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Maximum product of indexes of next greater on left and right

Last Updated : 24 Apr, 2025
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Given an array arr[1..n], for each element at position i (1 <= i <= n), define the following:

  • left(i) is the closest index j such that j < i and arr[j] > arr[i]. If no such j exists, then left(i) = 0.
  • right(i) is the closest index k such that k > i and arr[k] > arr[i]. If no such k exists, then right(i) = 0.
  • LRproduct(i) = left(i) * right(i).

The task is to find the maximum LR product of the given array.

Examples:

Input: arr[] = [1, 1, 1, 1, 0, 1, 1, 1, 1, 1]
Output: 24
Explanation: All elements are equal except 0. Only for 0, greater elements exist. On the left half, the closest greater element is at index 4 and on the right half, it is at index 6. The maximum product is 4 × 6 = 24.

Input: arr[] = [5, 4, 3, 4, 5]
Output: 8
Explanation: For [5, 4, 3, 4, 5]: left[] = [0, 1, 2, 1, 0], right[] = [0, 5, 4, 5, 0]
LRproduct[] = [0, 5, 8, 5, 0]. The maximum value in LRproduct is 8.

Please note indexes are considered from 1 to n and 0 is used for filling in cases when left(i) and/or right(i) do not exist.

Table of Content

  • Using Monotonic Stack With 2 Arrays - O(n) Time and O(n) Space
  • Using a Single Array - O(n) Time and O(n) Space

Using Monotonic Stack With 2 Arrays - O(n) Time and O(n) Space

The idea is to find the closest greater elements on both left and right for each element. Using a monotonic stack, we efficiently compute left(i) and right(i) in linear time. The key observation is that elements are processed in increasing order, ensuring each element is pushed and popped at most once. Finally, we compute LRproduct for all indices and return the maximum value.

