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Maximize product of a strictly increasing or decreasing subarray

  • Difficulty Level : Expert
  • Last Updated : 27 May, 2021

Given an array arr[] of size N, the task is to find the maximum product from any subarray consisting of elements in strictly increasing or decreasing order.

Examples: 

Input: arr[] = { 1, 2, 10, 8, 1, 100, 101 } 
Output: 10100 
Explanation: 
Increasing subarray with maximum product is {1, 100, 101}. 
Therefore, the required output is 1 * 100 * 101.

Input: arr[] = { 1, 5, 7, 2, 10, 12 } 
Output: 240

Naive Approach: The simplest approach to solve this problem is to generate all possible subarrays from the given array. For each subarray, check if the elements present in the subarray are either in a strictly increasing or decreasing order or not. If found to be true, then calculate the product of elements of the subarray. Finally, print the maximum product obtained. 



Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient approach The above approach can be optimized by traversing the array and for every increasing or decreasing subarray from the given array, find the maximum product using Kadane’s Algorithm. Finally, print the maximum of all products from increasing or decreasing subarrays. Follow the steps below to solve the problem: 

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum product of
// subarray in the array, arr[]
int maxSubarrayProduct(vector<int> arr, int n)
{
 
    // Maximum positive product
    // ending at the i-th index
    int max_ending_here = 1;
 
    // Minimum negative product ending
    // at the current index
    int min_ending_here = 1;
 
    // Maximum product up to
    // i-th index
    int max_so_far = 0;
 
    // Check if an array element
    // is positive or not
    int flag = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element
        // is positive
        if (arr[i] > 0) {
 
            // Update max_ending_here
            max_ending_here
                = max_ending_here * arr[i];
 
            // Update min_ending_here
            min_ending_here
                = min(min_ending_here * arr[i], 1);
 
            // Update flag
            flag = 1;
        }
 
        // If current element is 0, reset
        // the start index of subarray
        else if (arr[i] == 0) {
 
            // Update max_ending_here
            max_ending_here = 1;
 
            // Update min_ending_here
            min_ending_here = 1;
        }
 
        // If current element is negative
        else {
 
            // Stores max_ending_here
            int temp = max_ending_here;
 
            // Update max_ending_here
            max_ending_here
                = max(min_ending_here * arr[i], 1);
 
            // Update min_ending_here
            min_ending_here = temp * arr[i];
        }
 
        // Update max_so_far, if needed
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
 
    // If no array elements is positive
    // and max_so_far is 0
    if (flag == 0 && max_so_far == 0)
        return 0;
    return max_so_far;
}
 
// Function to find the maximum product of either
// increasing subarray or the decreasing subarray
int findMaxProduct(int* a, int n)
{
 
    // Stores start index of either increasing
    // subarray or the decreasing subarray
    int i = 0;
 
    // Initially assume maxProd to be 1
    int maxProd = -1e9;
 
    // Traverse the array
    while (i < n) {
 
        // Store the longest either increasing
        // subarray or the decreasing subarray
        // whose start index is i
        vector<int> v;
 
        v.push_back(a[i]);
 
        // Check for increasing subarray
        if (i < n - 1 && a[i] < a[i + 1]) {
 
            // Insert elements of
            // increasing subarray
            while (i < n - 1 && a[i] < a[i + 1]) {
                v.push_back(a[i + 1]);
                i += 1;
            }
        }
 
        // Check for decreasing subarray
        else if (i < n - 1 && a[i] > a[i + 1]) {
 
            // Insert elements of
            // decreasing subarray
            while (i < n - 1 && a[i] > a[i + 1]) {
                v.push_back(a[i + 1]);
                i += 1;
            }
        }
 
        // Stores maximum subarray product of
        // current increasing or decreasing
        // subarray
        int prod = maxSubarrayProduct(v, v.size());
 
        // Update maxProd
        maxProd = max(maxProd, prod);
 
        // Update i
        i++;
    }
 
    // Finally print maxProd
    return maxProd;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 10, 8, 1, 100, 101 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMaxProduct(arr, N);
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
class GFG
{
 
  // Function to find the maximum product of
  // subarray in the array, arr[]
  static int maxSubarrayProduct(Vector<Integer> arr, int n)
  {
 
