📜  最大长度的子序列,带有交替的符号和最大和

📅  最后修改于: 2021-05-08 17:37:59             🧑  作者: Mango

给定大小为n的数组arr [] ,它的正整数和负整数都排除零。任务是找到具有最大大小和最大和的交替符号的子序列,即在每个相邻元素的子序列符号中是相反的,例如,如果第一个为正,则第二个必须为负,然后为另一个正整数等等。

例子:

Input: arr[] = {2, 3, 7, -6, -4}
Output: 7 -4
Explanation:
Possible subsequences are [2, -6] [2, -4] [3, -6] [3, -4] [7, -6] [7, -4].
Out of these [7, -4] has the maximum sum. 

Input: arr[] = {-4, 9, 4, 11, -5, -17, 9, -3, -5, 2}
Output: -4 11 -5 9 -3 2  

方法:

解决上述问题的主要思想是包含相同符号的数组找到最大元素,这意味着我们必须从连续的正负元素中选择最大的元素。由于我们需要最大大小,因此我们将仅从每个段中获取一个元素,并且为了使总和最大化,我们需要获取每个段中的最大元素。

下面是上述方法的实现:

CPP
// C++ implementation to find the
// subsequence with alternating sign
// having maximum size and maximum sum.
  
#include 
using namespace std;
  
// Function to find the subsequence
// with alternating sign having
// maximum size and maximum sum.
void findSubsequence(int arr[], int n)
{
    int sign[n] = { 0 };
  
    // Find whether each element
    // is positive or negative
    for (int i = 0; i < n; i++) {
        if (arr[i] > 0)
            sign[i] = 1;
        else
            sign[i] = -1;
    }
  
    int k = 0;
    int result[n] = { 0 };
  
    // Find the required subsequence
    for (int i = 0; i < n; i++) {
  
        int cur = arr[i];
        int j = i;
  
        while (j < n && sign[i] == sign[j]) {
  
            // Find the maximum element
            // in the specified range
            cur = max(cur, arr[j]);
            ++j;
        }
  
        result[k++] = cur;
  
        i = j - 1;
    }
  
    // print the result
    for (int i = 0; i < k; i++)
        cout << result[i] << " ";
    cout << "\n";
}
  
// Driver code
int main()
{
    // array declaration
    int arr[] = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };
  
    // size of array
    int n = sizeof(arr) / sizeof(arr[0]);
  
    findSubsequence(arr, n);
  
    return 0;
}


Java
// Java implementation to find the
// subsequence with alternating sign
// having maximum size and maximum sum.
class GFG{
   
// Function to find the subsequence
// with alternating sign having
// maximum size and maximum sum.
static void findSubsequence(int arr[], int n)
{
    int sign[] = new int[n];
   
    // Find whether each element
    // is positive or negative
    for (int i = 0; i < n; i++) {
        if (arr[i] > 0)
            sign[i] = 1;
        else
            sign[i] = -1;
    }
   
    int k = 0;
    int result[] = new int[n];
   
    // Find the required subsequence
    for (int i = 0; i < n; i++) {
   
        int cur = arr[i];
        int j = i;
   
        while (j < n && sign[i] == sign[j]) {
   
            // Find the maximum element
            // in the specified range
            cur = Math.max(cur, arr[j]);
            ++j;
        }
   
        result[k++] = cur;
   
        i = j - 1;
    }
   
    // print the result
    for (int i = 0; i < k; i++)
        System.out.print(result[i]+ " ");
    System.out.print("\n");
}
   
// Driver code
public static void main(String[] args)
{
    // array declaration
    int arr[] = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };
   
    // size of array
    int n = arr.length;
   
    findSubsequence(arr, n);
}
}
  
// This code is contributed by Princi Singh


Python3
# Python3 implementation to find the
# subsequence with alternating sign
# having maximum size and maximum sum.
  
# Function to find the subsequence
# with alternating sign having
# maximum size and maximum sum.
def findSubsequence(arr, n):
    sign = [0]*n
  
    # Find whether each element
    # is positive or negative
    for i in range(n):
        if (arr[i] > 0):
            sign[i] = 1
        else:
            sign[i] = -1
  
    k = 0
    result = [0]*n
  
    # Find the required subsequence
    i = 0
    while i < n:
  
        cur = arr[i]
        j = i
  
        while (j < n and sign[i] == sign[j]):
  
            # Find the maximum element
            # in the specified range
            cur = max(cur, arr[j])
            j += 1
  
        result[k] = cur
        k += 1
  
        i = j - 1
        i += 1
  
    # print the result
    for i in range(k):
        print(result[i],end=" ")
  
# Driver code
if __name__ == '__main__':
    # array declaration
    arr=[-4, 9, 4, 11, -5, -17, 9, -3, -5, 2]
  
    # size of array
    n = len(arr)
  
    findSubsequence(arr, n)
  
# This code is contributed by mohit kumar 29


C#
// C# implementation to find the
// subsequence with alternating sign
// having maximum size and maximum sum.
using System;
  
public class GFG{
    
// Function to find the subsequence
// with alternating sign having
// maximum size and maximum sum.
static void findSubsequence(int []arr, int n)
{
    int []sign = new int[n];
    
    // Find whether each element
    // is positive or negative
    for (int i = 0; i < n; i++) {
        if (arr[i] > 0)
            sign[i] = 1;
        else
            sign[i] = -1;
    }
    
    int k = 0;
    int []result = new int[n];
    
    // Find the required subsequence
    for (int i = 0; i < n; i++) {
    
        int cur = arr[i];
        int j = i;
    
        while (j < n && sign[i] == sign[j]) {
    
            // Find the maximum element
            // in the specified range
            cur = Math.Max(cur, arr[j]);
            ++j;
        }
    
        result[k++] = cur;
    
        i = j - 1;
    }
    
    // print the result
    for (int i = 0; i < k; i++)
        Console.Write(result[i]+ " ");
    Console.Write("\n");
}
    
// Driver code
public static void Main(String[] args)
{
    // array declaration
    int []arr = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };
    
    // size of array
    int n = arr.Length;
    
    findSubsequence(arr, n);
}
}
// This code contributed by Rajput-Ji


输出:
-4 11 -5 9 -3 2

时间复杂度: O(N)