C++ 
// C++ program to find the maximum LR product  // using stack and 2 Arrays #include <bits/stdc++.h> using namespace std;  // Function to find the maximum LR product int findMaxLRProduct(vector<int>& arr) {          int n = arr.size();     vector<int> left(n, 0), right(n, 0);     stack<int> stk;      // Compute left(i) using a monotonic stack     for (int i = 0; i < n; i++) {          while (!stk.empty() && arr[stk.top()] <= arr[i]) {             stk.pop();         }          if (!stk.empty()) {                          // Store 1-based index             left[i] = stk.top() + 1;          }          stk.push(i);     }      // Clear stack to reuse for right(i)     while (!stk.empty()) {         stk.pop();     }      // Compute right(i) using a monotonic stack     for (int i = n - 1; i >= 0; i--) {          while (!stk.empty() && arr[stk.top()] <= arr[i]) {             stk.pop();         }          if (!stk.empty()) {                          // Store 1-based index             right[i] = stk.top() + 1;          }          stk.push(i);     }      int maxProduct = 0;      // Compute the maximum LR product     for (int i = 0; i < n; i++) {         maxProduct = max(maxProduct, left[i] * right[i]);     }      return maxProduct; }  // Driver code int main() {      vector<int> arr = {1, 1, 1, 1, 0, 1, 1, 1, 1, 1};      cout << findMaxLRProduct(arr) << endl;      return 0; }
Java 
// Java program to find the maximum LR product  // using stack and 2 Arrays import java.util.*;  class GfG {      // Function to find the maximum LR product     static int findMaxLRProduct(int[] arr) {                  int n = arr.length;         int[] left = new int[n], right = new int[n];         Stack<Integer> stk = new Stack<>();          // Compute left(i) using a monotonic stack         for (int i = 0; i < n; i++) {              while (!stk.isEmpty() && arr[stk.peek()] <= arr[i]) {                 stk.pop();             }              if (!stk.isEmpty()) {                                  // Store 1-based index                 left[i] = stk.peek() + 1;              }              stk.push(i);         }          // Clear stack to reuse for right(i)         stk.clear();          // Compute right(i) using a monotonic stack         for (int i = n - 1; i >= 0; i--) {              while (!stk.isEmpty() && arr[stk.peek()] <= arr[i]) {                 stk.pop();             }              if (!stk.isEmpty()) {                                  // Store 1-based index                 right[i] = stk.peek() + 1;              }              stk.push(i);         }          int maxProduct = 0;          // Compute the maximum LR product         for (int i = 0; i < n; i++) {             maxProduct = Math.max(maxProduct, left[i] * right[i]);         }          return maxProduct;     }      // Driver code     public static void main(String[] args) {          int[] arr = {1, 1, 1, 1, 0, 1, 1, 1, 1, 1};          System.out.println(findMaxLRProduct(arr));     } }
Python 
# Python program to find the maximum LR product  # using stack and 2 lists  def findMaxLRProduct(arr):          n = len(arr)     left = [0] * n     right = [0] * n     stk = []      # Compute left(i) using a monotonic stack     for i in range(n):          while stk and arr[stk[-1]] <= arr[i]:             stk.pop()          if stk:                          # Store 1-based index             left[i] = stk[-1] + 1           stk.append(i)      # Clear stack to reuse for right(i)     stk.clear()      # Compute right(i) using a monotonic stack     for i in range(n - 1, -1, -1):          while stk and arr[stk[-1]] <= arr[i]:             stk.pop()          if stk:                          # Store 1-based index             right[i] = stk[-1] + 1           stk.append(i)      max_product = 0      # Compute the maximum LR product     for i in range(n):         max_product = max(max_product, left[i] * right[i])      return max_product  # Driver code if __name__ == "__main__":      arr = [1, 1, 1, 1, 0, 1, 1, 1, 1, 1]      print(findMaxLRProduct(arr))
C# 
