最大化数组中 (a[i]+i)*(a[j]+j) 的值

给定一个输入大小为 n 的数组，求 (a[i] + i) * (a[j] + j) 的最大值，其中 i不等于j。
请注意， i 和 j 从0到n-1变化。
例子：

Input : a[] = [4,5,3,1,10]
Output : 84
Explanation:
We get the maximum value for i = 4 and j = 1
(10 + 4) * (5 + 1) = 84

Input : a[] = [10,0,0,0,-1]
Output : 30
Explanation:
We get the maximum value for i = 0 and j = 3
(10 + 0) * (0 + 3) = 30

天真的方法：最简单的方法是运行两个循环来考虑所有可能的对并跟踪表达式 (a[i]+i)*(a[j]+j) 的最大值。下面是这个想法的Python实现。时间复杂度为 O(n*n) ，其中 n 是输入大小。

C++

// C++ program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers. maxval()
// returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
#include
using namespace std;
 
int maxval(int a[], int n) {
 
        // at-least there must be two elements
        // in array
        if (n < 2) {
            return -99999;
        }
 
        // calculate maximum value
        int max = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int x = (a[i] + i) * (a[j] + j);
                if (max < x) {
                    max = x;
                }
            }
        }
 
        return max;
    }
 
        // test the function
    int main()
    {
        int arr[] = {4, 5, 3, 1, 10};
        int len = sizeof(arr)/sizeof(arr[0]);
        cout<<(maxval(arr, len));
    }
     
// This code is contributed by
// Shashank_Sharma

Java

// Java program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers. maxval()
// returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
 
public class GFG {
// Python
 
    static int maxval(int a[], int n) {
 
        // at-least there must be two elements
        // in array
        if (n < 2) {
            return -99999;
        }
 
        // calculate maximum value
        int max = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int x = (a[i] + i) * (a[j] + j);
                if (max < x) {
                    max = x;
                }
            }
        }
 
        return max;
    }
// test the function
 
    public static void main(String args[]) {
        int arr[] = {4, 5, 3, 1, 10};
        int len = arr.length;
        System.out.println(maxval(arr, len));
 
    }
}
 
/*This code is contributed by 29AjayKumar*/

Python3

# Python program to find maximum value (a[i]+i)*
# (a[j]+j) in an array of integers. maxval()
# returns maximum value of (a[i]+i)*(a[j]+j)
# where i is not equal to j
def maxval(a,n):
 
    # at-least there must be two elements
    # in array
    if (n < 2):
        return -99999
 
    # calculate maximum value
    max = 0
    for i in range(n):
        for j in range(i+1,n):
                x = (a[i]+i)*(a[j]+j)
                if max < x:
                    max = x
    return max
 
# test the function
print(maxval([4,5,3,1,10],5))

C#

// C# program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers. maxval()
// returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
using System;
public class GFG {
// Python
  
    static int maxval(int []a, int n) {
  
        // at-least there must be two elements
        // in array
        if (n < 2) {
            return -99999;
        }
  
        // calculate maximum value
        int max = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int x = (a[i] + i) * (a[j] + j);
                if (max < x) {
                    max = x;
                }
            }
        }
  
        return max;
    }
// test the function
  
    public static void Main() {
        int []arr = {4, 5, 3, 1, 10};
        int len = arr.Length;
        Console.Write(maxval(arr, len));
  
    }
}
  
/*This code is contributed by 29AjayKumar*/

Javascript

    // Javascript program to find maximum value (a[i]+i)*
    // (a[j]+j) in an array of integers. maxval()
    // returns maximum value of (a[i]+i)*(a[j]+j)
    // where i is not equal to j
     
    function maxval(a, n) {
    
        // at-least there must be two elements
        // in array
        if (n < 2) {
            return -99999;
        }
    
        // calculate maximum value
        let max = 0;
        for (let i = 0; i < n; i++) {
            for (let j = i + 1; j < n; j++) {
                let x = (a[i] + i) * (a[j] + j);
                if (max < x) {
                    max = x;
                }
            }
        }
    
        return max;
    }
     
    let arr = [4, 5, 3, 1, 10];
    let len = arr.length;
    document.write(maxval(arr, len));
                              

输出：

84

有效的方法：一种有效的方法是在数组中找到 a[i] + i 的最大值以及 a[i] + i 的第二个最大值。返回两个值的乘积。
查找最大值和第二个最大值可以在数组的单次遍历中完成。
因此，时间复杂度将为 O(n) 。
下面是这个想法的实现。

C++

// C++ program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers
// maxval() returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
#include
using namespace std;
#define MAX 5
 
int maxval(int a[MAX], int n)
{
 
    // there must be at-least two
    // elements in the array
    if (n < 2)
    {
        cout << "Invalid Input";
        return -9999;
    }
     
    // max1 will store the maximum value of
    // (a[i]+i)
    // max2 will store the second maximum value
    // of (a[i]+i)
    int max1 = 0, max2 = 0;
    for (int i = 0; i < n; i++)
    {
        int x = a[i] + i;
 
        // If current element x is greater than
        // first then update first and second
        if (x > max1)
        {
            max2 = max1;
            max1 = x;
        }
         
        // if x is in between max1 and
        // max2 then update max2
        else if (x > max2 & x != max1)
        {
            max2 = x;
        }
    }
    return (max1 * max2);
}
 
// Driver Code
int main()
{
    int arr[] = {4, 5, 3, 1, 10};
    int len = sizeof(arr)/arr[0];
    cout << maxval(arr, len);
}
 
