给定阵列中所有对的最小LCM

给定大小为N的数组arr [] ，任务是查找给定数组中所有唯一对的最小LCM(最小公倍数)，其中1 <= N <= 10 ^5，1 <= arr [i] < = 10 ⁵ 。
例子：

Input: arr[] = {2, 4, 3} 
Output: 4 
Explanation 
LCM (2, 4) = 4 
LCM (2, 3) = 6 
LCM (4, 3) = 12 
Minimum possible LCM is 4.
Input: arr [] ={1, 5, 2, 2, 6} 
Output: 2 

天真的方法

1. 生成所有可能的对，并为每个唯一对计算LCM。
2. 从所有唯一对中找到最小的LCM。

下面是上述方法的实现：

C++

// C++ program to find
// minimum possible lcm
// from any pair
 
#include 
using namespace std;
 
// function to compute
// GCD of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// function that return
// minimum possible lcm
// from any pair
int minLCM(int arr[], int n)
{
    int ans = INT_MAX;
    for (int i = 0; i < n; i++) {
 
        // fix the ith element and
        // iterate over all the array
        // to find minimum LCM
        for (int j = i + 1; j < n; j++) {
 
            int g = gcd(arr[i], arr[j]);
            int lcm = arr[i] / g * arr[j];
            ans = min(ans, lcm);
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 3, 6, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minLCM(arr, n) << endl;
    return 0;
}

Java

// Java program to find minimum
// possible lcm from any pair
import java.io.*;
import java.util.*;
 
class GFG {
     
// Function to compute
// GCD of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function that return minimum
// possible lcm from any pair
static int minLCM(int arr[], int n)
{
    int ans = Integer.MAX_VALUE;
     
    for(int i = 0; i < n; i++)
    {
         
       // Fix the ith element and
       // iterate over all the array
       // to find minimum LCM
       for(int j = i + 1; j < n; j++)
       {
          int g = gcd(arr[i], arr[j]);
          int lcm = arr[i] / g * arr[j];
          ans = Math.min(ans, lcm);
       }
    }
    return ans;
}
     
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 3, 6, 5 };
    int n = arr.length;
     
    System.out.println(minLCM(arr,n));
}
}
 
// This code is contributed by coder001

Python3

# Python3 program to find minimum
# possible lcm from any pair
import sys
 
# Function to compute
# GCD of two numbers
def gcd(a, b):
     
    if (b == 0):
        return a;
    return gcd(b, a % b);
 
# Function that return minimum
# possible lcm from any pair
def minLCM(arr, n):
     
    ans = 1000000000;
    for i in range(n):
 
        # Fix the ith element and
        # iterate over all the
        # array to find minimum LCM
        for j in range(i + 1, n):
             
            g = gcd(arr[i], arr[j]);
            lcm = arr[i] / g * arr[j];
            ans = min(ans, lcm);
             
    return ans;
 
# Driver code
arr = [ 2, 4, 3, 6, 5 ];
 
print(minLCM(arr, 5))
 
# This code is contributed by grand_master

C#

// C# program to find minimum
// possible lcm from any pair
using System;
class GFG{
     
// Function to compute
// GCD of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function that return minimum
// possible lcm from any pair
static int minLCM(int []arr, int n)
{
    int ans = Int32.MaxValue;
     
    for(int i = 0; i < n; i++)
    {
         
    // Fix the ith element and
    // iterate over all the array
    // to find minimum LCM
    for(int j = i + 1; j < n; j++)
    {
        int g = gcd(arr[i], arr[j]);
        int lcm = arr[i] / g * arr[j];
        ans = Math.Min(ans, lcm);
    }
    }
    return ans;
}
     
// Driver code
public static void Main()
{
    int []arr = { 2, 4, 3, 6, 5 };
    int n = arr.Length;
     
    Console.Write(minLCM(arr,n));
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

 
    // Javascript program to find
    // minimum possible lcm
    // from any pair
     
    // function to compute
    // GCD of two numbers
    function gcd(a, b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
 
    // function that return
    // minimum possible lcm
    // from any pair
    function minLCM(arr, n)
    {
        let ans = Number.MAX_VALUE;
        for (let i = 0; i < n; i++) {
 
