给定N个正整数和一个正整数M的数组arr [] ，任务是查找最长子集的长度，该子集的最小缺失整数为M。如果不存在这样的子集，则打印“ -1” 。
例子：
Input: arr[] = {1, 2, 4}, M = 3
Output: 3
Explanation:
Possible subsets of the given array are {1}, {2}, {3}, {1, 2}, {2, 4}, {1, 4} and {1, 2, 4}.
Among these, the subsets containing elements in the range [1, M – 1] but not M is {1, 2} and {1, 2, 4}.
These subsets contain 3 as the smallest missing positive integer.
Therefore, subset {1, 2, 4} is the longest required subset with length 3.
Input: arr[] = {2, 2, 3}, M = 4
Output: -1
Explanation:
The smallest missing positive integer in the array is 1.
Therefore, all possible subsets of the array will have 1 as the smallest missing positive integer.
Therefore, no subset exists with 4 as the minimum missing integer.
天真的方法：最简单的方法是生成给定数组的所有可能子集，并在最小缺失整数为M的那些子集中存储最大长度。最后，打印最大长度作为答案。
时间复杂度： O(N * 2 N )
辅助空间： O(N)
高效方法：
要优化上述方法，请考虑以下观察。如果数组中没有缺少小于M的元素，则最大子集的大小将为1，由除M之外的所有数组元素组成。否则，没有最小丢失数M的子集是不可能的。
请按照以下步骤操作：
- 将数组中除M之外的所有元素插入集合。
- 在数组arr []中找到不等于M的元素(例如， cnt )的频率。
- 在集合中找到最小的缺失数，如果它等于M ，则打印cnt的值作为子集的最大大小。
- 否则，请打印“ -1” ，因为没有任何子集，最小缺失数为M。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
#define ll long long int
using namespace std;
// Function to find and return the
// length of the longest subset
// whose smallest missing value is M
ll findLengthOfMaxSubset(int arr[],
int n, int m)
{
// Initialize a set
set s;
int answer = 0;
for (int i = 0; i < n; i++) {
int tmp = arr[i];
// If array element is not
// equal to M
if (tmp != m) {
// Insert into set
s.insert(tmp);
// Increment frequency
answer++;
}
}
// Stores minimum missing
// number
int min = 1;
// Iterate to find the
// minimum missing
// integer
while (s.count(min)) {
min++;
}
// If minimum obtained
// is less than M
if (min != m) {
// Update answer
answer = -1;
}
// Return answer
return answer;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
int M = 3;
cout << findLengthOfMaxSubset(
arr, N, M);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find and return the
// length of the longest subset
// whose smallest missing value is M
static int findLengthOfMaxSubset(int arr[],
int n, int m)
{
// Initialize a set
Set s = new HashSet<>();
int answer = 0;
for(int i = 0; i < n; i++)
{
int tmp = arr[i];
// If array element is not
// equal to M
if (tmp != m)
{
// Insert into set
s.add(tmp);
// Increment frequency
answer++;
}
}
// Stores minimum missing
// number
int min = 1;
// Iterate to find the
// minimum missing
// integer
while (s.contains(min))
{
min++;
}
// If minimum obtained
// is less than M
if (min != m)
{
// Update answer
answer = -1;
}
// Return answer
return answer;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 4 };
int N = arr.length;
int M = 3;
System.out.print(findLengthOfMaxSubset(
arr, N, M));
}
}
// This code is contributed by offbeat Python3
# Python3 program to implement
# the above approach
# Function to find and return the
# length of the longest subset
# whose smallest missing value is M
def findLengthOfMaxSubset(arr, n, m):
# Initialize a set
s = [];
answer = 0;
for i in range(n):
tmp = arr[i];
# If array element is not
# equal to M
if (tmp != m):
# Insert into set
s.append(tmp);
# Increment frequency
answer += 1;
# Stores minimum missing
# number
min = 1;
# Iterate to find the
# minimum missing
# integer
while (s.count(min)):
min += 1;
# If minimum obtained
# is less than M
if (min != m):
# Update answer
answer = -1;
# Return answer
return answer;
# Driver Code
if __name__ == "__main__":
arr = [ 1, 2, 4 ];
N = len(arr);
M = 3;
print(findLengthOfMaxSubset(arr, N, M));
# This code is contributed by AnkitRai01C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find and return the
// length of the longest subset
// whose smallest missing value is M
static int findLengthOfMaxSubset(int []arr,
int n, int m)
{
// Initialize a set
HashSet s = new HashSet();
int answer = 0;
for(int i = 0; i < n; i++)
{
int tmp = arr[i];
// If array element is not
// equal to M
if (tmp != m)
{
// Insert into set
s.Add(tmp);
// Increment frequency
answer++;
}
}
// Stores minimum missing
// number
int min = 1;
// Iterate to find the
// minimum missing
// integer
while (s.Contains(min))
{
min++;
}
// If minimum obtained
// is less than M
if (min != m)
{
// Update answer
answer = -1;
}
// Return answer
return answer;
}
// Driver code
public static void Main (string[] args)
{
int []arr = { 1, 2, 4 };
int N = arr.Length;
int M = 3;
Console.Write(findLengthOfMaxSubset(arr, N, M));
}
}
// This code is contributed by rutvik_56 3
时间复杂度： O(N)
辅助空间： O(N)