左右下一个较大的索引的最大乘积

给定一个数组 a[1..N]。对于位置 i (1 <= i <= N) 处的每个元素。在哪里

1. L(i)被定义为最接近的索引 j，使得 j < i 并且 a[j] > a[i]。如果不存在这样的 j 则L(i) = 0 。
2. R(i)被定义为最接近的索引 k，使得 k > i 并且 a[k] > a[i]。如果不存在这样的 k 则R(i) = 0 。

LRProduct(i) = L(i)*R(i) 。
我们需要找到一个具有最大 LRProduct 的索引

例子：

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. 

注意：以起始索引为 1 查找 LRproduct。

这个问题基于 Next Greater Element。
从当前位置开始，我们需要在其左右两侧找到最近的较大元素。
所以为了找到下一个更大的元素，我们使用了从左到右的堆栈。简单地说，我们正在检查哪个元素更大并将它们的索引存储在指定的位置。
1-如果堆栈为空，则推送当前索引。
2-如果堆栈不为空
....a）如果当前元素大于顶部元素，则将当前元素的索引存储在顶部元素的索引上。
这样做，一次从左侧遍历数组元素，一次从右侧遍历数组元素并形成左右数组，然后将它们相乘以找到最大乘积值。

C++

// C++ program to find the max
// LRproduct[i] among all i
#include 
using namespace std;
#define MAX 1000
 
// function to find just next greater
// element in left side
vector nextGreaterInLeft(int a[], int n)
{
    vector left_index(MAX, 0);
    stack 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 nextGreaterInRight(int a[], int n)
{
    vector right_index(MAX, 0);
    stack 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 left = nextGreaterInLeft(arr, n);
 
    // for each element storing the index of just
    // greater element in right side
    vector 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 s = new Stack();
     
        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 s = new Stack();
        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 s = new Stack();
     
        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 s = new Stack();
        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

Javascript

    // Javascript program to find the max LRproduct[i] among all i
     
    let MAX = 1000;
      
    // function to find just next greater
    // element in left side
    function nextGreaterInLeft(a, n)
    {
        let left_index = new Array(MAX);
        left_index.fill(0);
        let s = [];
      
        for (let i = n - 1; i >= 0; i--) {
      
            // checking if current element is
            // greater than top
            while (s.length != 0 && a[i] > a[s[s.length - 1] - 1])
            {
                let r = s[s.length - 1];
                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
    function nextGreaterInRight(a, n)
    {
        let right_index = new Array(MAX);
        right_index.fill(0);
        let s = [];
        for (let i = 0; i < n; ++i) {
      
            // checking if current element is
            // greater than top
            while (s.length != 0 && a[i] > a[s[s.length - 1] - 1])
            {
                let r = s[s.length - 1];
                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
    function LRProduct(arr, n)
    {
          
        // for each element storing the index of just
        // greater element in left side
        let left = nextGreaterInLeft(arr, n);
      
        // for each element storing the index of just
        // greater element in right side
        let right = nextGreaterInRight(arr, n);
        let ans = -1;
        for (let i = 1; i <= n; i++) {
      
            // finding the max index product
            ans = Math.max(ans, left[i] * right[i]);
        }
      
        return ans;
    }
     
    let arr = [ 5, 4, 3, 4, 5 ];
    let n = arr.length;
 
    document.write(LRProduct(arr, n));
     
    // This code is contributed by suresh07.

---

输出：

8

方法二：减少使用的空间，只用一个数组存储左右最大值。

方法：

先决条件：https://www.geeksforgeeks.org/next-greater-element/

• 为了找到左边的下一个更大的元素，我们使用了从左边开始的堆栈，并且相同的堆栈用于将右侧最大元素索引与左侧最大元素索引相乘。
• 函数maxProduct( ) 用于通过迭代结果数组来返回最大产品。

C++

// C++ program to find max LR product
#include 
using namespace std;
 
stack mystack;
      
// To find greater element to left
void nextGreaterToLeft(int arr[], int res[], int N) {
    mystack.push(0);
    res[0] = 0;
      
    // iterate through the array
    for(int i = 1; i < N; i++) {
        while(mystack.size() > 0  &&  arr[mystack.top()] <= arr[i])
            mystack.pop();
          
        // place the index to the left in the resultant array
        res[i] = (mystack.size() == 0) ? 0 : mystack.top()+1;
        mystack.push(i);
    }
}
  
//// To find greater element to right
void nextGreaterToRight(int arr[], int res[], int N) {
    mystack = stack();
      
    int n = N;
    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.size() > 0  &&  arr[mystack.top()] <= arr[i])
            mystack.pop();
          
        //multiply the index to the right with the index to the left
        //in the resultant array
        res[i] = (mystack.size() == 0) ? res[i]*0 : res[i]*(mystack.top()+1);
        mystack.push(i);
    }
}
  
//function to return the max value in the resultant array
int maxProduct(int arr[], int res[], int N) {
    nextGreaterToLeft(arr,res, N);        //to find left max
    nextGreaterToRight(arr,res, N);    //to find right max
 
    int Max = res[0];
    for(int i = 1; i < N; i++){
        Max = max(Max, res[i]);
    }
    return Max;
}
     
int main()
{
    int arr[] = {5, 4, 3, 4, 5};
    int N = sizeof(arr) / sizeof(arr[0]);
    int res[N];
      
    int maxprod = maxProduct(arr, res, N);
    cout << maxprod << endl;
 
    return 0;
}
 
// This code is contributed by decode2207.