// This code is contributed
// by Akanksha Rai

C

// C program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers
// maxval() returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
#include
#include
#define MAX 5
 
int maxval(int a[MAX], int n) {
 
    // there must be at-least two elements in
    // the array
    if (n < 2) {
        printf("Invalid Input");
        return -9999;
    }
    // max1 will store the maximum value of
    //      (a[i]+i)
    // max2 will store the second maximum value
    //      of (a[i]+i)
    int max1 = 0, max2 = 0;
    for (int i = 0; i < n; i++) {
        int x = a[i] + i;
 
        // If current element x is greater than
        // first then update first and second
        if (x > max1) {
            max2 = max1;
            max1 = x;
        }// if x is in between max1 and
            // max2 then update max2
        else if (x > max2 & x != max1) {
            max2 = x;
        }
    }
    return (max1 * max2);
 
    // test the function
}
 
int main() {
    int arr[] = {4, 5, 3, 1, 10};
    int len = sizeof(arr)/arr[0];
    printf("%d",maxval(arr, len));
}
// This code is contributed by 29AjayKumar

Java

// Java program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers
// maxval() returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
 
class GFG {
 
    static int maxval(int[] a, int n) {
 
        // there must be at-least two elements in
        // the array
        if (n < 2) {
            System.out.print("Invalid Input");
            return -9999;
        }
        // max1 will store the maximum value of
        //      (a[i]+i)
        // max2 will store the second maximum value
        //      of (a[i]+i)
        int max1 = 0, max2 = 0;
        for (int i = 0; i < n; i++) {
            int x = a[i] + i;
 
            // If current element x is greater than
            // first then update first and second
            if (x > max1) {
                max2 = max1;
                max1 = x;
            } // if x is in between max1 and
            // max2 then update max2
            else if (x > max2 & x != max1) {
                max2 = x;
            }
        }
        return (max1 * max2);
 
// test the function
    }
 
    public static void main(String[] args) {
        int arr[] = {4, 5, 3, 1, 10};
        int len = arr.length;
        System.out.println(maxval(arr, len));
    }
}
// This code is contributed by Rajput-Ji

Python3

# Python program to find maximum value (a[i]+i)*
# (a[j]+j) in an array of integers
# maxval() returns maximum value of (a[i]+i)*(a[j]+j)
# where i is not equal to j
def maxval(a,n):
 
    # there must be at-least two elements in
    # the array
    if (n < 2):
        print("Invalid Input")
        return -9999
 
    # max1 will store the maximum value of
    #      (a[i]+i)
    # max2 will store the second maximum value
    #      of (a[i]+i)
    (max1, max2) = (0, 0)
    for i in range(n):
        x = a[i] + i
 
        # If current element x is greater than
        # first then update first and second
        if (x > max1):
            max2 = max1
            max1 = x
 
        # if x is in between max1 and
        # max2 then update max2
        elif (x > max2 and x != max1):
             max2 = x
    return(max1*max2)
 
# test the function
print(maxval([4,5,3,1,10],5))

C＃

// C# program to find maximum value (a[i]+i)*
// (a[j]+j) in an array of integers
// maxval() returns maximum value of (a[i]+i)*(a[j]+j)
// where i is not equal to j
using System;   
public class GFG {
  
    static int maxval(int[] a, int n) {
  
        // there must be at-least two elements in
        // the array
        if (n < 2) {
            Console.WriteLine("Invalid Input");
            return -9999;
        }
        // max1 will store the maximum value of
        //      (a[i]+i)
        // max2 will store the second maximum value
        //      of (a[i]+i)
        int max1 = 0, max2 = 0;
        for (int i = 0; i < n; i++) {
            int x = a[i] + i;
  
            // If current element x is greater than
            // first then update first and second
            if (x > max1) {
                max2 = max1;
                max1 = x;
            } // if x is in between max1 and
            // max2 then update max2
            else if (x > max2 & x != max1) {
                max2 = x;
            }
        }
        return (max1 * max2);
  
// test the function
    }
  
    public static void Main() {
        int []arr = {4, 5, 3, 1, 10};
        int len = arr.Length;
        Console.WriteLine(maxval(arr, len));
    }
}
// This code is contributed by PrinciRaj1992

PHP

 $max1)
        {
            $max2 = $max1;
            $max1 = $x;
        } 
         
        // if x is in between max1 and
        // max2 then update max2
        else if (($x > $max2) & ($x != $max1))
        {
            $max2 = $x;
        }
    }
    return ($max1 * $max2);
}
 
// Driver Code
$arr = array(4, 5, 3, 1, 10);
$len = count($arr);
echo maxval($arr, $len);
 
// This code is contributed by ajit.
?>

Javascript

 
    // Javascript program to find maximum value (a[i]+i)*
    // (a[j]+j) in an array of integers
    // maxval() returns maximum value of (a[i]+i)*(a[j]+j)
    // where i is not equal to j
     
    function maxval(a, n)
    {
   
        // there must be at-least two elements in
        // the array
        if (n < 2) {
            document.write("Invalid Input");
            return -9999;
        }
        // max1 will store the maximum value of
        //      (a[i]+i)
        // max2 will store the second maximum value
        //      of (a[i]+i)
        let max1 = 0, max2 = 0;
        for (let i = 0; i < n; i++) {
            let x = a[i] + i;
   
            // If current element x is greater than
            // first then update first and second
            if (x > max1)
            {
                max2 = max1;
                max1 = x;
            }
            // if x is in between max1 and
            // max2 then update max2
            else if (x > max2 & x != max1) {
                max2 = x;
            }
        }
        return (max1 * max2);
   
        // test the function
    }
     
    let arr = [4, 5, 3, 1, 10];
    let len = arr.length;
    document.write(maxval(arr, len));
     

输出：

84