Maximum product of indexes of next greater on left and right
Given an array a[1..N]. For each element at position i (1 <= i <= N). Where
- 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.
- 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; } |
chevron_right
filter_none
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) |
chevron_right
filter_none
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 // (manishshaw1) |
chevron_right
filter_none
Output:
8
Recommended Posts:
- Maximum difference between first and last indexes of an element in array
- Closest greater or same value on left side for every element in array
- Maximum product from array such that frequency sum of all repeating elements in product is less than or equal to 2 * k
- Maximum Product Subarray | Added negative product case
- Number of pairs in an array such that product is greater than sum
- Minimum element whose n-th power is greater than product of an array of size n
- Find maximum path sum in a 2D matrix when exactly two left moves are allowed
- Find maximum difference between nearest left and right smaller elements
- Count smaller elements on right side and greater elements on left side using Binary Index Tree
- Count of subarrays whose maximum element is greater than k
- Find minimum value to assign all array elements so that array product becomes greater
- Maximum Product Subarray
- Maximum Product Subarray | Set 3
- Maximum Product Cutting | DP-36
- Maximum element in an array such that its previous and next element product is maximum
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
- Minimum increments to convert to an array of consecutive integers
- Maximum sum in an array such that every element has exactly one adjacent element to it
- Maximum possible sub-array sum after at most X swaps
- Sum of maximum of all subarrays | Divide and Conquer
- Change one element in the given array to make it an Arithmetic Progression