Java

//java program to find max LR product
 
import java.util.*;
 
public class GFG {
    Stack 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=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 value 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

Python3

# Python3 program to find max LR product
mystack = []
       
# To find greater element to left
def nextGreaterToLeft(arr, res):
    mystack.append(0)
    res[0] = 0
       
    # iterate through the array
    for i in range(1, len(arr)):
        while(len(mystack) > 0  and arr[mystack[-1]] <= arr[i]):
            mystack.pop()
           
        # place the index to the left in the resultant array
        if (len(mystack) == 0):
            res[i] = 0
        else:
            res[i] = mystack[-1]+1
        mystack.append(i)
   
# To find greater element to right
def nextGreaterToRight(arr, res):
    mystack = []
       
    n = len(arr)
    mystack.append(n-1)
    res[n-1] *= 0
       
    # iterate through the array in the reverse order
    for i in range(n - 2, -1, -1):
        while(len(mystack) > 0  and arr[mystack[-1]] <= arr[i]):
            mystack.pop()
           
        # multiply the index to the right with the index to the left
        # in the resultant array
        if (len(mystack) == 0):
            res[i] = res[i]*0
        else :
            res[i] = res[i]*(mystack[-1]+1)
        mystack.append(i)
   
# function to return the max value in the resultant array
def maxProduct(arr, res):
    nextGreaterToLeft(arr,res)       #to find left max
    nextGreaterToRight(arr,res)      #to find right max
 
    Max = res[0]
    for i in range(1, len(res)):
        Max = max(Max, res[i])
    return Max
 
  # Driver code
arr = [5, 4, 3, 4, 5]
res = [0]*(len(arr))
 
maxprod = maxProduct(arr, res)
print(maxprod)
 
# This code is contributed by mukesh07.

C＃

// C# program to find max LR product
using System;
using System.Collections;
class GFG {
 
    static Stack mystack = new Stack();
      
    //To find greater element to left
    static void nextGreaterToLeft(int[] arr,int[] res) {
        mystack.Push(0);
        res[0] = 0;
          
        //iterate through the array
        for(int i=1;i 0  &&  arr[(int)mystack.Peek()] <= arr[i])
                mystack.Pop();
              
            //place the index to the left in the resultant array
            res[i] = (mystack.Count == 0) ? 0 : (int)mystack.Peek()+1;
            mystack.Push(i);
        }
    }
      
    ////To find greater element to right
    static void nextGreaterToRight(int[] arr,int[] res) {
        mystack = new Stack();
          
        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.Count == 0  &&  arr[(int)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.Count == 0) ? res[i]*0 : res[i]*((int)mystack.Peek()+1);
            mystack.Push(i);
        }
    }
      
    //function to return the max value in the resultant array
    static 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;
    }
     
  static void Main() {
    int[] arr = {5, 4, 3, 4, 5};
    int[] res = new int[arr.Length];
      
    int maxprod = maxProduct(arr, res);
    Console.WriteLine(maxprod);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

    // Javascript program to find max LR product
     
    let mystack = [];
      
    // To find greater element to left
    function nextGreaterToLeft(arr, res) {
        mystack.push(0);
        res[0] = 0;
          
        // iterate through the array
        for(let i = 1; i < arr.length; i++) {
            while(mystack.length > 0  &&  arr[mystack[mystack.length - 1]] <= arr[i])
                mystack.pop();
              
            // place the index to the left in the resultant array
            res[i] = (mystack.length == 0) ? 0 : mystack[mystack.length - 1]+1;
            mystack.push(i);
        }
    }
      
    //// To find greater element to right
    function nextGreaterToRight(arr, res) {
        mystack = [];
          
        let n = arr.length;
        mystack.push(n-1);
        res[n-1] *= 0;
          
        // iterate through the array in the reverse order
        for(let i = n - 2; i >= 0; i--) {
            while(mystack.length > 0  &&  arr[mystack[mystack.length - 1]] <= arr[i])
                mystack.pop();
              
            // multiply the index to the right with the index to the left
            // in the resultant array
            res[i] = (mystack.length == 0) ? res[i]*0 : res[i]*(mystack[mystack.length - 1]+1);
            mystack.push(i);
        }
    }
      
    // function to return the max value in the resultant array
    function maxProduct(arr, res) {
        nextGreaterToLeft(arr,res);        //to find left max
        nextGreaterToRight(arr,res);    //to find right max
  
        let max = res[0];
        for(let i = 1;i<res.length;i++){
            max = Math.max(max, res[i]);
        }
        return max;
    }
     
    let arr = [5, 4, 3, 4, 5];
    let res = new Array(arr.length);
 
    let maxprod = maxProduct(arr, res);
    document.write(maxprod);
     
    // This code is contributed by divyesh072019.

输出

8

时间复杂度： O(n)