给定一个最大为n的自然数数组 arr ,找出数组arr可以分成两个相等大小的数组的最大大小,这样第一个数组包含所有相同的元素,而第二个数组包含所有不同的元素。
例子:
Input :n = 8, arr[] ={7, 3, 7, 1, 7, 7}
Output :
Maximum size is : 3
arr1[] ={7, 7, 7}
arr2[] ={1, 3, 7}
Explanation :
It is possible to construct two arrays of size 3.
The first array is [7, 7, 7] and the second array is [1, 3, 7].
Input : n = 7, arr[] ={1, 2, 1, 5, 1, 6, 7, 2}
Output :
Maximum size is : 3
arr1[] ={1, 1, 1}
arr2[] ={2, 5, 6}
方法:
为了解决上面提到的问题,主要思想是使用散列来找到数组中每个元素的频率。
- 使用哈希向量 v 查找数组 arr[] 中出现的频率最高的元素。
- 查找数组 arr[] 中存在的所有唯一元素。
- 有两种情况与最大频率的元件:最大频率元件将进入第一阵列然后在阵列的尺寸至多DIFF1 – 1和MAX1相应。否则,最大频率中的至少一个元素进行到所述第二阵列和所述尺寸是至多DIFF1和MAX1? 1相应。然后找到可以将数组拆分为max(min(diff1 ? 1, max1), min(diff1, max1 ? 1)) 的最大大小。
- 使用 max_size 和具有最大频率 max1 的元素查找相似元素的第一个数组。
- 使用 max_size 和哈希向量 v 找到第二个唯一元素数组。
下面是上述方法的实现:
C++
// C++ program to find the max-size to which
// an array can be divided into 2 equal parts
// such that one part contains unique elements
// while another contains similar elements
#include
using namespace std;
// Function to find the max-size to which an array
// can be divided into 2 equal parts
void Solve(int arr[], int size, int n)
{
vector v(n + 1);
// Vector to find the frequency of
// each element of array
for (int i = 0; i < size; i++)
v[arr[i]]++;
// Find the maximum frequency
// element present in array arr[]
int max1 = (max_element(v.begin(), v.end())
- v.begin());
// Find total unique elements
// present in array arr[]
int diff1 = n + 1 - count(v.begin(), v.end(), 0);
// Find the Max-Size to which
// an array arr[] can be splitted
int max_size = max(min(v[max1] - 1, diff1),
min(v[max1], diff1 - 1));
cout << "Maximum size is :" << max_size << "\n";
// Find the first array
// containing same elements
cout << "The First Array Is : \n";
for (int i = 0; i < max_size; i++) {
cout << max1 << " ";
v[max1] -= 1;
}
cout << "\n";
// Find the second array
// containing unique elements
cout << "The Second Array Is : \n";
for (int i = 0; i < (n + 1); i++) {
if (v[i] > 0) {
cout << i << " ";
max_size--;
}
if (max_size < 1)
break;
}
cout << "\n";
}
// Driver code
int main()
{
// initialise n
int n = 7;
// array declaration
int arr[] = { 1, 2, 1, 5, 1, 6, 7, 2 };
// size of array
int size = sizeof(arr) / sizeof(arr[0]);
Solve(arr, size, n);
return 0;
}
Java
// Java program to find the
// max-size to which an array
// can be divided into 2 equal parts
// such that one part contains unique
// elements while another contains
// similar elements
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to find the max-size to
// which an array can be divided into
// 2 equal parts
static void Solve(int arr[],
int size, int n)
{
int[] v = new int[n + 1];
// Array to find the frequency of
// each element of array
for (int i = 0; i < size; i++)
v[arr[i]]++;
// Find the index maximum frequency
// element present in array arr[]
int max1 = -1, mx = -1;
for (int i = 0; i < v.length; i++)
{
if (v[i] > mx)
{
mx = v[i];
max1 = i;
}
}
// Find total unique elements
// present in array arr[]
int cnt = 0;
for (int i : v)
{
if (i == 0)
++cnt;
}
int diff1 = n + 1 - cnt;
// Find the Max-Size to which
// an array arr[] can be splitted
int max_size = Math.max(Math.min(v[max1] - 1,
diff1),
Math.min(v[max1],
diff1 - 1));
System.out.println("Maximum size is: " +
max_size);
// Find the first array
// containing same elements
System.out.println("First Array is");
for (int i = 0; i < max_size; i++)
{
System.out.print(max1 + " ");
v[max1] -= 1;
}
System.out.println();
// Find the second array
// containing unique elements