// C# program to find the maximum LR product  // using stack and 2 Arrays using System; using System.Collections.Generic;  class GfG {      // Function to find the maximum LR product     public static int FindMaxLRProduct(int[] arr) {                  int n = arr.Length;         int[] left = new int[n], right = new int[n];         Stack<int> stk = new Stack<int>();          // Compute left(i) using a monotonic stack         for (int i = 0; i < n; i++) {              while (stk.Count > 0 && arr[stk.Peek()] <= arr[i]) {                 stk.Pop();             }              if (stk.Count > 0) {                                  // Store 1-based index                 left[i] = stk.Peek() + 1;              }              stk.Push(i);         }          // Clear stack to reuse for right(i)         stk.Clear();          // Compute right(i) using a monotonic stack         for (int i = n - 1; i >= 0; i--) {              while (stk.Count > 0 && arr[stk.Peek()] <= arr[i]) {                 stk.Pop();             }              if (stk.Count > 0) {                                  // Store 1-based index                 right[i] = stk.Peek() + 1;              }              stk.Push(i);         }          int maxProduct = 0;          // Compute the maximum LR product         for (int i = 0; i < n; i++) {             maxProduct = Math.Max(maxProduct, left[i] * right[i]);         }          return maxProduct;     }      // Driver code     public static void Main() {          int[] arr = {1, 1, 1, 1, 0, 1, 1, 1, 1, 1};          Console.WriteLine(FindMaxLRProduct(arr));     } }
JavaScript 
// JavaScript program to find the maximum LR product  // using stack and 2 Arrays  function findMaxLRProduct(arr) {          let n = arr.length;     let left = new Array(n).fill(0);     let right = new Array(n).fill(0);     let stk = [];      // Compute left(i) using a monotonic stack     for (let i = 0; i < n; i++) {          while (stk.length > 0 && arr[stk[stk.length - 1]] <= arr[i]) {             stk.pop();         }          if (stk.length > 0) {                          // Store 1-based index             left[i] = stk[stk.length - 1] + 1;          }          stk.push(i);     }      // Clear stack to reuse for right(i)     stk = [];      // Compute right(i) using a monotonic stack     for (let i = n - 1; i >= 0; i--) {          while (stk.length > 0 && arr[stk[stk.length - 1]] <= arr[i]) {             stk.pop();         }          if (stk.length > 0) {                          // Store 1-based index             right[i] = stk[stk.length - 1] + 1;          }          stk.push(i);     }      let maxProduct = 0;      // Compute the maximum LR product     for (let i = 0; i < n; i++) {         maxProduct = Math.max(maxProduct, left[i] * right[i]);     }      return maxProduct; }  // Driver code let arr = [1, 1, 1, 1, 0, 1, 1, 1, 1, 1];  console.log(findMaxLRProduct(arr));

Output
24

Using a Single Array - O(n) Time and O(n) Space

The idea is 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.

