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Maximum product of indexes of next greater on left and right
  • Difficulty Level : Medium
  • Last Updated : 15 Apr, 2021
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Given an array a[1..N]. For each element at position i (1 <= i <= N). Where 
 

  1. L(i) is defined as closest index j such that j < i and a[j] > a[i]. If no such j exists then L(i) = 0.
  2. R(i) is defined as closest index k such that k > i and a[k] > a[i]. If no such k exists then R(i) = 0.

LRProduct(i) = L(i)*R(i)
We need to find an index with maximum LRProduct
Examples: 
 

Input : 1 1 1 1 0 1 1 1 1 1 
Output : 24 
For {1, 1, 1, 1, 0, 1, 1, 1, 1, 1} all element are same except 0. So only for zero their exist greater element and for others it will be zero. for zero, on left 4th element is closest and greater than zero and on right 6th element is closest and greater. so maximum 
product will be 4*6 = 24.
Input : 5 4 3 4 5 
Output : 8 
For {5, 4, 3, 4, 5}, L[] = {0, 1, 2, 1, 0} and R[] 
= {0, 5, 4, 5, 0}, 
LRProduct = {0, 5, 8, 5, 0} and max in this is 8. 
 

Note: Taking starting index as 1 for finding LRproduct.
 

This problem is based on Next Greater Element
From the current position, we need to find the closest greater element on its left and right side. 
So to find next greater element, we used stack one from left and one from right.simply we are checking which element is greater and storing their index at specified position. 
1- if stack is empty, push current index. 
2- if stack is not empty 
….a) if current element is greater than top element then store the index of current element on index of top element. 
Do this, once traversing array element from left and once from right and form the left and right array, then, multiply them to find max product value. 
 



C++




// C++ program to find the max
// LRproduct[i] among all i
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
 
// function to find just next greater
// element in left side
vector<int> nextGreaterInLeft(int a[], int n)
{
    vector<int> left_index(MAX, 0);
    stack<int> s;
 
    for (int i = n - 1; i >= 0; i--) {
 
        // checking if current element is greater than top
        while (!s.empty() && a[i] > a[s.top() - 1]) {
            int r = s.top();
            s.pop();
 
            // on index of top store the current element
            // index which is just greater than top element
            left_index[r - 1] = i + 1;
        }
 
        // else push the current element in stack
        s.push(i + 1);
    }
    return left_index;
}
 
// function to find just next greater element
// in right side
vector<int> nextGreaterInRight(int a[], int n)
{
    vector<int> right_index(MAX, 0);
    stack<int> s;
    for (int i = 0; i < n; ++i) {
 
        // checking if current element is greater than top
        while (!s.empty() && a[i] > a[s.top() - 1]) {
            int r = s.top();
            s.pop();
 
            // on index of top store the current element
            // index which is just greater than top element
            // stored index should be start with 1
            right_index[r - 1] = i + 1;
        }
 
        // else push the current element in stack
        s.push(i + 1);
    }
    return right_index;
}
 
// Function to find maximum LR product
int LRProduct(int arr[], int n)
{
    // for each element storing the index of just
    // greater element in left side
    vector<int> left = nextGreaterInLeft(arr, n);
 
    // for each element storing the index of just
    // greater element in right side
    vector<int> right = nextGreaterInRight(arr, n);
    int ans = -1;
    for (int i = 1; i <= n; i++) {
 
        // finding the max index product
        ans = max(ans, left[i] * right[i]);
    }
 
    return ans;
}
 
// Drivers code
int main()
{
    int arr[] = { 5, 4, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[1]);
 
    cout << LRProduct(arr, n);
 
    return 0;
}

Java




// Java program to find the
// max LRproduct[i] among all i
import java.io.*;
import java.util.*;
 
class GFG
{
    static int MAX = 1000;
     
    // function to find just next
    // greater element in left side
    static int[] nextGreaterInLeft(int []a,
                                   int n)
    {
        int []left_index = new int[MAX];
        Stack<Integer> s = new Stack<Integer>();
     
        for (int i = n - 1; i >= 0; i--)
        {
     
            // checking if current
            // element is greater than top
            while (s.size() != 0 &&
                     a[i] > a[s.peek() - 1])
            {
                int r = s.peek();
                s.pop();
     
