给定由N个正整数组成的数组arr [] ，任务是查找给定数组的所有唯一元素的LCM。如果数组不包含任何唯一元素，则打印“ -1 ”。
例子：
Input: arr[] = {1, 2, 1, 3, 3, 4}
Output: 4
Explanation:
The unique elements of the given array are: 2 and 4. Therefore, the LCM of (2, 4) is 4.
Input: arr[] = {1, 1, 2, 2, 3, 3}
Output: -1
天真的方法：解决给定问题的最简单方法是遍历每个数组元素arr [i]的给定数组arr [ ]并检查arr [i]是否唯一。如果发现为真，则将其存储在另一个数组中。遍历所有元素后，打印新创建的数组中存在的元素的LCM。
时间复杂度： O(N 2 + N * log(M))，其中M是数组arr []的最大元素。
辅助空间： O(N)
高效方法：还可以通过使用哈希在数组中查找唯一元素来优化上述方法。请按照以下步骤解决问题：
- 初始化一个变量，将ans表示为1，以存储数组唯一元素的LCM。
- 遍历数组arr []并将每个元素的频率存储在Map中。
- 现在，遍历地图，并且如果任何元素的计数等于1，则ANS的值更新为ANS的LCM和当前元素。
- 完成上述步骤后，如果ans的值为1，则打印“ -1” 。否则，输出ans的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find GCD of two numbers
int findGCD(int a, int b)
{
// Base Case
if (b == 0)
return a;
// Recursively find the GCD
return findGCD(b, a % b);
}
// Function to find LCM of two numbers
int findLCM(int a, int b)
{
return (a * b) / findGCD(a, b);
}
// Function to find LCM of unique elements
// present in the array
int uniqueElementsLCM(int arr[], int N)
{
// Stores the frequency of each
// number of the array
unordered_map freq;
// Store the frequency of each
// element of the array
for (int i = 0; i < N; i++) {
freq[arr[i]]++;
}
// Store the required result
int lcm = 1;
// Traverse the map freq
for (auto i : freq) {
// If the frequency of the
// current element is 1, then
// update ans
if (i.second == 1) {
lcm = findLCM(lcm, i.first);
}
}
// If there is no unique element,
// set lcm to -1
if (lcm == 1)
lcm = -1;
// Print the result
cout << lcm;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 1, 3, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
uniqueElementsLCM(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.HashMap;
import java.util.Map.Entry;
class GFG{
// Function to find GCD of two numbers
static int findGCD(int a, int b)
{
// Base Case
if (b == 0)
return a;
// Recursively find the GCD
return findGCD(b, a % b);
}
// Function to find LCM of two numbers
static int findLCM(int a, int b)
{
return (a * b) / findGCD(a, b);
}
// Function to find LCM of unique elements
// present in the array
static void uniqueElementsLCM(int arr[], int N)
{
// Stores the frequency of each
// number of the array
HashMap freq = new HashMap();
// Store the frequency of each
// element of the array
for(int i = 0; i < N; i++)
{
freq.put(arr[i],
freq.getOrDefault(arr[i], 0) + 1);
}
// Store the required result
int lcm = 1;
// Traverse the map freq
for(Entry i : freq.entrySet())
{
// If the frequency of the
// current element is 1, then
// update ans
if (i.getValue() == 1)
{
lcm = findLCM(lcm, i.getKey());
}
}
// If there is no unique element,
// set lcm to -1
if (lcm == 1)
lcm = -1;
// Print the result
System.out.print(lcm);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 1, 3, 3, 4 };
int N = arr.length;
// Function Call
uniqueElementsLCM(arr, N);
}
}
// This code is contributed by abhinavjain194 Python3
# Python3 program for the above approach
from collections import defaultdict
# Function to find GCD of two numbers
def findGCD(a, b):
# Base Case
if (b == 0):
return a
# Recursively find the GCD
return findGCD(b, a % b)
# Function to find LCM of two numbers
def findLCM(a, b):
return (a * b) // findGCD(a, b)
# Function to find LCM of unique elements
# present in the array
def uniqueElementsLCM(arr, N):
# Stores the frequency of each
# number of the array
freq = defaultdict(int)
# Store the frequency of each
# element of the array
for i in range(N):
freq[arr[i]] += 1
# Store the required result
lcm = 1
# Traverse the map freq
for i in freq:
# If the frequency of the
# current element is 1, then
# update ans
if (freq[i] == 1):
lcm = findLCM(lcm, i)
# If there is no unique element,
# set lcm to -1
if (lcm == 1):
lcm = -1
# Print the result
print(lcm)
# Driver Code
if __name__ == "__main__":
arr = [ 1, 2, 1, 3, 3, 4 ]
N = len(arr)
# Function Call
uniqueElementsLCM(arr, N)
# This code is contributed by ukaspC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find GCD of two numbers
static int findGCD(int a, int b)
{
// Base Case
if (b == 0)
return a;
// Recursively find the GCD
return findGCD(b, a % b);
}
// Function to find LCM of two numbers
static int findLCM(int a, int b)
{
return (a * b) / findGCD(a, b);
}
// Function to find LCM of unique elements
// present in the array
static void uniqueElementsLCM(int []arr, int N)
{
// Stores the frequency of each
// number of the array
Dictionary freq = new Dictionary();
// Store the frequency of each
// element of the array
for (int i = 0; i < N; i++) {
if(freq.ContainsKey(arr[i]))
freq[arr[i]]++;
else
freq.Add(arr[i],1);
}
// Store the required result
int lcm = 1;
// Traverse the map freq
foreach(KeyValuePair kvp in freq) {
// If the frequency of the
// current element is 1, then
// update ans
if (kvp.Value == 1) {
lcm = findLCM(lcm, kvp.Key);
}
}
// If there is no unique element,
// set lcm to -1
if (lcm == 1)
lcm = -1;
// Print the result
Console.Write(lcm);
}
// Driver Code
public static void Main()
{
int []arr = {1, 2, 1, 3, 3, 4 };
int N = arr.Length;
// Function Call
uniqueElementsLCM(arr, N);
}
}
// This code is contributed by ipg2016107. 输出：
4时间复杂度： O(N * log(M))，其中M是数组arr []的最大元素
辅助空间： O(N)