C++ 
// C++ program to find the maximum LR product  // using a single stack #include <bits/stdc++.h> using namespace std;  // Function to find the maximum LR product int findMaxLRProduct(vector<int>& arr) {          int n = arr.size();     vector<int> left(n, 0);     stack<int> stk;     int maxProduct = 0;      // Compute left(i) and right(i) using a single stack     for (int i = 0; i < n; i++) {          while (!stk.empty() && arr[stk.top()] <= arr[i]) {             stk.pop();         }          if (!stk.empty()) {                          // Store 1-based index for left             left[i] = stk.top() + 1;         }          stk.push(i);     }      // Clear stack to reuse for right(i)     while (!stk.empty()) {         stk.pop();     }      // Compute right(i) and calculate max LR product     for (int i = n - 1; i >= 0; i--) {          while (!stk.empty() && arr[stk.top()] <= arr[i]) {             stk.pop();         }          if (!stk.empty()) {                          // Compute LR product on the fly             maxProduct = max(maxProduct, left[i] * (stk.top() + 1));         }          stk.push(i);     }      return maxProduct; }  // Driver code int main() {      vector<int> arr = {1, 1, 1, 1, 0, 1, 1, 1, 1, 1};      cout << findMaxLRProduct(arr) << endl;      return 0; }
Java 
// Java program to find the maximum LR product  // using a single stack import java.util.*;  class GfG {      // Function to find the maximum LR product     static int findMaxLRProduct(int[] arr) {                  int n = arr.length;         int[] left = new int[n];         Stack<Integer> stk = new Stack<>();         int maxProduct = 0;          // Compute left(i) and right(i) using a single stack         for (int i = 0; i < n; i++) {              while (!stk.isEmpty() && arr[stk.peek()] <= arr[i]) {                 stk.pop();             }              if (!stk.isEmpty()) {                                  // Store 1-based index for left                 left[i] = stk.peek() + 1;             }              stk.push(i);         }          // Clear stack to reuse for right(i)         stk.clear();          // Compute right(i) and calculate max LR product         for (int i = n - 1; i >= 0; i--) {              while (!stk.isEmpty() && arr[stk.peek()] <= arr[i]) {                 stk.pop();             }              if (!stk.isEmpty()) {                                  // Compute LR product on the fly                 maxProduct = Math.max(maxProduct, left[i] * (stk.peek() + 1));             }              stk.push(i);         }          return maxProduct;     }      // Driver code     public static void main(String[] args) {          int[] arr = {1, 1, 1, 1, 0, 1, 1, 1, 1, 1};          System.out.println(findMaxLRProduct(arr));     } }
Python 
# Python program to find the maximum LR product  # using a single stack  def find_max_lr_product(arr):          n = len(arr)     left = [0] * n     stk = []     max_product = 0      # Compute left(i) and right(i) using a single stack     for i in range(n):          while stk and arr[stk[-1]] <= arr[i]:             stk.pop()          if stk:                          # Store 1-based index for left             left[i] = stk[-1] + 1          stk.append(i)      # Clear stack to reuse for right(i)     stk.clear()      # Compute right(i) and calculate max LR product     for i in range(n - 1, -1, -1):          while stk and arr[stk[-1]] <= arr[i]:             stk.pop()          if stk:                          # Compute LR product on the fly             max_product = max(max_product, left[i] * (stk[-1] + 1))          stk.append(i)      return max_product  # Driver code if __name__ == "__main__":          arr = [1, 1, 1, 1, 0, 1, 1, 1, 1, 1]      print(find_max_lr_product(arr))
C# 
// C# program to find the maximum LR product  // using a single stack using System; using System.Collections.Generic;  class GfG {      // Function to find the maximum LR product     public static int FindMaxLRProduct(int[] arr) {                  int n = arr.Length;         int[] left = new int[n];         Stack<int> stk = new Stack<int>();         int maxProduct = 0;          // Compute left(i) and right(i) using a single stack         for (int i = 0; i < n; i++) {              while (stk.Count > 0 && arr[stk.Peek()] <= arr[i]) {                 stk.Pop();             }              if (stk.Count > 0) {                                  // Store 1-based index for left                 left[i] = stk.Peek() + 1;             }              stk.Push(i);         }          // Clear stack to reuse for right(i)         stk.Clear();          // Compute right(i) and calculate max LR product         for (int i = n - 1; i >= 0; i--) {              while (stk.Count > 0 && arr[stk.Peek()] <= arr[i]) {                 stk.Pop();             }              if (stk.Count > 0) {                                  // Compute LR product on the fly                 maxProduct = Math.Max(maxProduct, left[i] * (stk.Peek() + 1));             }              stk.Push(i);         }          return maxProduct;     }      // Driver code     public static void Main() {          int[] arr = {1, 1, 1, 1, 0, 1, 1, 1, 1, 1};          Console.WriteLine(FindMaxLRProduct(arr));     } }
JavaScript 
// JavaScript program to find the maximum LR product  // using a single stack  function findMaxLRProduct(arr) {          let n = arr.length;     let left = new Array(n).fill(0);     let stk = [];     let maxProduct = 0;      // Compute left(i) and right(i) using a single stack     for (let i = 0; i < n; i++) {          while (stk.length > 0 && arr[stk[stk.length - 1]] <= arr[i]) {             stk.pop();         }          if (stk.length > 0) {                          // Store 1-based index for left             left[i] = stk[stk.length - 1] + 1;         }          stk.push(i);     }      // Clear stack to reuse for right(i)     stk = [];      // Compute right(i) and calculate max LR product     for (let i = n - 1; i >= 0; i--) {          while (stk.length > 0 && arr[stk[stk.length - 1]] <= arr[i]) {             stk.pop();         }          if (stk.length > 0) {                          // Compute LR product on the fly             maxProduct = Math.max(maxProduct, left[i] * (stk[stk.length - 1] + 1));         }          stk.push(i);     }      return maxProduct; }  // Driver code let arr = [1, 1, 1, 1, 0, 1, 1, 1, 1, 1];  console.log(findMaxLRProduct(arr));

