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# Maximum product of bitonic subsequence of size 3

Given an array arr[] of positive integers of size N, the task is to find the maximum product of bitonic subsequence of size 3.
Bitonic Subsequence: subsequence in which elements are first in the increasing order and then decreasing order. Elements in the subsequence are follow this order arr[i] < arr[j] > arr[k] for i < j < k where i, j, k are the index of the given array.
Note: If no such element is found then print -1.

Examples:

Input: arr[] = {1, 8, 3, 7, 5, 6, 7}
Output: 126
Explanation:
Bitonic subsequences of size 3 are
{1, 8, 3}, {1, 8, 7}, {1, 8, 5}, {1, 8, 6}, {1, 7, 6}, {3, 7, 6}, {1, 7, 5}, {3, 7, 5}.
Hence the maximum product of bitonic subsequence is 3*7*6 = 126

Input: arr[] = {1, 8, 3, 7}
Output: 56
Explanation:
Bitonic subsequences of size 3 are
{1, 8, 3}, {1, 8, 7}, {1, 7, 3}.
Hence the maximum product of bitonic subsequence is 1*8*7 = 56

Naive Approach:A simple solution is to find the product of all the bitonic subsequences of size 3 and take the maximum among them.

Algorithm:

• Initialize ans to -1, such that if there is no such subsequence then the output will be -1.
• Iterate over the Array with three nested loops with loop variables as i, j, and k for choosing three elements of the array.
• Check if arr[j] > arr[i] and arr[j] > arr[k] then update the ans with the maximum value between ans and arr[i] * arr[j] * arr[k].

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the// maximum product of the bitonic// subsequence of size 3 #include using namespace std; // Function to find the maximum// product of bitonic subsequence// of size 3int maxProduct(int arr[], int n){         // Initialize ans to -1 if no such    // subsequence exist in the array    int ans = -1;         // Nested loops to choose the three    // elements of the array    for (int i = 0; i < n - 2; i++) {        for (int j = i + 1; j < n - 1; j++) {            for (int k = j + 1; k < n; k++) {                                 // Condition to check if                // they form a bitonic subsequence                if (arr[i] < arr[j] &&                      arr[j] > arr[k])                    ans = max(                       ans, arr[i] * arr[j] * arr[k]                       );            }        }    }    return ans;} // Driver Codeint main(){    int arr[] = { 1, 8, 3, 7 };     int n = sizeof(arr) / sizeof(arr[0]);     // Function call    cout << maxProduct(arr, n) << endl;   }

## Java

 // Java implementation to find the// maximum product of the bitonic// subsequence of size 3import java.util.*; class GFG{  // Function to find the maximum// product of bitonic subsequence// of size 3static int maxProduct(int arr[], int n){          // Initialize ans to -1 if no such    // subsequence exist in the array    int ans = -1;          // Nested loops to choose the three    // elements of the array    for (int i = 0; i < n - 2; i++) {        for (int j = i + 1; j < n - 1; j++) {            for (int k = j + 1; k < n; k++) {                                  // Condition to check if                // they form a bitonic subsequence                if (arr[i] < arr[j] &&                      arr[j] > arr[k])                    ans = Math.max(                       ans, arr[i] * arr[j] * arr[k]                       );            }        }    }    return ans;}  // Driver Codepublic static void main(String[] args){    int arr[] = { 1, 8, 3, 7 };      int n = arr.length;      // Function call    System.out.print(maxProduct(arr, n) +"\n");   }} // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to find the# maximum product of the bitonic# subsequence of size 3 # Function to find the maximum# product of bitonic subsequence# of size 3def maxProduct(arr, n):     # Initialize ans to -1 if no such    # subsequence exist in the array    ans = -1     # Nested loops to choose the three    # elements of the array    for i in range(n - 2):        for j in range(i + 1, n - 1):            for k in range(j + 1, n):                 # Condition to check if                # they form a bitonic subsequence                if (arr[i] < arr[j] and arr[j] > arr[k]):                    ans = max(ans, arr[i] * arr[j] * arr[k])     return ans # Driver Codeif __name__ == '__main__':    arr= [ 1, 8, 3, 7]     n = len(arr)     # Function call    print(maxProduct(arr, n)) # This code is contributed by mohit kumar 29

## C#

 // C# implementation to find the// maximum product of the bitonic// subsequence of size 3using System; class GFG {      // Function to find the maximum     // product of bitonic subsequence    // of size 3    static int maxProduct(int[] arr, int n)    {        // Initialize ans to -1 if no such        // subsequence exist in the array        int ans = -1;                  // Nested loops to choose the three        // elements of the array        for (int i = 0; i < n - 2; i++) {            for (int j = i + 1; j < n - 1; j++) {                for (int k = j + 1; k < n; k++) {                                          // Condition to check if                    // they form a bitonic subsequence                    if (arr[i] < arr[j] &&                          arr[j] > arr[k])                        ans = Math.Max(ans, arr[i] * arr[j] * arr[k]                           );                }            }        }        return ans;    }         // Driver code    static void Main()    {        int[] arr = new int[] { 1, 8, 3, 7 };        int n = arr.Length;             // Function call to find product        Console.Write(maxProduct(arr, n));    }} // This code is contributed by shubhamsingh

## Javascript



Output:

56

Performance Analysis:

• Time Complexity: As in the above approach, there is three nested loops to find the maximum product of the bitonic subsequence of size 3, hence the Time Complexity will be O(N3).
• Auxiliary Space: As in the above approach, there is no extra space used, hence the auxiliary space will be O(1).