    // Maximum positive product
    // ending at the i-th index
    int max_ending_here = 1;
 
    // Minimum negative product ending
    // at the current index
    int min_ending_here = 1;
 
    // Maximum product up to
    // i-th index
    int max_so_far = 0;
 
    // Check if an array element
    // is positive or not
    int flag = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++)
    {
 
      // If current element
      // is positive
      if (arr.get(i) > 0)
      {
 
        // Update max_ending_here
        max_ending_here
          = max_ending_here * arr.get(i);
 
        // Update min_ending_here
        min_ending_here
          = Math.min(min_ending_here * arr.get(i), 1);
 
        // Update flag
        flag = 1;
      }
 
      // If current element is 0, reset
      // the start index of subarray
      else if (arr.get(i) == 0)
      {
 
        // Update max_ending_here
        max_ending_here = 1;
 
        // Update min_ending_here
        min_ending_here = 1;
      }
 
      // If current element is negative
      else
      {
 
        // Stores max_ending_here
        int temp = max_ending_here;
 
        // Update max_ending_here
        max_ending_here
          = Math.max(min_ending_here * arr.get(i), 1);
 
        // Update min_ending_here
        min_ending_here = temp * arr.get(i);
      }
 
      // Update max_so_far, if needed
      if (max_so_far < max_ending_here)
        max_so_far = max_ending_here;
    }
 
    // If no array elements is positive
    // and max_so_far is 0
    if (flag == 0 && max_so_far == 0)
      return 0;
    return max_so_far;
  }
 
  // Function to find the maximum product of either
  // increasing subarray or the decreasing subarray
  static int findMaxProduct(int[] a, int n)
  {
 
    // Stores start index of either increasing
    // subarray or the decreasing subarray
    int i = 0;
 
    // Initially assume maxProd to be 1
    int maxProd = 1;
 
    // Traverse the array
    while (i < n) {
 
      // Store the longest either increasing
      // subarray or the decreasing subarray
      // whose start index is i
      Vector<Integer> v = new Vector<>();
 
      v.add(a[i]);
 
      // Check for increasing subarray
      if (i < n - 1 && a[i] < a[i + 1]) {
 
        // Insert elements of
        // increasing subarray
        while (i < n - 1 && a[i] < a[i + 1]) {
          v.add(a[i + 1]);
          i += 1;
        }
      }
 
      // Check for decreasing subarray
      else if (i < n - 1 && a[i] > a[i + 1]) {
 
        // Insert elements of
        // decreasing subarray
        while (i < n - 1 && a[i] > a[i + 1]) {
          v.add(a[i + 1]);
          i += 1;
        }
      }
 
      // Stores maximum subarray product of
      // current increasing or decreasing
      // subarray
      int prod = maxSubarrayProduct(v, v.size());
 
      // Update maxProd
      maxProd = Math.max(maxProd, prod);
 
      // Update i
      i++;
    }
 
    // Finally print maxProd
    return maxProd;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 1, 2, 10, 8, 1, 100, 101 };
    int N = arr.length;
 
    System.out.print(findMaxProduct(arr, N));
  }
}
 
// This code contributed by gauravrajput1

Python3




# Python3 program to implement
# the above approach
 
# Function to find the maximum product
# of subarray in the array, arr[]
def maxSubarrayProduct(arr, n):
     
    # Maximum positive product
    # ending at the i-th index
    max_ending_here = 1
 
    # Minimum negative product ending
    # at the current index
    min_ending_here = 1
 
    # Maximum product up to
    # i-th index
    max_so_far = 0
 
    # Check if an array element
    # is positive or not
    flag = 0
 
    # Traverse the array
    for i in range(n):
         
        # If current element
        # is positive
        if (arr[i] > 0):
             
            # Update max_ending_here
            max_ending_here = max_ending_here * arr[i]
 
            # Update min_ending_here
            min_ending_here = min(
                min_ending_here * arr[i], 1)
 
            # Update flag
            flag = 1
 
        # If current element is 0, reset
        # the start index of subarray
        elif (arr[i] == 0):
             
            # Update max_ending_here
            max_ending_here = 1
             
            # Update min_ending_here
            min_ending_here = 1
             
        # If current element is negative
        else:
             
            # Stores max_ending_here
            temp = max_ending_here
 
            # Update max_ending_here
            max_ending_here = max(
                min_ending_here * arr[i], 1)
 
            # Update min_ending_here
            min_ending_here = temp * arr[i]
 
        # Update max_so_far, if needed
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here
 
    # If no array elements is positive
    # and max_so_far is 0
    if (flag == 0 and max_so_far == 0):
        return 0
         
    return max_so_far
 
# Function to find the maximum product
# of either increasing subarray or the
# decreasing subarray
def findMaxProduct(a, n):
     