            // fix the ith element and
            // iterate over all the array
            // to find minimum LCM
            for (let j = i + 1; j < n; j++) {
 
                let g = gcd(arr[i], arr[j]);
                let lcm = arr[i] / g * arr[j];
                ans = Math.min(ans, lcm);
            }
        }
 
        return ans;
    }
     
    let arr = [ 2, 4, 3, 6, 5 ];
    let n = arr.length;
    document.write(minLCM(arr, n));
 
 

输出：

4

时间复杂度： O(N ² )
高效方法：此方法取决于以下公式：

Product of two number = LCM of two number * GCD of two number 

1. 在LCM公式中，分母是两个数字的GCD，两个数字的GCD永远不会大于数字本身。
2. 因此，对于固定GCD，找到给定数组中存在的固定GCD的最小两个倍数。
3. 仅存储每个GCD的最小的两个倍数，因为选择数组中存在的更大的GCD的倍数，无论如何，它将永远不会给出最小的答案。
4. 最后，使用筛子查找最小的两个数字，该数字是所选GCD的倍数。

下面是上述方法的实现：

C++

// C++ program to find the
// pair having minimum LCM
 
#include 
using namespace std;
 
// function that return
// pair having minimum LCM
int minLCM(int arr[], int n)
{
    int mx = 0;
    for (int i = 0; i < n; i++) {
 
        // find max element in the array as
        // the gcd of two elements from the
        // array can't greater than max element.
        mx = max(mx, arr[i]);
    }
 
    // created a 2D array to store minimum
    // two multiple of any particular i.
    vector > mul(mx + 1);
 
    for (int i = 0; i < n; i++) {
        if (mul[arr[i]].size() > 1) {
            // we already found two
            // smallest multiple
            continue;
        }
        mul[arr[i]].push_back(arr[i]);
    }
 
    // iterating over all gcd
    for (int i = 1; i <= mx; i++) {
 
        // iterating over its multiple
        for (int j = i + i; j <= mx; j += i) {
 
            if (mul[i].size() > 1) {
 
                // if we already found the
                // two smallest multiple of i
                break;
            }
            for (int k : mul[j]) {
                if (mul[i].size() > 1)
                    break;
                mul[i].push_back(k);
            }
        }
    }
 
    int ans = INT_MAX;
    for (int i = 1; i <= mx; i++) {
 
        if (mul[i].size() <= 1)
            continue;
 
        // choosing smallest two multiple
        int a = mul[i][0], b = mul[i][1];
 
        // calculating lcm
        int lcm = (a * b) / i;
 
        ans = min(ans, lcm);
    }
 
    // return final answer
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 3, 6, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minLCM(arr, n) << endl;
    return 0;
}

Java

// Java program to find the
// pair having minimum LCM
import java.util.Vector;
class GFG{
 
// Function that return
// pair having minimum LCM
static int minLCM(int arr[],
                  int n)
{
  int mx = 0;
  for (int i = 0; i < n; i++)
  {
    // Find max element in the
    // array as the gcd of two
    // elements from the array
    // can't greater than max element.
    mx = Math.max(mx, arr[i]);
  }
 
  // Created a 2D array to store minimum
  // two multiple of any particular i.
  Vector []mul = new Vector[mx + 1];
   
  for (int i = 0; i < mul.length; i++)
    mul[i] = new Vector();
  for (int i = 0; i < n; i++)
  {
    if (mul[arr[i]].size() > 1)
    {
      // We already found two
      // smallest multiple
      continue;
    }
    mul[arr[i]].add(arr[i]);
  }
 
  // Iterating over all gcd
  for (int i = 1; i <= mx; i++)
  {
    // Iterating over its multiple
    for (int j = i + i; j <= mx; j += i)
    {
      if (mul[i].size() > 1)
      {
        // If we already found the
        // two smallest multiple of i
        break;
      }     
      for (int k : mul[j])
      {
        if (mul[i].size() > 1)
          break;
        mul[i].add(k);
      }
    }
  }
 
  int ans = Integer.MAX_VALUE;
  for (int i = 1; i <= mx; i++)
  {
    if (mul[i].size() <= 1)
      continue;
 