Efficient approach: The idea is to find the largest value on the left side and right side of each index which are smaller than the element present at the current index, to do this use a Self Balancing BST and then for every element find the maximum product that can be formed and take the maximum out of those products.
Self-Balancing BST is implemented as set in C++ and TreeSet in Java.

Algorithm:

• Declare a self-balancing BST (say s).
• Declare two new arrays left[] and right[] to store the lower bound for arr[i] in left of that element in left[i] and lower bound of arr[i] in right of that element in right[i].
• Run a loop from 0 to length – 1 to find the lower bound of arr[i] is left of that element and store it in the left[i].
• Run a loop from length -1 to 0 to find the lower bound of arr[i] in the right of that element and store it in the right[i].
• Run a loop from 0 to length – 1 to find the bitonic subsequence that can be formed using that element to get the maximum product using the left[] and right[] array. That is for every element maximum product bitonic subsequence that can be formed is left[i] * right[i] * arr[i].

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the// maximum product of the bitonic// subsequence of size 3 #include using namespace std; // Function to find the maximum// product of bitonic subsequence// of size 3int maxProduct(int arr[], int n){         // Self Balancing BST    set s;    set::iterator it;         // Left array to store the    // maximum smallest value for    // every element in left of it    int Left[n];     // Right array to store the    // maximum smallest value for    // every element in right of it    int Right[n];     // Loop to find the maximum    // smallest element in left of    // every element in array    for (int i = 0; i < n; i++) {        s.insert(arr[i]);        it = s.lower_bound(arr[i]);                 // Condition to check if there        // is a maximum smallest element        if (it != s.begin()) {            it--;            Left[i] = *it;        }        else {            Left[i] = -1;        }    }    // Clear Set    s.clear();         // Loop to find the maximum    // smallest element in right of    // every element in array    for (int i = n - 1; i >= 0; i--) {        s.insert(arr[i]);        it = s.lower_bound(arr[i]);                 // Condition to check if there        // is such element exists        if (it != s.begin()) {            it--;            Right[i] = *it;        }                 // If no such element exists.        else {            Right[i] = -1;        }    }    int ans = -1;         // Loop to find the maximum product    // bitonic subsequence of size 3    for (int i = 0; i < n; i++) {        if (Left[i] > 0 and Right[i] > 0)            ans = max(ans, arr[i] * Left[i] * Right[i]);    }     if (ans < 0) {        return -1;    }    else {        return ans;    }} // Driver Codeint main(){    int arr[] = { 1, 8, 3, 7, 5, 6, 7 };    int n = sizeof(arr) / sizeof(arr[0]);         // Function Call    cout << maxProduct(arr, n);}

## Java

 // Java implementation to find the// maximum product of the bitonic// subsequence of size 3import java.util.*;import java.lang.System; class GFG{  public static int maxProduct(int arr[],int n) {    // Self Balancing BST    TreeSet ts = new TreeSet();     // Left array to store the    // maximum smallest value for    // every element in left of it    int Left[] = new int[n];      // Right array to store the    // maximum smallest value for    // every element in right of it    int Right[] = new int[n];     // Loop to find the maximum    // smallest element in left of    // every element in array    for(int i = 0; i < n; i++)    {        ts.add(arr[i]);         if(ts.lower(arr[i]) == null)            Left[i] = -1;        else            Left[i] = ts.lower(arr[i]);    }     ts.clear();     // Loop to find the maximum    // smallest element in right of    // every element in array    for (int i = n-1; i >= 0; i--)    {        ts.add(arr[i]);         if(ts.lower(arr[i]) == null)            Right[i] = -1;        else            Right[i] = ts.lower(arr[i]);    }     // Loop to find the maximum product    // bitonic subsequence of size 3    int ans = 0;     for(int i = 0; i < n; i++)    {        //Condition to check whether a sequence is bitonic or not        if(Left[i] != -1 && Right[i] != -1)            ans = Math.max(ans, Left[i] * arr[i] * Right[i]);    }     return ans; }  // Driver Code public static void main(String args[]){    int arr[] = {1, 8, 3, 7, 5, 6, 7 };     int n = arr.length;     int maximum_product = maxProduct(arr,n);     System.out.println(maximum_product);}} // This code is contributed by Siddhi.

