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📜  通过从给定数组的子数组的所有元素中减去相同的值来最大可能地求和

📅  最后修改于: 2021-04-28 17:30:41             🧑  作者: Mango

给定一个由N个整数组成的数组a [] ,任务是找到可以通过从子数组所有元素中减去任意值(例如X)来实现的最大和。

例子:

方法:
请按照以下步骤解决问题:

  • 遍历数组
  • 对于每一个元素,发现这是最近在其左侧,其右侧最接近的较小的元素。
  • 通过该元素计算current_element *(j – i – 1)来计算可能的总和,其中ji分别是左侧和右侧最接近的较小数字的索引。
  • 在所有这些中找到最大可能的和。

下面是上述方法的实现:

C++
// C++ program to implement
// the above approach
#include 
using namespace std;
  
// Function to generate previous smaller 
// element for each array element
vector findPrevious(vector a, int n)
{
    vector ps(n);
  
    // The first element has no
    // previous smaller
    ps[0] = -1;
  
    // Stack to keep track of elements
    // that have occurred previously
    stack Stack;
  
    // Push the first index
    Stack.push(0);
      
    for(int i = 1; i < n; i++)
    {
          
        // Pop all the elements until the previous
        // element is smaller than current element
        while (Stack.size() > 0 && 
             a[Stack.top()] >= a[i])
            Stack.pop();
  
        // Store the previous smaller element
        ps[i] = Stack.size() > 0 ? 
                Stack.top() : -1;
  
        // Push the index of the current element
        Stack.push(i);
    }
  
    // Return the array
    return ps;
}
  
// Function to generate next smaller element
// for each array element
vector findNext(vector a, int n)
{
    vector ns(n);
  
    ns[n - 1] = n;
  
    // Stack to keep track of elements
    // that have occurring next
    stack Stack;
    Stack.push(n - 1);
  
    // Iterate in reverse order
    // for calculating next smaller
    for(int i = n - 2; i >= 0; i--)
    {
          
        // Pop all the elements until the
        // next element is smaller
        // than current element
        while (Stack.size() > 0 && 
             a[Stack.top()] >= a[i])
            Stack.pop();
  
        // Store the next smaller element
        ns[i] = Stack.size() > 0 ? 
                Stack.top() : n;
  
        // Push the index of the current element
        Stack.push(i);
    }
  
    // Return the array
    return ns;
}
  
// Function to find the maximum sum by
// subtracting same value from all
// elements of a Subarray
int findMaximumSum(vector a, int n)
{
      
    // Stores previous smaller element
    vector prev_smaller = findPrevious(a, n);
  
    // Stores next smaller element
    vector next_smaller = findNext(a, n);
  
    int max_value = 0;
    for(int i = 0; i < n; i++)
    {
          
        // Calculate contribution
        // of each element
        max_value = max(max_value, a[i] * 
                       (next_smaller[i] - 
                        prev_smaller[i] - 1));
    }
  
    // Return answer
    return max_value;
}
  
// Driver Code    
int main()
{
    int n = 3;
    vector a{ 80, 48, 82 };
      
    cout << findMaximumSum(a, n);
      
    return 0;
}
  
// This code is contributed by divyeshrabadiya07


Java
// Java Program to implement
// the above approach
import java.util.*;
  
public class GFG {
  
    // Function to find the maximum sum by
    // subtracting same value from all
    // elements of a Subarray
    public static int findMaximumSum(int[] a, int n)
    {
        // Stores previous smaller element
        int prev_smaller[] = findPrevious(a, n);
  
        // Stores next smaller element
        int next_smaller[] = findNext(a, n);
  
        int max_value = 0;
        for (int i = 0; i < n; i++) {
  
            // Calculate contribution
            // of each element
            max_value
                = Math.max(max_value,
                        a[i] * (next_smaller[i]
                                - prev_smaller[i] - 1));
        }
  
        // Return answer
        return max_value;
    }
  
    // Function to generate previous smaller element
    // for each array element
    public static int[] findPrevious(int[] a, int n)
    {
        int ps[] = new int[n];
  
        // The first element has no
        // previous smaller
        ps[0] = -1;
  
        // Stack to keep track of elements
        // that have occurred previously
        Stack stack = new Stack<>();
  
        // Push the first index
        stack.push(0);
        for (int i = 1; i < a.length; i++) {
  
            // Pop all the elements until the previous
            // element is smaller than current element
            while (stack.size() > 0
                && a[stack.peek()] >= a[i])
                stack.pop();
  
            // Store the previous smaller element
            ps[i] = stack.size() > 0 ? stack.peek() : -1;
  
            // Push the index of the current element
            stack.push(i);
        }
  
        // Return the array
        return ps;
    }
  
    // Function to generate next smaller element
    // for each array element
    public static int[] findNext(int[] a, int n)
    {
        int ns[] = new int[n];
  
        ns[n - 1] = n;
  
        // Stack to keep track of elements
        // that have occurring next
        Stack stack = new Stack<>();
        stack.push(n - 1);
  
        // Iterate in reverse order
        // for calculating next smaller
        for (int i = n - 2; i >= 0; i--) {
  
