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📜  使用给定的操作缩小数组后,找到最后一个元素的最大值

📅  最后修改于: 2021-05-08 17:38:34             🧑  作者: Mango

给定一个由N个元素组成的数组arr [] ,您必须对给定的数组执行以下操作,直到将数组简化为单个元素为止,

  1. 选择两个索引ij ,使i!= j
  2. arr [i]替换为arr [i] – arr [j],然后从数组中删除arr [j]

任务是最大化并打印数组最后剩余元素的值。

例子:

方法:为了最大化最后剩余元素的值,有以下三种情况:

  1. 数组同时具有负数和正数:首先,我们将从负数中减去所有正数(一个除外)。此后,我们将只剩下一个正数和一个负数。现在,我们将从正数中减去该负数,最终将得到一个正数。因此,在这种情况下,结果是数组元素的绝对值之和。
  2. 数组仅包含正数:首先我们找到最小的数,然后从中减去除一个正数外的所有正数。此后,我们只得到一个正数和一个负数,现在我们将从该正数中减去负数,最后将得到一个正数。在这里我们可以看到最小的
    number消失了,并且该值基本上也从与情况1不同的下一个更大的元素中切出。因此,在这种情况下,结果是数组元素的绝对值之和– 2 *最小元素。
  3. 数组仅包含负数:首先,我们找到最大数,然后从中减去除一个负数外的所有负数。此后,我们仅得到一个负数和一个正数,现在我们将从该正数中减去负数,最后将得到一个正数。在这里,我们可以观察到最大数目消失了,并且该值基本上也从与情况1不同的下一个更大的元素中切出。因此,在这种情况下,结果是数组元素的绝对值之和– 2 *绝对值最大的元素。在这里,我们以最大值为绝对数,在负数的情况下,最大值为最小值。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximized value
int find_maximum_value(int a[], int n)
{
    int sum = 0;
    int minimum = INT_MAX;
    int pos = 0, neg = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Overall minimum absolute value
        // of some element from the array
        minimum = min(minimum, abs(a[i]));
  
        // Add all absolute values
        sum += abs(a[i]);
  
        // Count positive and negative elements
        if (a[i] >= 0)
            pos += 1;
        else
            neg += 1;
    }
  
    // Both positive and negative
    // values are present
    if (pos > 0 && neg > 0)
        return sum;
  
    // Only positive or negative
    // values are present
    return (sum - 2 * minimum);
}
  
// Driver code
int main()
{
    int a[] = { 5, 4, 6, 2 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << find_maximum_value(a, n);
  
    return 0;
}


Java
// Java implementation of the approach 
import java.io.*;
  
class GFG 
{
      
    // Function to return the maximized value 
    static int find_maximum_value(int a[], int n) 
    { 
        int sum = 0; 
        int minimum = Integer.MAX_VALUE; 
        int pos = 0, neg = 0; 
      
        for (int i = 0; i < n; i++) 
        { 
      
            // Overall minimum absolute value 
            // of some element from the array 
            minimum = Math.min(minimum, Math.abs(a[i])); 
      
            // Add all absolute values 
            sum += Math.abs(a[i]); 
      
            // Count positive and negative elements 
            if (a[i] >= 0) 
                pos += 1; 
            else
                neg += 1; 
        } 
      
        // Both positive and negative 
        // values are present 
        if (pos > 0 && neg > 0) 
            return sum; 
      
        // Only positive or negative 
        // values are present 
        return (sum - 2 * minimum); 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    {
          
        int []a = { 5, 4, 6, 2 }; 
        int n = a.length; 
      
        System.out.println(find_maximum_value(a, n)); 
    }
}
  
// This code is contributed by ajit


Python
# Python3 implementation of the approach
  
# Function to return the maximized value
def find_maximum_value(a, n):
      
    sum = 0
    minimum = 10**9
    pos = 0
    neg = 0
  
    for i in range(n):
  
        # Overall minimum absolute value
        # of some element from the array
        minimum = min(minimum, abs(a[i]))
  
        # Add all absolute values
        sum += abs(a[i])
  
        # Count positive and negative elements
        if (a[i] >= 0):
            pos += 1
        else:
            neg += 1
  
    # Both positive and negative
    # values are present
    if (pos > 0 and neg > 0):
        return sum
  
    # Only positive or negative
    # values are present
    return (sum - 2 * minimum)
  
# Driver code
  
a= [5, 4, 6, 2]
n = len(a)
  
print(find_maximum_value(a, n))
  
# This code is contributed by mohit kumar 29


C#
// C# implementation of the approach 
using System;
  
class GFG
{
  
    // Function to return the maximized value 
    static int find_maximum_value(int []a, int n) 
    { 
        int sum = 0; 
        int minimum = int.MaxValue; 
        int pos = 0, neg = 0; 
      
        for (int i = 0; i < n; i++) 
        { 
      
            // Overall minimum absolute value 
            // of some element from the array 
            minimum = Math.Min(minimum, Math.Abs(a[i])); 
      
            // Add all absolute values 
            sum += Math.Abs(a[i]); 
      
            // Count positive and negative elements 
            if (a[i] >= 0) 
                pos += 1; 
            else
                neg += 1; 
        } 
      
        // Both positive and negative 
        // values are present 
        if (pos > 0 && neg > 0) 
            return sum; 
      
        // Only positive or negative 
        // values are present 
        return (sum - 2 * minimum); 
    } 
      
    // Driver code 
    static public void Main ()
    {
        int []a = { 5, 4, 6, 2 }; 
        int n = a.Length; 
      
        Console.WriteLine(find_maximum_value(a, n)); 
    }
}
  
// This code is contributed by AnkitRai01


输出:
13

时间复杂度: O(N)