给定长度为N≥2的数组arr [] 。任务是从给定的数组中删除一个元素，以使删除后的数组的LCM最小化。
例子：
Input: arr[] = {18, 12, 24}
Output: 24
Remove 12: LCM(18, 24) = 72
Remove 18: LCM(12, 24) = 24
Remove 24: LCM(12, 18) = 36
Input: arr[] = {5, 15, 9, 36}
Output: 45
方法：
- 想法是找到所有长度为(N – 1)的子序列的LCM值，并删除该LCM子序列中不存在的元素。找到的最小LCM就是答案。
- 为了最佳地找到子序列的LCM，请使用单状态动态编程维护一个prefixLCM []和一个suffixLCM []数组。
- LCM(prefixLCM [i – 1]，后缀LCM [i + 1])的最小值是必需的答案。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to return the LCM of two numbers
int lcm(int a, int b)
{
int GCD = __gcd(a, b);
return (a * b) / GCD;
}
// Function to return the minimum LCM
// after removing a single element
// from the given array
int MinLCM(int a[], int n)
{
// Prefix and Suffix arrays
int Prefix[n + 2];
int Suffix[n + 2];
// Single state dynamic programming relation
// for storing LCM of first i elements
// from the left in Prefix[i]
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1) {
Prefix[i] = lcm(Prefix[i - 1], a[i - 1]);
}
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing LCM of all the elements having
// index greater than or equal to i in Suffix[i]
for (int i = n - 1; i >= 1; i -= 1) {
Suffix[i] = lcm(Suffix[i + 1], a[i - 1]);
}
// If first or last element of
// the array has to be removed
int ans = min(Suffix[2], Prefix[n - 1]);
// If any other element is replaced
for (int i = 2; i < n; i += 1) {
ans = min(ans, lcm(Prefix[i - 1], Suffix[i + 1]));
}
// Return the minimum LCM
return ans;
}
// Driver code
int main()
{
int a[] = { 5, 15, 9, 36 };
int n = sizeof(a) / sizeof(a[0]);
cout << MinLCM(a, n);
return 0;
} Java
// Java implementation of the above approach
class GFG
{
// Function to return the LCM of two numbers
static int lcm(int a, int b)
{
int GCD = __gcd(a, b);
return (a * b) / GCD;
}
// Function to return the minimum LCM
// after removing a single element
// from the given array
static int MinLCM(int a[], int n)
{
// Prefix and Suffix arrays
int []Prefix = new int[n + 2];
int []Suffix = new int[n + 2];
// Single state dynamic programming relation
// for storing LCM of first i elements
// from the left in Prefix[i]
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1)
{
Prefix[i] = lcm(Prefix[i - 1],
a[i - 1]);
}
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing LCM of all the elements having
// index greater than or equal to i in Suffix[i]
for (int i = n - 1; i >= 1; i -= 1)
{
Suffix[i] = lcm(Suffix[i + 1],
a[i - 1]);
}
// If first or last element of
// the array has to be removed
int ans = Math.min(Suffix[2],
Prefix[n - 1]);
// If any other element is replaced
for (int i = 2; i < n; i += 1)
{
ans = Math.min(ans, lcm(Prefix[i - 1],
Suffix[i + 1]));
}
// Return the minimum LCM
return ans;
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver code
public static void main(String []args)
{
int a[] = { 5, 15, 9, 36 };
int n = a.length;
System.out.println(MinLCM(a, n));
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation of
# the above approach
from math import gcd
# Function to return the LCM
# of two numbers
def lcm(a, b) :
GCD = gcd(a, b);
return (a * b) // GCD;
# Function to return the minimum LCM
# after removing a single element
# from the given array
def MinLCM(a, n) :
# Prefix and Suffix arrays
Prefix = [0] * (n + 2);
Suffix = [0] * (n + 2);
# Single state dynamic programming relation
# for storing LCM of first i elements
# from the left in Prefix[i]
Prefix[1] = a[0];
for i in range(2, n + 1) :
Prefix[i] = lcm(Prefix[i - 1],
a[i - 1]);
# Initializing Suffix array
Suffix[n] = a[n - 1];
# Single state dynamic programming relation
# for storing LCM of all the elements having
# index greater than or equal to i in Suffix[i]
for i in range(n - 1, 0, -1) :
Suffix[i] = lcm(Suffix[i + 1], a[i - 1]);
# If first or last element of
# the array has to be removed
ans = min(Suffix[2], Prefix[n - 1]);
# If any other element is replaced
for i in range(2, n) :
ans = min(ans, lcm(Prefix[i - 1],
Suffix[i + 1]));
# Return the minimum LCM
return ans;
# Driver code
if __name__ == "__main__" :
a = [ 5, 15, 9, 36 ];
n = len(a);
print(MinLCM(a, n));
# This code is contributed by AnkitRai01C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the LCM of two numbers
static int lcm(int a, int b)
{
int GCD = __gcd(a, b);
return (a * b) / GCD;
}
// Function to return the minimum LCM
// after removing a single element
// from the given array
static int MinLCM(int []a, int n)
{
// Prefix and Suffix arrays
int []Prefix = new int[n + 2];
int []Suffix = new int[n + 2];
// Single state dynamic programming relation
// for storing LCM of first i elements
// from the left in Prefix[i]
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1)
{
Prefix[i] = lcm(Prefix[i - 1],
a[i - 1]);
}
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing LCM of all the elements having
// index greater than or equal to i in Suffix[i]
for (int i = n - 1; i >= 1; i -= 1)
{
Suffix[i] = lcm(Suffix[i + 1],
a[i - 1]);
}
// If first or last element of
// the array has to be removed
int ans = Math.Min(Suffix[2],
Prefix[n - 1]);
// If any other element is replaced
for (int i = 2; i < n; i += 1)
{
ans = Math.Min(ans, lcm(Prefix[i - 1],
Suffix[i + 1]));
}
// Return the minimum LCM
return ans;
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver code
public static void Main(String []args)
{
int []a = { 5, 15, 9, 36 };
int n = a.Length;
Console.WriteLine(MinLCM(a, n));
}
}
// This code is contributed by PrinciRaj1992输出：
45
时间复杂度： O(N * log(M))其中M是数组中的最大元素。