    # Stores start index of either
    # increasing subarray or the
    # decreasing subarray
    i = 0
 
    # Initially assume maxProd to be 1
    maxProd = -10**9
 
    # Traverse the array
    while (i < n):
         
        # Store the longest either increasing
        # subarray or the decreasing subarray
        # whose start index is i
        v = []
 
        v.append(a[i])
 
        # Check for increasing subarray
        if i < n - 1 and a[i] < a[i + 1]:
             
            # Insert elements of
            # increasing subarray
            while (i < n - 1 and a[i] < a[i + 1]):
                v.append(a[i + 1])
                i += 1
 
        # Check for decreasing subarray
        elif (i < n - 1 and a[i] > a[i + 1]):
             
            # Insert elements of
            # decreasing subarray
            while (i < n - 1 and a[i] > a[i + 1]):
                v.append(a[i + 1])
                i += 1
 
        # Stores maximum subarray product of
        # current increasing or decreasing
        # subarray
        prod = maxSubarrayProduct(v, len(v))
 
        # Update maxProd
        maxProd = max(maxProd, prod)
 
        # Update i
        i += 1
 
    # Finally prmaxProd
    return maxProd
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 10, 8, 1, 100, 101 ]
    N = len(arr)
 
    print (findMaxProduct(arr, N))
 
# This code is contributed by mohit kumar 29

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // Function to find the maximum product of
  // subarray in the array, []arr
  static int maxSubarrayProduct(List<int> arr, int n)
  {
 
    // Maximum positive product
    // ending at the i-th index
    int max_ending_here = 1;
 
    // Minimum negative product ending
    // at the current index
    int min_ending_here = 1;
 
    // Maximum product up to
    // i-th index
    int max_so_far = 0;
 
    // Check if an array element
    // is positive or not
    int flag = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++)
    {
 
      // If current element
      // is positive
      if (arr[i] > 0)
      {
 
        // Update max_ending_here
        max_ending_here
          = max_ending_here * arr[i];
 
        // Update min_ending_here
        min_ending_here
          = Math.Min(min_ending_here * arr[i], 1);
 
        // Update flag
        flag = 1;
      }
 
      // If current element is 0, reset
      // the start index of subarray
      else if (arr[i] == 0)
      {
 
        // Update max_ending_here
        max_ending_here = 1;
 
        // Update min_ending_here
        min_ending_here = 1;
      }
 
      // If current element is negative
      else
      {
 
        // Stores max_ending_here
        int temp = max_ending_here;
 
        // Update max_ending_here
        max_ending_here
          = Math.Max(min_ending_here * arr[i], 1);
 
        // Update min_ending_here
        min_ending_here = temp * arr[i];
      }
 
      // Update max_so_far, if needed
      if (max_so_far < max_ending_here)
        max_so_far = max_ending_here;
    }
 
    // If no array elements is positive
    // and max_so_far is 0
    if (flag == 0 && max_so_far == 0)
      return 0;
    return max_so_far;
  }
 
  // Function to find the maximum product of either
  // increasing subarray or the decreasing subarray
  static int findMaxProduct(int[] a, int n)
  {
 
    // Stores start index of either increasing
    // subarray or the decreasing subarray
    int i = 0;
 
    // Initially assume maxProd to be 1
    int maxProd = 1;
 
    // Traverse the array
    while (i < n) {
 
      // Store the longest either increasing
      // subarray or the decreasing subarray
      // whose start index is i
      List<int> v = new List<int>();
 
      v.Add(a[i]);
 