            // Pop all the elements until the
            // next element is smaller
            // than current element
            while (stack.size() > 0
                && a[stack.peek()] >= a[i])
                stack.pop();
  
            // Store the next smaller element
            ns[i] = stack.size() > 0 ? stack.peek()
                                    : a.length;
  
            // Push the index of the current element
            stack.push(i);
        }
  
        // Return the array
        return ns;
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int n = 3;
        int a[] = { 80, 48, 82 };
        System.out.println(findMaximumSum(a, n));
    }
}


Python3
# Python3 program to implement 
# the above approach 
  
# Function to find the maximum sum by 
# subtracting same value from all 
# elements of a Subarray 
def findMaximumSum(a, n):
      
    # Stores previous smaller element 
    prev_smaller = findPrevious(a, n)
      
    # Stores next smaller element
    next_smaller = findNext(a, n)
      
    max_value = 0
    for i in range(n):
          
        # Calculate contribution 
        # of each element 
        max_value = max(max_value, a[i] *
                    (next_smaller[i] -
                        prev_smaller[i] - 1))
          
    # Return answer
    return max_value
  
# Function to generate previous smaller 
# element for each array element 
def findPrevious(a, n):
      
    ps = [0] * n
      
    # The first element has no 
    # previous smaller 
    ps[0] = -1
      
    # Stack to keep track of elements 
    # that have occurred previously 
    stack = []
      
    # Push the first index
    stack.append(0)
      
    for i in range(1, n):
          
        # Pop all the elements until the previous 
        # element is smaller than current element 
        while len(stack) > 0 and a[stack[-1]] >= a[i]:
            stack.pop()
              
        # Store the previous smaller element 
        ps[i] = stack[-1] if len(stack) > 0 else -1
          
        # Push the index of the current element
        stack.append(i)
          
    # Return the array 
    return ps
  
# Function to generate next smaller 
# element for each array element 
def findNext(a, n):
      
    ns = [0] * n
    ns[n - 1] = n
      
    # Stack to keep track of elements 
    # that have occurring next 
    stack = []
    stack.append(n - 1)
      
    # Iterate in reverse order 
    # for calculating next smaller
    for i in range(n - 2, -1, -1):
          
        # Pop all the elements until the 
        # next element is smaller 
        # than current element 
        while (len(stack) > 0 and
                a[stack[-1]] >= a[i]):
            stack.pop()
          
        # Store the next smaller element 
        ns[i] = stack[-1] if len(stack) > 0 else n
          
        # Push the index of the current element
        stack.append(i)
          
    # Return the array
    return ns
  
# Driver code
n = 3
a = [ 80, 48, 82 ]
  
print(findMaximumSum(a, n))
  
# This code is contributed by Stuti Pathak


C#
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
  
// Function to find the maximum sum by
// subtracting same value from all
// elements of a Subarray
public static int findMaximumSum(int[] a, int n)
{
    // Stores previous smaller element
    int []prev_smaller = findPrevious(a, n);
  
    // Stores next smaller element
    int []next_smaller = findNext(a, n);
  
    int max_value = 0;
    for (int i = 0; i < n; i++)
    {
  
    // Calculate contribution
    // of each element
    max_value = Math.Max(max_value,
                a[i] * (next_smaller[i] - 
                        prev_smaller[i] - 1));
    }
  
    // Return answer
    return max_value;
}
  
// Function to generate previous smaller element
// for each array element
public static int[] findPrevious(int[] a, int n)
{
    int []ps = new int[n];
  
    // The first element has no
    // previous smaller
    ps[0] = -1;
  
    // Stack to keep track of elements
    // that have occurred previously
    Stack stack = new Stack();
  
    // Push the first index
    stack.Push(0);
    for (int i = 1; i < a.Length; i++)
    {
  
    // Pop all the elements until the previous
    // element is smaller than current element
    while (stack.Count > 0 && 
            a[stack.Peek()] >= a[i])
        stack.Pop();
  
    // Store the previous smaller element
    ps[i] = stack.Count > 0 ? stack.Peek() : -1;
  
    // Push the index of the current element
    stack.Push(i);
    }
  
    // Return the array
    return ps;
}
  
// Function to generate next smaller element
// for each array element
public static int[] findNext(int[] a, int n)
{
    int []ns = new int[n];
  
    ns[n - 1] = n;
  
    // Stack to keep track of elements
    // that have occurring next
    Stack stack = new Stack();
    stack.Push(n - 1);
  
    // Iterate in reverse order
    // for calculating next smaller
    for (int i = n - 2; i >= 0; i--) 
    {
  
    // Pop all the elements until the
    // next element is smaller
    // than current element
    while (stack.Count > 0 && 
            a[stack.Peek()] >= a[i])
        stack.Pop();
  
    // Store the next smaller element
    ns[i] = stack.Count > 0 ? stack.Peek()
        : a.Length;
  
    // Push the index of the current element
    stack.Push(i);
    }
  
    // Return the array
    return ns;
}
  
// Driver Code
public static void Main(String []args)
{
    int n = 3;
    int []a = { 80, 48, 82 };
    Console.WriteLine(findMaximumSum(a, n));
}
}
  
// This code is contributed by Amit Katiyar


输出:
144

时间复杂度: O(N)
辅助空间: O(N)