    //  Choosing smallest
    // two multiple
    int a = mul[i].get(0),
        b = mul[i].get(1);
 
    // Calculating lcm
    int lcm = (a * b) / i;
 
    ans = Math.min(ans, lcm);
  }
 
  // Return final answer
  return ans;
}
 
// Driver code
public static void main(String[] args)
{
  int arr[] = {2, 4, 3, 6, 5};
  int n = arr.length;
  System.out.print(minLCM(arr, n) + "\n");
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program to find the
# pair having minimum LCM
import sys
 
# function that return
# pair having minimum LCM
def minLCM(arr, n) :
    mx = 0
    for i in range(n) :
 
        # find max element in the array as
        # the gcd of two elements from the
        # array can't greater than max element.
        mx = max(mx, arr[i])
 
    # created a 2D array to store minimum
    # two multiple of any particular i.
    mul = [[] for i in range(mx + 1)]
 
    for i in range(n) :
        if (len(mul[arr[i]]) > 1) :
           
            # we already found two
            # smallest multiple
            continue
         
        mul[arr[i]].append(arr[i])
 
    # iterating over all gcd
    for i in range(1, mx + 1) :
 
        # iterating over its multiple
        for j in range(i + i, mx + 1, i) :
 
            if (len(mul[i]) > 1) :
 
                # if we already found the
                # two smallest multiple of i
                break
     
            for k in mul[j] :
                if (len(mul[i]) > 1) :
                    break
                mul[i].append(k)
 
    ans = sys.maxsize
    for i in range(1, mx + 1) :
        if (len(mul[i]) <= 1) :
            continue
 
        # choosing smallest two multiple
        a, b = mul[i][0], mul[i][1]
 
        # calculating lcm
        lcm = (a * b) // i
        ans = min(ans, lcm)
 
    # return final answer
    return ans
 
# Driver code
arr = [ 2, 4, 3, 6, 5 ]
n = len(arr)
print(minLCM(arr, n))
 
# This code is contributed by divyesh072019

C＃

// C# program to find the
// pair having minimum LCM
using System;
using System.Collections.Generic;
class GFG{
 
// Function that return
// pair having minimum LCM
static int minLCM(int []arr,
                  int n)
{
  int mx = 0;
   
  for (int i = 0; i < n; i++)
  {
    // Find max element in the
    // array as the gcd of two
    // elements from the array
    // can't greater than max element.
    mx = Math.Max(mx, arr[i]);
  }
 
  // Created a 2D array to store minimum
  // two multiple of any particular i.
  List []mul = new List[mx + 1];
   
  for (int i = 0; i < mul.Length; i++)
    mul[i] = new List();
   
  for (int i = 0; i < n; i++)
  {
    if (mul[arr[i]].Count > 1)
    {
      // We already found two
      // smallest multiple
      continue;
    }
    mul[arr[i]].Add(arr[i]);
  }
 
  // Iterating over all gcd
  for (int i = 1; i <= mx; i++)
  {
    // Iterating over its multiple
    for (int j = i + i; j <= mx; j += i)
    {
      if (mul[i].Count > 1)
      {
        // If we already found the
        // two smallest multiple of i
        break;
      }     
      foreach (int k in mul[j])
      {
        if (mul[i].Count > 1)
          break;
        mul[i].Add(k);
      }
    }
  }
 
  int ans = int.MaxValue;
  for (int i = 1; i <= mx; i++)
  {
    if (mul[i].Count <= 1)
      continue;
 
    //  Choosing smallest
    // two multiple
    int a = mul[i][0],
        b = mul[i][1];
 
    // Calculating lcm
    int lcm = (a * b) / i;
 
    ans = Math.Min(ans, lcm);
  }
 
  // Return readonly answer
  return ans;
}
 
// Driver code
public static void Main(String[] args)
{
  int []arr = {2, 4, 3, 6, 5};
  int n = arr.Length;
  Console.Write(minLCM(arr, n) + "\n");
}
}
 
// This code is contributed by Princi Singh

输出：

4

时间复杂度： O((N + M)* log(M))
辅助空间： O(M) ，其中M是数组中的最大元素。