查找给定数组中作为剩余元素总和因子的元素

给定一个大小为N的数组A[] ，任务是找到数组中的元素，这些元素是剩余元素之和的因子。因此，只需从数组中选择一个元素并获取剩余元素的总和，然后检查总和是否可以被所选元素完全整除。如果它是可分的，则返回该元素。

例子：

Input: A[] = {2, 4, 6, 8, 10, 12, 14}
Output: [2, 4, 8, 14]
Explanation: 
1. Take sum for remaining element except selected one.
2. For element 2, sum of remaining element is 4+6+8+10+12+14=54 
3. Similarly for complete array: [54, 52, 50, 48, 46, 44, 42]
3. 54/2, 52/4, 48/8, 42/14 are perfectly divisible so resultant elements are [2, 4, 8, 14]

Input: A[]= {3, 6, 8, 10, 7, 15}
Output: [7]

朴素的方法：从数组中获取所有元素的总和。 现在从 sum 中逐个减去每个元素，并将其附加到新数组p[]。将每个总和除以给定数组中的相应索引元素，并将其附加到新数组q[]。将数组A[]和数组q[]中的相应元素相乘，并将其与数组p[]中类似的索引元素进行比较。如果它们相等，则将其附加到新数组z[ ] 。如果没有找到这样的元素，则返回-1。

下面是上述方法的实现。

C++

// c++ program for the above approach
#include 
using namespace std;
 
// Function to find element
vector Factor(vector A)
{
 
  // Sum of all element
  int s = 0;
 
  for (int i = 0; i < A.size(); i++)
  {
    s += A[i];
  }
 
  // Subtract each element from sum
  vector p;
  for (int i : A)
    p.push_back(s - i);
 
  // Divide corresponding element
  // from array p and l
  vector q;
  for (int i = 0; i < A.size(); i++)
    q.push_back(p[i] / A[i]);
 
  // Check sum is divisible by
  // corresponding element or not
  vector z;
  for (int i = 0; i < q.size(); i++)
  {
 
    // First we divided element now multiple
    // to check perfect divisibility of element
    if (q[i] * A[i] == p[i])
      z.push_back(A[i]);
  }
 
  return z;
}
 
// Driver code
int main()
{
  vector A = {2, 4, 6, 8, 10, 12, 14};
 
  // Calling function
  vector b = Factor(A);
 
  // Print required array
  for (auto i : b)
  {
    cout << i << " ";
  }
}
 
// This code is contributed by amreshkumar3.

Java

// Java program for the above approach
import java.util.ArrayList;
 
class GFG{
 
// Function to find element
static ArrayList Factor(int[] A)
{
     
    // Sum of all element
    int s = 0;
 
    for(int i = 0; i < A.length; i++)
    {
        s += A[i];
    }
 
    // Subtract each element from sum
    ArrayList p = new ArrayList<>();
    for(int i : A)
        p.add(s - i);
 
    // Divide corresponding element
    // from array p and l
    ArrayList q = new ArrayList();
    for(int i = 0; i < A.length; i++)
        q.add((int) Math.floor(p.get(i) / A[i]));
 
    // Check sum is divisible by
    // corresponding element or not
    ArrayList z = new ArrayList();
    for(int i = 0; i < q.size(); i++)
    {
         
        // First we divided element now multiple
        // to check perfect divisibility of element
        if (q.get(i) * A[i] == p.get(i))
            z.add(A[i]);
    }
     
    // If no such element found return -1
    if (z.size() == 0)
        return new ArrayList();
         
    return z;
}
 
// Driver code
public static void main(String args[])
{
    int[] A = { 2, 4, 6, 8, 10, 12, 14 };
 
    // Calling function
    ArrayList b = Factor(A);
 
    // Print required array
    System.out.println(b);
}
}
 
// This code is contributed by gfgking

Python3

# Python program for the above approach
 
# Function to find element
def Factor(A):
   
    # Sum of all element
    s = sum(A)
     
    # Subtract each element from sum
    p =[]
    for i in A:
        p.append(s-i)
 
    # Divide corresponding element
    # from array p and l
    q =[]
    for i in range(len(A)):
        q.append(p[i]//A[i])
 