Output
24


We can further optimize this using a single for loop. Please refer Largest Area in a Histogram and try to solve this problem using single stack and single loop,



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    Design a stack with the following operations. push(Stack s, x): Adds an item x to stack s pop(Stack s): Removes the top item from stack s merge(Stack s1, Stack s2): Merge contents of s2 into s1. Time Complexity of all above operations should be O(1). If we use array implementation of the stack, then
    8 min read
    The Stock Span Problem
    The stock span problem is a financial problem where we have a series of daily price quotes for a stock denoted by an array arr[] and the task is to calculate the span of the stock's price for all days. The span of the stock's price on ith day represents the maximum number of consecutive days leading
    11 min read
    Next Greater Element (NGE) for every element in given Array
    Given an array arr[] of integers, the task is to find the Next Greater Element for each element of the array in order of their appearance in the array. Note: The Next Greater Element for an element x is the first greater element on the right side of x in the array. Elements for which no greater elem
    9 min read
    Next Greater Frequency Element
    Given an array, for each element find the value of the nearest element to the right which is having a frequency greater than that of the current element. If there does not exist an answer for a position, then make the value '-1'.Examples: Input: arr[] = [2, 1, 1, 3, 2, 1]Output: [1, -1, -1, 2, 1, -1
    9 min read
    Maximum product of indexes of next greater on left and right
    Given an array arr[1..n], for each element at position i (1 <= i <= n), define the following:left(i) is the closest index j such that j < i and arr[j] > arr[i]. If no such j exists, then left(i) = 0.right(i) is the closest index k such that k > i and arr[k] > arr[i]. If no such k e
    11 min read
    Iterative Tower of Hanoi
    The Tower of Hanoi is a mathematical puzzle with three poles and stacked disks of different sizes. The goal is to move all disks from the source pole to the destination pole using an auxiliary pole, following two rules:Only one disk can be moved at a time.A larger disk cannot be placed on a smaller
    6 min read
    Sort a stack using a temporary stack
    Given a stack of integers, sort it in ascending order using another temporary stack.Examples: Input: [34, 3, 31, 98, 92, 23]Output: [3, 23, 31, 34, 92, 98]Explanation: After Sorting the given array it would be look like as [3, 23, 31, 34, 92, 98]Input: [3, 5, 1, 4, 2, 8]Output: [1, 2, 3, 4, 5, 8] Ap
    6 min read
    Reverse a stack without using extra space in O(n)
    Reverse a Stack without using recursion and extra space. Even the functional Stack is not allowed. Examples: Input : 1->2->3->4 Output : 4->3->2->1 Input : 6->5->4 Output : 4->5->6 We have discussed a way of reversing a stack in the below post.Reverse a Stack using Recu
    6 min read
    Delete middle element of a stack
    Given a stack with push(), pop(), and empty() operations, The task is to delete the middle element of it without using any additional data structure.Input: s = [10, 20, 30, 40, 50]Output: [50, 40, 20, 10]Explanation: The bottom-most element will be 10 and the top-most element will be 50. Middle elem
    8 min read
    Check if a queue can be sorted into another queue using a stack
    Given a Queue consisting of first n natural numbers (in random order). The task is to check whether the given Queue elements can be arranged in increasing order in another Queue using a stack. The operation allowed are: Push and pop elements from the stack Pop (Or Dequeue) from the given Queue. Push
    9 min read
    Check if an array is stack sortable
    Given an array arr[] of n distinct elements, where each element is between 1 and n (inclusive), determine if it is stack-sortable.Note: An array a[] is considered stack-sortable if it can be rearranged into a sorted array b[] using a temporary stack stk with the following operations:Remove the first
    6 min read
    Largest Rectangular Area in a Histogram
    Given a histogram represented by an array arr[], where each element of the array denotes the height of the bars in the histogram. All bars have the same width of 1 unit.Task is to find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of c
    15+ min read
    Maximum of minimums of every window size in a given array
    Given an integer array arr[] of size n, the task is to find the maximum of the minimums for every window size in the given array, where the window size ranges from 1 to n.Example:Input: arr[] = [10, 20, 30]Output: [30, 20, 10]Explanation: First element in output indicates maximum of minimums of all
    14 min read
    Find index of closing bracket for a given opening bracket in an expression
    Given a string with brackets. If the start index of the open bracket is given, find the index of the closing bracket. Examples: Input : string = [ABC[23]][89] index = 0 Output : 8 The opening bracket at index 0 corresponds to closing bracket at index 8.Recommended PracticeClosing bracket indexTry It
    7 min read
    Maximum difference between nearest left and right smaller elements
    Given an array of integers, the task is to find the maximum absolute difference between the nearest left and the right smaller element of every element in the array. Note: If there is no smaller element on right side or left side of any element then we take zero as the smaller element. For example f
    12 min read
    Delete consecutive same words in a sequence
    Given an array of n strings arr[]. The task is to determine the number of words remaining after pairwise destruction. If two consecutive words in the array are identical, they cancel each other out. This process continues until no more eliminations are possible. Examples: Input: arr[] = ["gfg", "for
    7 min read
    Check mirror in n-ary tree
    Given two n-ary trees, determine whether they are mirror images of each other. Each tree is described by e edges, where e denotes the number of edges in both trees. Two arrays A[]and B[] are provided, where each array contains 2*e space-separated values representing the edges of both trees. Each edg
    11 min read
    Reverse a number using stack
    Given a number , write a program to reverse this number using stack.Examples: Input : 365Output : 563Input : 6899Output : 9986We have already discussed the simple method to reverse a number in this post. In this post we will discuss about how to reverse a number using stack.The idea to do this is to
    5 min read
    Reversing the first K elements of a Queue
    Given an integer k and a queue of integers, The task is to reverse the order of the first k elements of the queue, leaving the other elements in the same relative order.Only following standard operations are allowed on the queue. enqueue(x): Add an item x to rear of queuedequeue(): Remove an item fr
    13 min read
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