                // on index of top store
                // the current element
                // index which is just
                // greater than top element
                left_index[r - 1] = i + 1;
            }
     
            // else push the current
            // element in stack
            s.push(i + 1);
        }
        return left_index;
    }
     
    // function to find just next
    // greater element in right side
    static int[] nextGreaterInRight(int []a,
                                    int n)
    {
        int []right_index = new int[MAX];
        Stack<Integer> s = new Stack<Integer>();
        for (int i = 0; i < n; ++i) {
     
            // checking if current element
            // is greater than top
            while (s.size() != 0 &&
                        a[i] > a[s.peek() - 1])
            {
                int r = s.peek();
                s.pop();
     
                // on index of top store
                // the current element index
                // which is just greater than
                // top element stored index
                // should be start with 1
                right_index[r - 1] = i + 1;
            }
     
            // else push the current
            // element in stack
            s.push(i + 1);
        }
        return right_index;
    }
     
    // Function to find
    // maximum LR product
    static int LRProduct(int []arr, int n)
    {
         
        // for each element storing
        // the index of just greater
        // element in left side
        int []left = nextGreaterInLeft(arr, n);
     
        // for each element storing
        // the index of just greater
        // element in right side
        int []right = nextGreaterInRight(arr, n);
        int ans = -1;
        for (int i = 1; i <= n; i++)
        {
     
            // finding the max
            // index product
            ans = Math.max(ans, left[i] *
                                right[i]);
        }
     
        return ans;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int []arr = new int[]{ 5, 4, 3, 4, 5 };
        int n = arr.length;
     
        System.out.print(LRProduct(arr, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Python3




# Python3 program to find the 
# max LRproduct[i] among all i
 
# Method to find the next greater
# value in left side
def nextGreaterInLeft(a):
     
    left_index = [0] * len(a)
    s = []
     
    for i in range(len(a)):
         
        # Checking if current
        # element is greater than top
        while len(s) != 0 and a[i] >= a[s[-1]]:
             
            # Pop the element till we can't
            # get the larger value then
            # the current value
            s.pop()
             
        if len(s) != 0:
            left_index[i] = s[-1]
        else:
            left_index[i] = 0
 
        # Else push the element in the stack
        s.append(i)
 
    return left_index
         
# Method to find the next
# greater value in right
def nextGreaterInRight(a):
     
    right_index = [0] * len(a)
    s = []
     
    for i in range(len(a) - 1, -1, -1):
         
        # Checking if current element
        # is greater than top
        while len(s) != 0 and a[i] >= a[s[-1]]:
             
            # Pop the element till we can't
            # get the larger value then
            # the current value
            s.pop()
             
        if len(s) != 0:
            right_index[i] = s[-1]
        else:
            right_index[i] = 0
             
        # Else push the element in the stack
        s.append(i)
 
    return right_index
         
def LRProduct(arr):
     
    # For each element storing
    # the index of just greater
    # element in left side
    left = nextGreaterInLeft(arr)
 
    # For each element storing
    # the index of just greater
    # element in right side
    right = nextGreaterInRight(arr)
 
    ans = -1
 
    # As we know the answer will
    # belong to the range from
    # 1st index to second last index.
    # Because for 1st index left
    # will be 0 and for last
    # index right will be 0
    for i in range(1, len(left) - 1):
         
        if left[i] == 0 or right[i] == 0:
             
            # Finding the max index product
            ans = max(ans, 0)
        else:
            temp = (left[i] + 1) * (right[i] + 1)
             
            # Finding the max index product
            ans = max(ans, temp)
 
    return ans
 
# Driver Code
arr = [ 5, 4, 3, 4, 5 ]
 
print(LRProduct(arr))
 
# This code is contributed by Mohit Pathneja

C#




// C# program to find the max LRproduct[i]
// among all i
using System;
using System.Collections.Generic;
 
class GFG {
     
    static int MAX = 1000;
     
    // function to find just next greater
    // element in left side
    static int[] nextGreaterInLeft(int []a, int n)
    {
        int []left_index = new int[MAX];
        Stack<int> s = new Stack<int>();
     
        for (int i = n - 1; i >= 0; i--) {
     
            // checking if current element is
            // greater than top
            while (s.Count != 0 && a[i] > a[s.Peek() - 1])
            {
                int r = s.Peek();
                s.Pop();
     