## Python3

 import sysfrom bisect import bisect_left # Function to find the maximum# product of bitonic subsequence# of size 3  def maxProduct(arr, n):     # Left array to store the    # maximum smallest value for    # every element in left of it    Left = [-1 for i in range(n)]     # Right array to store the    # maximum smallest value for    # every element in right of it    Right = [-1 for i in range(n)]     # Loop to find the maximum    # smallest element in left of    # every element in array    for i in range(1, n):        max_value = -sys.maxsize        for j in range(i):            if arr[j] < arr[i]:                max_value = max(max_value, arr[j])        Left[i] = max_value     # Loop to find the maximum    # smallest element in right of    # every element in array    for i in range(n-2, -1, -1):        max_value = -sys.maxsize        for j in range(i+1, n):            if arr[j] < arr[i]:                max_value = max(max_value, arr[j])        Right[i] = max_value     ans = -sys.maxsize     # Loop to find the maximum product    # bitonic subsequence of size 3    for i in range(n):        if Left[i] > 0 and Right[i] > 0:            ans = max(ans, arr[i] * Left[i] * Right[i])     if ans == -sys.maxsize:        return -1    else:        return ans  # Driver Codearr = [1, 8, 3, 7, 5, 6, 7]n = len(arr) # Function Callprint(maxProduct(arr, n))

## Javascript

 // Javascript implementation to find the// maximum product of the bitonic// subsequence of size 3  function lower_bound(s, x){         let arr = Array.from(s);    arr.sort();         let l = 0;    let h = arr.length - 1;         while(l <= h){        let m = Math.floor((l + h)/2);                 if(arr[m] > x) h = m - 1;        else if(arr[m] < x) l = m + 1;        else{            return m;        }    }         return l;} // Function to find the maximum// product of bitonic subsequence// of size 3function maxProduct(arr, n){         // Self Balancing BST    let s = new Set();         // Left array to store the    // maximum smallest value for    // every element in left of it    let Left = new Array(n);     // Right array to store the    // maximum smallest value for    // every element in right of it    let Right = new Array(n);     // Loop to find the maximum    // smallest element in left of    // every element in array    for (let i = 0; i < n; i++) {        s.add(arr[i]);        it = lower_bound(s, arr[i]);                 // Condition to check if there        // is a maximum smallest element        let temp = Array.from(s);        temp.sort();        if (it != 0) {            Left[i] = temp[it-1];        }        else {            Left[i] = -1;        }    }    // Clear Set    s.clear();         // Loop to find the maximum    // smallest element in right of    // every element in array    for (let i = n - 1; i >= 0; i--) {        s.add(arr[i]);        it = lower_bound(s, arr[i]);                          // Condition to check if there        // is such element exists        let temp = Array.from(s);        temp.sort();        if (it != 0) {            Right[i] = temp[it-1];        }                 // If no such element exists.        else {            Right[i] = -1;        }    }    let ans = -1;         // Loop to find the maximum product    // bitonic subsequence of size 3    for (let i = 0; i < n; i++) {        if (Left[i] > 0 && Right[i] > 0)            ans = Math.max(ans, arr[i] * Left[i] * Right[i]);    }     if (ans < 0) {        return -1;    }    else {        return ans;    }} // Driver Code let arr = [ 1, 8, 3, 7, 5, 6, 7 ];let n = arr.length; // Function Callconsole.log(maxProduct(arr, n)); // The code is contributed by Nidhi goel.

## C#

 using System; class GFG {     // Function to find the maximum    // product of bitonic subsequence    // of size 3    static int maxProduct(int[] arr, int n)    {         // Left array to store the        // maximum smallest value for        // every element in left of it        int[] Left = new int[n];        for (int i = 0; i < n; i++)            Left[i] = -1;         // Right array to store the        // maximum smallest value for        // every element in right of it        int[] Right = new int[n];        for (int i = 0; i < n; i++)            Right[i] = -1;         // Loop to find the maximum        // smallest element in left of        // every element in array        for (int i = 1; i < n; i++) {            int max_value = int.MinValue;            for (int j = 0; j < i; j++) {                if (arr[j] < arr[i])                    max_value = Math.Max(max_value, arr[j]);            }            Left[i] = max_value;        }         // Loop to find the maximum        // smallest element in right of        // every element in array        for (int i = n - 2; i >= 0; i--) {            int max_value = int.MinValue;            for (int j = i + 1; j < n; j++) {                if (arr[j] < arr[i])                    max_value = Math.Max(max_value, arr[j]);            }            Right[i] = max_value;        }         int ans = int.MinValue;         // Loop to find the maximum product        // bitonic subsequence of size 3        for (int i = 0; i < n; i++) {            if (Left[i] > 0 && Right[i] > 0)                ans = Math.Max(ans,                               arr[i] * Left[i] * Right[i]);        }         if (ans == int.MinValue)            return -1;        else            return ans;    }     // Driver Code    public static void Main()    {        int[] arr = { 1, 8, 3, 7, 5, 6, 7 };        int n = arr.Length;         // Function Call        Console.WriteLine(maxProduct(arr, n));    }}

Output:

126

Performance Analysis:

• Time Complexity: O(NlogN).
• Auxiliary Space: O(N).