Maximum product of indexes of next greater on left and right

Software Development
#1

Hello Everyone,

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.

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++ 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;

}

Output:

8

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

Approach:

  • 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);

}

}

Output

8

Time Complexity: O(n)