给定一个由N个整数组成的数组arr [] ，任务是找到最小的数，该数将数组中元素的最小数量相除。
例子：
Input: arr[] = {2, 12, 6}
Output: 5
Here, 1 divides 3 elements
2 divides 3 elements
3 divides 2 elements
4 divides 1 element
5 divides no element
6 divides 2 elements
7 divides no element
8 divides no element
9 divides no element
10 divides no element
11 divides no element
12 divides 1 element
5 is the smallest number not dividing any
number in the array. Thus, ans = 5
Input: arr[] = {1, 7, 9}
Output: 2
方法：让我们先观察一些细节。将零元素相除的数字已经存在，即max(arr)+ 1 。现在，我们只需要找到将数组中的零数相除的最小数即可。
在本文中，将讨论使用筛网(M = max(arr))在O(M * log(M)+ N)时间内解决此问题的方法。
- 首先，在数组中找到最大元素M ，并创建一个长度为M +1的频率表freq [] ，以存储1到M之间的数字的频率。
- 迭代数组，并为每个索引i将freq []更新为freq [arr [i]] ++ 。
- 现在，应用筛网算法。在1到M +1之间的所有元素之间进行迭代。
- 假设我们要迭代数字X。
- 创建一个临时变量cnt 。
- 对于X与M之间的每个X倍数{X，2X，3X…。}，将cnt更新为cnt = cnt + freq [kX] 。
- 如果CNT = 0，那么答案将是X否则继续迭代为X的下一个值。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the smallest number
// that divides minimum number of elements
// in the given array
int findMin(int* arr, int n)
{
// m stores the maximum in the array
int m = 0;
for (int i = 0; i < n; i++)
m = max(m, arr[i]);
// Frequency array
int freq[m + 2] = { 0 };
for (int i = 0; i < n; i++)
freq[arr[i]]++;
// Sieve
for (int i = 1; i <= m + 1; i++) {
int j = i;
int cnt = 0;
// Incrementing j
while (j <= m) {
cnt += freq[j];
j += i;
}
// If no multiples of j are
// in the array
if (!cnt)
return i;
}
return m + 1;
}
// Driver code
int main()
{
int arr[] = { 2, 12, 6 };
int n = sizeof(arr) / sizeof(int);
cout << findMin(arr, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return the smallest number
// that divides minimum number of elements
// in the given array
static int findMin(int arr[], int n)
{
// m stores the maximum in the array
int m = 0;
for (int i = 0; i < n; i++)
m = Math.max(m, arr[i]);
// Frequency array
int freq [] = new int[m + 2];
for (int i = 0; i < n; i++)
freq[arr[i]]++;
// Sieve
for (int i = 1; i <= m + 1; i++)
{
int j = i;
int cnt = 0;
// Incrementing j
while (j <= m)
{
cnt += freq[j];
j += i;
}
// If no multiples of j are
// in the array
if (cnt == 0)
return i;
}
return m + 1;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 2, 12, 6 };
int n = arr.length;
System.out.println(findMin(arr, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
# Function to return the smallest number
# that divides minimum number of elements
# in the given array
def findMin(arr, n):
# m stores the maximum in the array
m = 0
for i in range(n):
m = max(m, arr[i])
# Frequency array
freq = [0] * (m + 2)
for i in range(n):
freq[arr[i]] += 1
# Sieve
for i in range(1, m + 2):
j = i
cnt = 0
# Incrementing j
while (j <= m):
cnt += freq[j]
j += i
# If no multiples of j are
# in the array
if (not cnt):
return i
return m + 1
# Driver code
arr = [2, 12, 6]
n = len(arr)
print(findMin(arr, n))
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the smallest number
// that divides minimum number of elements
// in the given array
static int findMin(int []arr, int n)
{
// m stores the maximum in the array
int m = 0;
for (int i = 0; i < n; i++)
m = Math.Max(m, arr[i]);
// Frequency array
int []freq = new int[m + 2];
for (int i = 0; i < n; i++)
freq[arr[i]]++;
// Sieve
for (int i = 1; i <= m + 1; i++)
{
int j = i;
int cnt = 0;
// Incrementing j
while (j <= m)
{
cnt += freq[j];
j += i;
}
// If no multiples of j are
// in the array
if (cnt == 0)
return i;
}
return m + 1;
}
// Driver code
public static void Main ()
{
int []arr = { 2, 12, 6 };
int n = arr.Length;
Console.WriteLine(findMin(arr, n));
}
}
// This code is contributed by AnkitRai01输出 ：
5
时间复杂度： O(Mlog(M)+ N)