    # Check sum is divisible by
    # corresponding element or not
    z =[]
    for i in range(len(q)):
       
          # First we divided element now multiple
        # to check perfect divisibility of element
        if q[i]*A[i]== p[i]:
            z.append(A[i])
             
    # If no such element found return -1
    if len(z)== 0:
      return -1
    return z
 
A = [2, 4, 6, 8, 10, 12, 14]
 
# Calling function
b = Factor(A)
 
# Print required array
print(b)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Function to find element
  static List Factor(int[] A)
  {
 
    // Sum of all element
    int s = 0;
 
    for(int i = 0; i < A.Length; i++)
    {
      s += A[i];
    }
 
    // Subtract each element from sum
    List p = new List();
    foreach(int i in A)
      p.Add(s - i);
 
    // Divide corresponding element
    // from array p and l
    List q = new List();
    for(int i = 0; i < A.Length; i++)
      q.Add((int) Math.Floor((double)p[i] / A[i]));
 
    // Check sum is divisible by
    // corresponding element or not
    List z = new List();
    for(int i = 0; i < q.Count; i++)
    {
 
      // First we divided element now multiple
      // to check perfect divisibility of element
      if (q[i] * A[i] == p[i])
        z.Add(A[i]);
    }
 
    // If no such element found return -1
    if (z.Count == 0)
      return new List();
 
    return z;
  }
 
  // Driver code
  public static void Main(String []args)
  {
    int[] A = { 2, 4, 6, 8, 10, 12, 14 };
 
    // Calling function
    List b = Factor(A);
 
    // Print required array
    foreach(int i in b)
      Console.Write(i+", ");
  }
}
 
 
// This code is contributed by 29AjayKumar

Javascript

      // JavaScript code for the above approach
 
      // Function to find element
      function Factor(A) {
 
          // Sum of all element
          let s = 0;
 
          for (let i = 0; i < A.length; i++) {
              s += A[i]
          }
 
          // Subtract each element from sum
          p = []
          for (i of A)
              p.push(s - i)
 
          // Divide corresponding element
          // from array p and l
          q = []
          for (i = 0; i < A.length; i++)
              q.push(Math.floor(p[i] / A[i]))
 
          // Check sum is divisible by
          // corresponding element or not
          z = []
          for (let i = 0; i < q.length; i++) {
 
              // First we divided element now multiple
              // to check perfect divisibility of element
              if (q[i] * A[i] == p[i])
                  z.push(A[i])
          }
          // If no such element found return -1
          if (z.length == 0)
              return -1
          return z
      }
      A = [2, 4, 6, 8, 10, 12, 14]
 
      // Calling function
      b = Factor(A)
 
      // Print required array
      document.write(b)
 
// This code is contributed by Potta Lokesh
  

输出

[2, 4, 8, 14]

时间复杂度： O(N)
辅助空间： O(N)

高效方法：在这种方法中，不需要使用多个循环和多个数组。所以空间复杂度和时间复杂度会降低。在这种情况下，所有的减法、除法、乘法运算都在一个循环中执行。请按照以下步骤解决问题：

• 将变量s初始化为数组A[] 的和。
• 初始化数组z[]以存储结果。
• 使用变量i遍历范围[0, len(A))并执行以下任务：
  • 将变量a初始化为sl[i]，将 b初始化为a/A[i]。
  • 如果b*A[i]等于a，则将A[i]附加到z[] 中。
• 执行上述步骤后，如果结果数组为空，则打印-1 ，否则打印数组z[]的元素作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find sum of all elements of an array
int sum(vector& A)
{
  int res = 0;
  for (auto it : A)
    res += it;
  return res;
}
 
// Function to find element
vector Factor(vector& A)
{
 
  // Sum of all element
  int s = sum(A);
  vector z;
 
  // Loop to find the factors of sum.
  for (int i = 0; i < A.size(); ++i) {
 
    // a is sum of remaining elements.
    int a = s - A[i];
 
    // b is integer value or factor of b.
    int b = a / A[i];
 
    // Check the divisibility
    if (b * A[i] == a)
      z.push_back(A[i]);
  }
 
  // If no element found return -1
  if (z.size() == 0)
    return { -1 };
 
  return z;
}
 
// Drive Code
 
int main()
{
  vector A = { 2, 4, 6, 8, 10, 12, 14 };
 
  // Calling function
  vector b = Factor(A);
 