                // on index of top store the current
                // element index which is just greater
                // than top element
                left_index[r - 1] = i + 1;
            }
     
            // else push the current element in stack
            s.Push(i + 1);
        }
        return left_index;
    }
     
    // function to find just next greater element
    // in right side
    static int[] nextGreaterInRight(int []a, int n)
    {
        int []right_index = new int[MAX];
        Stack<int> s = new Stack<int>();
        for (int i = 0; i < n; ++i) {
     
            // checking if current element is
            // greater than top
            while (s.Count != 0 && a[i] > a[s.Peek() - 1])
            {
                int r = s.Peek();
                s.Pop();
     
                // on index of top store the current
                // element index which is just greater
                // than top element stored index should
                // be start with 1
                right_index[r - 1] = i + 1;
            }
     
            // else push the current element in stack
            s.Push(i + 1);
        }
        return right_index;
    }
     
    // Function to find maximum LR product
    static int LRProduct(int []arr, int n)
    {
         
        // for each element storing the index of just
        // greater element in left side
        int []left = nextGreaterInLeft(arr, n);
     
        // for each element storing the index of just
        // greater element in right side
        int []right = nextGreaterInRight(arr, n);
        int ans = -1;
        for (int i = 1; i <= n; i++) {
     
            // finding the max index product
            ans = Math.Max(ans, left[i] * right[i]);
        }
     
        return ans;
    }
     
    // Drivers code
    static void Main()
    {
        int []arr = new int[]{ 5, 4, 3, 4, 5 };
        int n = arr.Length;
     
        Console.Write(LRProduct(arr, n));
    }
}
 
// This code is contributed by Manish Shaw
Output: 
8

 

Method 2: Reducing the space used by using only one array to store both left and right max.

Approach:

Prerequisite: https://www.geeksforgeeks.org/next-greater-element/

  • To find the next greater element to left, we used a stack from the left, and the same stack is used for multiplying the right greatest element index with the left greatest element index.
  • Function maxProduct( ) is used for returning the max product by iterating the resultant array.

Java




//java program to find max LR product
 
import java.util.*;
 
public class GFG {
    Stack<Integer> mystack = new Stack<>();
     
    //To find greater element to left
    void nextGreaterToLeft(int[] arr,int[] res) {
        mystack.push(0);
        res[0] = 0;
         
        //iterate through the array
        for(int i=1;i<arr.length;i++) {
            while(!mystack.isEmpty()  &&  arr[mystack.peek()] <= arr[i])
                mystack.pop();
             
            //place the index to the left in the resultant array
            res[i] = (mystack.isEmpty()) ? 0 : mystack.peek()+1;
            mystack.push(i);
        }
    }
     
    ////To find greater element to right
    void nextGreaterToRight(int[] arr,int[] res) {
        mystack.clear();
         
        int n = arr.length;
        mystack.push(n-1);
        res[n-1] *= 0;
         
        //iterate through the array in the reverse order
        for(int i=n-2;i>=0;i--) {
            while(!mystack.isEmpty()  &&  arr[mystack.peek()] <= arr[i])
                mystack.pop();
             
            //multiply the index to the right with the index to the left
            //in the resultant array
            res[i] = (mystack.isEmpty()) ? res[i]*0 : res[i]*(mystack.peek()+1);
            mystack.push(i);
        }
    }
     
    //function to return the max vaue in the resultant array
    int maxProduct(int[] arr,int[] res) {
        nextGreaterToLeft(arr,res);        //to find left max
        nextGreaterToRight(arr,res);    //to find right max
 
        int max = res[0];
        for(int i = 1;i<res.length;i++){
            max = Math.max(max, res[i]);
        }
        return max;
    }
     
    //Driver function
    public static void main(String args[]) {
        GFG obj = new GFG();
        int arr[] = {5, 4, 3, 4, 5};
        int res[] = new int[arr.length];
         
        int maxprod = obj.maxProduct(arr, res);
        System.out.println(maxprod);
    }
}
 
//this method is contributed by Likhita AVL
Output
8

Time Complexity: O(n)

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