      // Check for increasing subarray
      if (i < n - 1 && a[i] < a[i + 1]) {
 
        // Insert elements of
        // increasing subarray
        while (i < n - 1 && a[i] < a[i + 1]) {
          v.Add(a[i + 1]);
          i += 1;
        }
      }
 
      // Check for decreasing subarray
      else if (i < n - 1 && a[i] > a[i + 1]) {
 
        // Insert elements of
        // decreasing subarray
        while (i < n - 1 && a[i] > a[i + 1]) {
          v.Add(a[i + 1]);
          i += 1;
        }
      }
 
      // Stores maximum subarray product of
      // current increasing or decreasing
      // subarray
      int prod = maxSubarrayProduct(v, v.Count);
 
      // Update maxProd
      maxProd = Math.Max(maxProd, prod);
 
      // Update i
      i++;
    }
 
    // Finally print maxProd
    return maxProd;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 1, 2, 10, 8, 1, 100, 101 };
    int N = arr.Length;
 
    Console.Write(findMaxProduct(arr, N));
  }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
//Javascript program for the above approach
 
// Function to find the maximum product of
// subarray in the array, arr[]
function maxSubarrayProduct(arr, n)
{
 
    // Maximum positive product
    // ending at the i-th index
    var max_ending_here = 1;
 
    // Minimum negative product ending
    // at the current index
    var min_ending_here = 1;
 
    // Maximum product up to
    // i-th index
    var max_so_far = 0;
 
    // Check if an array element
    // is positive or not
    var flag = 0;
 
    // Traverse the array
    for (var i = 0; i < n; i++) {
 
        // If current element
        // is positive
        if (arr[i] > 0) {
 
            // Update max_ending_here
            max_ending_here
                = max_ending_here * arr[i];
 
            // Update min_ending_here
            min_ending_here
                = Math.min(min_ending_here * arr[i], 1);
 
            // Update flag
            flag = 1;
        }
 
        // If current element is 0, reset
        // the start index of subarray
        else if (arr[i] == 0) {
 
            // Update max_ending_here
            max_ending_here = 1;
 
            // Update min_ending_here
            min_ending_here = 1;
        }
 
        // If current element is negative
        else {
 
            // Stores max_ending_here
            var temp = max_ending_here;
 
            // Update max_ending_here
            max_ending_here
                = Math.max(min_ending_here * arr[i], 1);
 
            // Update min_ending_here
            min_ending_here = temp * arr[i];
        }
 
        // Update max_so_far, if needed
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
 
    // If no array elements is positive
    // and max_so_far is 0
    if (flag == 0 && max_so_far == 0)
        return 0;
    return max_so_far;
}
 
// Function to find the maximum product of either
// increasing subarray or the decreasing subarray
function findMaxProduct(a, n)
{
 
    // Stores start index of either increasing
    // subarray or the decreasing subarray
    var i = 0;
 
    // Initially assume maxProd to be 1
    var maxProd = -1e9;
 
    // Traverse the array
    while (i < n) {
 
        // Store the longest either increasing
        // subarray or the decreasing subarray
        // whose start index is i
        var v=[];
 
        v.push(a[i]);
 
        // Check for increasing subarray
        if (i < n - 1 && a[i] < a[i + 1]) {
 
            // Insert elements of
            // increasing subarray
            while (i < n - 1 && a[i] < a[i + 1]) {
                v.push(a[i + 1]);
                i += 1;
            }
        }
 
        // Check for decreasing subarray
        else if (i < n - 1 && a[i] > a[i + 1]) {
 
            // Insert elements of
            // decreasing subarray
            while (i < n - 1 && a[i] > a[i + 1]) {
                v.push(a[i + 1]);
                i += 1;
            }
        }
 
        // Stores maximum subarray product of
        // current increasing or decreasing
        // subarray
        var prod = maxSubarrayProduct(v, v.length);
 
        // Update maxProd
        maxProd = Math.max(maxProd, prod);
 
        // Update i
        i++;
    }
 
    // Finally print maxProd
    return maxProd;
}
 
    var arr = [ 1, 2, 10, 8, 1, 100, 101 ];
    var N = arr.length;
    document.write(findMaxProduct(arr, N));
 
//This code is contributed by SoumikMondal
</script>
Output: 
10100

 

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

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