System.out.println("The Second Array Is :");
for (int i = 0; i < (n + 1); i++)
{
if (v[i] > 0)
{
System.out.print(i + " ");
max_size--;
}
if (max_size < 1)
break;
}
System.out.println();
}
// Driver Code
public static void main(String[] args)
{
// initialise n
int n = 7;
// array declaration
int arr[] = new int[] {1, 2, 1, 5,
1, 6, 7, 2};
// size of array
int size = arr.length;
Solve(arr, size, n);
}
}
// This code is contributed by Sri_srajit
Python3
# Python3 program to find the max-size to which
# an array can be divided into 2 equal parts
# such that one part contains unique elements
# while another contains similar elements
# Function to find the max-size to which an
# array can be divided into 2 equal parts
def Solve(arr, size, n):
v = [0] * (n + 1);
# Vector to find the frequency of
# each element of list
for i in range(size):
v[arr[i]] += 1
# Find the maximum frequency
# element present in list arr
max1 = max(set(arr), key = v.count)
# Find total unique elements
# present in list arr
diff1 = n + 1 - v.count(0)
# Find the Max-Size to which
# an array arr[] can be splitted
max_size = max(min(v[max1] - 1, diff1),
min(v[max1], diff1 - 1))
print("Maximum size is :", max_size)
# Find the first array
# containing same elements
print("The First Array Is : ")
for i in range(max_size):
print(max1, end = " ")
v[max1] -= 1
print()
# Find the second array
# containing unique elements
print("The Second Array Is : ")
for i in range(n + 1):
if (v[i] > 0):
print(i, end = " ")
max_size -= 1
if (max_size < 1):
break
print()
# Driver code
if __name__ == "__main__":
# Initialise n
n = 7
# Array declaration
arr = [ 1, 2, 1, 5, 1, 6, 7, 2 ]
# Size of array
size = len(arr)
Solve(arr, size, n)
# This code is contributed by chitranayal
C#
// C# program to find the max-size
// to which an array can be divided
// into 2 equal parts such that one
// part contains unique elements
// while another contains similar
// elements
using System;
class GFG{
// Function to find the max-size to
// which an array can be divided into
// 2 equal parts
static void Solve(int []arr,
int size, int n)
{
int[] v = new int[n + 1];
// Array to find the frequency of
// each element of array
for(int i = 0; i < size; i++)
v[arr[i]]++;
// Find the index maximum frequency
// element present in array arr[]
int max1 = -1, mx = -1;
for(int i = 0; i < v.Length; i++)
{
if (v[i] > mx)
{
mx = v[i];
max1 = i;
}
}
// Find total unique elements
// present in array arr[]
int cnt = 0;
foreach(int i in v)
{
if (i == 0)
++cnt;
}
int diff1 = n + 1 - cnt;
// Find the Max-Size to which
// an array arr[] can be splitted
int max_size = Math.Max(Math.Min(v[max1] - 1,
diff1),
Math.Min(v[max1],
diff1 - 1));
Console.Write("Maximum size is :" +
max_size + "\n");
// Find the first array
// containing same elements
Console.Write("The First Array Is :\n");
for(int i = 0; i < max_size; i++)
{
Console.Write(max1 + " ");
v[max1] -= 1;
}
Console.Write("\n");
// Find the second array
// containing unique elements
Console.Write("The Second Array Is :\n");
for(int i = 0; i < (n + 1); i++)
{
if (v[i] > 0)
{
Console.Write(i + " ");
max_size--;
}
if (max_size < 1)
break;
}
Console.Write("\n");
}
// Driver Code
public static void Main(string[] args)
{
// Initialise n
int n = 7;
// Array declaration
int []arr = new int[] { 1, 2, 1, 5,
1, 6, 7, 2 };
// Size of array
int size = arr.Length;
Solve(arr, size, n);
}
}
// This code is contributed by rutvik_56
Javascript
输出:
Maximum size is :3
The First Array Is :
1 1 1
The Second Array Is :
2 5 6
时间复杂度: O(N)