# Maximum product of indexes of next greater on left and right

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)