else {
                    int rightEnd = leftEnd;
                    while (rightEnd < n
                           && X[rightEnd] == element) {
                        rightEnd = rightEnd + 1;
                    }

                    adjacent_frequency.Add(rightEnd
                                           - leftEnd);
                    leftEnd = rightEnd + 1;
                }
            }

            // Variable initialized to hold
            // maximum suffix length
            // for distinct element
            int max = int.MinValue;

            // Loop for traversing over
            // frequency Array
            for (int k = 0; k < adjacent_frequency.Count;
                 k++) {

                // Updating max with max
                // adjacent frequency
                max = adjacent_frequency[k] > max
                          ? adjacent_frequency[k]
                          : max;
            }
            bool flag = false;

            // Condition when an element is
            // not present at first
            // index of X[]
            if (X[0] != element) {
                Console.WriteLine("Element : " + element
                                  + " Max suffix length : "
                                  + max);
                flag = true;
                max_all = max > max_all ? max : max_all;
            }

            // This will execute when an
            // element for which we are
            // finding max  suffix length,
            // is not present at first
            // index of X[]
            else if (flag != true) {
                for (int j = 1;
                     j < adjacent_frequency.Count; j++) {
                    max = (adjacent_frequency[0]
                           + adjacent_frequency[j])
                                  > max
                              ? (adjacent_frequency[0]
                                 + adjacent_frequency[j])
                              : max;
                    max_all = max > max_all ? max : max_all;
                }

                // Printing the max length
                // for an element
                Console.WriteLine("element : " + element
                                  + " Max suffix length : "
                                  + max);
            }
        }

        // Printing maximum length among
        // all distinct element
        Console.WriteLine("Maximum length among all : "
                          + max_all);
    }

    // Driver Function
    static public void Main()
    {

        // Input Array
        List<int> X = new List<int>{ 2, 2, 5, 5, 5, 3,
                                     5, 5, 3, 3, 2 };

        // Reversing and updating X[]
        X = Reverse_Array(X);

        // Variable to hold length of X[]
        int N = X.Count;

        // List for storing distinct
        // elements present in X[]
        List<int> set = Set(N, X);

        // Function call for applying
        // approach on X[]
        approach(set, N, X);
    }
}
// This code is contributed by Prasad Kandekar(prasad264)

Javascript

// JavaScript code to implement the approach

// function for reversing array
function reverseArray(X)
{
    let start = 0;
    let end = X.length - 1;
    while (start < end) {
        let temp = X[start];
        X[start] = X[end];
        X[end] = temp;
        start++;
        end--;
    }
    return X;
}

// Method for getting distinct
// elements from X[]
function getDistinctElements(X)
{
    let s = [];
    for (let i = 0; i < X.length; i++) {
        if (s.includes(X[i])) {
            continue;
        }
        else {
            s.push(X[i]);
        }
    }

    return s;
}

// function for implementing approach
function approach(s, X)
{
    // Variable initialized to hold
    // maximum suffix length among
    // distinct element
    let max_all = Number.MIN_SAFE_INTEGER;

    // Loop for traversing over set
    // containing distinct elements
    for (let i = 0; i < s.length; i++) {

        // Array initialized to
        // count the frequency
        // of same adjacent element
        let adjacent_frequency = [];

        let leftEnd = 0;

        // Loop for initializing
        // frequency in
        // adjacent_frequency list
        while (leftEnd < X.length) {
            if (X[leftEnd] != s[i]) {
                leftEnd++;
            }
            else {
                let rightEnd = leftEnd;
                while (rightEnd < X.length
                       && X[rightEnd] == s[i]) {
                    rightEnd++;
                }

                adjacent_frequency.push(rightEnd - leftEnd);
                leftEnd = rightEnd + 1;
            }
        }

        // maximum suffix length
        // for distinct element
        let max_val = Number.MIN_SAFE_INTEGER;

        // Loop for traversing over
        // frequency Array
        for (let k = 0; k < adjacent_frequency.length;
             k++) {

            // Updating max with max
            // adjacent frequency
            max_val
                = Math.max(adjacent_frequency[k], max_val);
        }

        let flag = false;

        // Condition when an element is
        // not present at first
        // index of X[]
        if (X[0] != s[i]) {
            console.log("Element :", s[i],
                        "Max suffix length :", max_val);
            flag = true;
            max_all = Math.max(max_val, max_all);

            // This will execute when an
            // element for which we are
            // finding max  suffix length,
            // is not present at first
            // index of X[]
        }
        else if (!flag) {
            for (let j = 1; j < adjacent_frequency.length;
                 j++) {
                max_val
                    = Math.max(adjacent_frequency[0]
                                   + adjacent_frequency[j],
                               max_val);
                max_all = Math.max(max_val, max_all);
            }

            // Printing the max length
            // for an element
            console.log("Element :", s[i],
                        "Max suffix length :", max_val);
        }
    }

    // Printing maximum length among
    // all distinct element
    console.log("Maximum length among all :", max_all);
}

// Driver code

// Input Array
let X = [ 2, 2, 5, 5, 5, 3, 5, 5, 3, 3, 2 ];

// Reversing and updating X[]
X = reverseArray(X);

// List for storing distinct
// elements present in X[]
let s = getDistinctElements(X);

// Function call for applying
// approach on X[]
approach(s, X)

Output

element : 2 Max suffix length :  3
Element : 3 Max suffix length :  2
Element : 5 Max suffix length :  3
Maximum length among all : 3

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