  // Print resultant element
  for (auto it : b)
    cout << it << " ";
 
  return 0;
}
 
// This code is contributed by rakeshsahni

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find sum of all elements of an array
  static int sum(int[] A)
  {
    int res = 0;
    for (int i = 0; i < A.length; i++) {
      res += A[i];
    }
    return res;
  }
 
  // Function to find element
  static ArrayList Factor(int[] A)
  {
 
    // Sum of all element
    int s = sum(A);
    ArrayList z = new ArrayList();
 
    // Loop to find the factors of sum.
    for (int i = 0; i < A.length; ++i) {
 
      // a is sum of remaining elements.
      int a = s - A[i];
 
      // b is integer value or factor of b.
      int b = a / A[i];
 
      // Check the divisibility
      if (b * A[i] == a){
        z.add(A[i]);
      }
    }
 
    // If no element found return -1
    if (z.size() == 0){
      ArrayList l1 = new ArrayList();
      l1.add(-1);
      return l1;
    }
 
    return z;
  }
 
  // Drive Code
  public static void main (String[] args)
  {
    int A[] = new int[] { 2, 4, 6, 8, 10, 12, 14 };
 
    // Calling function
    ArrayList b = Factor(A);
 
    // Print resultant element
    System.out.println(b);
  }
}
 
// This code is contributed by Shubham Singh

Python3

# Python program for the above approach
 
# Function to find element
def Factor(A):
 
    # Sum of all element
    s = sum(A)
    z = []
 
    # Loop to find the factors of sum.
    for i in range(len(A)):
 
        # a is sum of remaining elements.
        a = s-A[i]
 
        # b is integer value or factor of b.
        b = a//A[i]
 
        # Check the divisibility
        if b * A[i] == a:
            z.append(A[i])
 
    # If no element found return -1
    if len(z) == 0:
        return -1
 
    return z
 
 
A = [2, 4, 6, 8, 10, 12, 14]
 
# Calling function
b = Factor(A)
 
# Print resultant element
print(b)

C＃

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find sum of all elements of an array
  static int sum(List A)
  {
    int res = 0;
    foreach(int it in A) res += it;
    return res;
  }
 
  // Function to find element
  static List Factor(List A)
  {
 
    // Sum of all element
    int s = sum(A);
    List z = new List();
 
    // Loop to find the factors of sum.
    for (int i = 0; i < A.Count; ++i) {
 
      // a is sum of remaining elements.
      int a = s - A[i];
 
      // b is integer value or factor of b.
      int b = a / A[i];
 
      // Check the divisibility
      if (b * A[i] == a)
        z.Add(A[i]);
    }
 
    // If no element found return -1
    if (z.Count == 0)
      return new List() { -1 };
 
    return z;
  }
 
  // Drive Code
 
  public static void Main()
  {
    List A
      = new List() { 2, 4, 6, 8, 10, 12, 14 };
 
    // Calling function
    List b = Factor(A);
 
    // Print resultant element
    Console.Write("[ ");
    int it;
    for (it = 0; it < b.Count - 1; it++) {
      Console.Write(b[it] + ", ");
    }
    Console.Write(b[it] + " ]");
  }
}
 
// This code is contributed by ukasp.

Javascript

// JavaScript program for the above approach
 
// Function to find sum of all elements of an array
function sum(A)
{
  var res = 0;
  for (var it of A){
    res += parseInt(it);}
  return res;
}
 
// Function to find element
function Factor(A)
{
 
  // Sum of all element
  var s = sum(A);
   
  var z =[];
 
  // Loop to find the factors of sum.
  for (var i = 0; i < A.length; ++i) {
 
    // a is sum of remaining elements.
    var a = s - A[i];
 
    // b is integer value or factor of b.
    var b = parseInt(a / A[i]);
 
    // Check the divisibility
    if (b * A[i] == a)
      z.push(A[i]);
  }
 
  // If no element found return -1
  if (z.length == 0)
    return [ -1 ];
 
  return z;
}
 
// Drive Code
A = [ 2, 4, 6, 8, 10, 12, 14 ];
 
// Calling function
b = Factor(A);
 
// Print resultant element
for (var it of b)
document.write(it + " ");
 
//This code is contributed by Shubham Singh

输出

[2, 4, 8, 14]

时间复杂度： O(N)
辅助空间： O(1)