给定大小为N的数组arr [] ，任务是找到需要插入到数组中的元素的最小数量，以使数组中存在所有可能的对的绝对差。
例子：
Input: arr[] = { 3, 5 }
Output: 3
Explanation:
Inserting 2 into the array modifies arr[] to { 2, 3, 5 }
Inserting 1 into the array modifies arr[] to { 1, 2, 3, 5 }
Inserting 4 into the array modifies arr[] to { 1, 2, 3, 4, 5 }
Since absolute difference of all possible pairs from the array are present in the array, the required output is 3.
Input: arr[] = { 2, 4 }
Output: 0
Explanation:
Since absolute difference of all possible pairs array are already present in the array, the required output is 0.
天真的方法：解决此问题的最简单方法是从数组中找到每个可能的对的绝对差，然后检查所获得的差是否存在于数组中。如果不存在，则插入获得的差异。否则，打印插入数组的元素数。
时间复杂度： O(N * X)，其中X是数组中的最大元素。
辅助空间： O(X)
高效的方法：可以基于以下观察来优化上述方法：
Since the absolute difference of all possible pairs of final array must be present in the array.
Therefore, the final array must be in the form of X, 2 * X, 3 * X, 4 * X, …
From the above sequence of the final array, it can be observed the value of X must be the GCD of the given array.
请按照以下步骤解决问题：
- 遍历数组并计算给定数组X的GCD。
- 找到数组中最大的元素，例如Max 。
- 通过插入所有可能的对的绝对差值，数组中元素的总数为(Max)/ X。
- 因此，插入数组的元素总数等于((Max / X)– N) 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the GCD of an array
int findGcd(int arr[], int N)
{
// Stores GCD of an array
int gcd = arr[0];
// Traverse the array
for (int i = 1; i < N; i++) {
// Update gcd
gcd = __gcd(gcd, arr[i]);
}
return gcd;
}
// Function to find minimum count of elements
// inserted into the array such that absolute
// difference of pairs present in the array
void findMax(int arr[], int N)
{
// Stores gcd of the array
int gcd = findGcd(arr, N);
// Stores the largest elemetn
// in the array
int Max = INT_MIN;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update Max
Max = max(Max, arr[i]);
}
// Stores minimum count of elements inserted
// into the array such that absolute difference
// of pairs present in the array
int ans = (Max / gcd) - N;
cout << ans;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 3, 5 };
// Size of the array
int N = (sizeof(arr) / (sizeof(arr[0])));
// Function Call
findMax(arr, N);
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Recursive function to return gcd of a and b
static int gcdd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcdd(a - b, b);
return gcdd(a, b - a);
}
// Function to find the GCD of an array
static int findGcd(int arr[], int N)
{
// Stores GCD of an array
int gcd = arr[0];
// Traverse the array
for (int i = 1; i < N; i++)
{
// Update gcd
gcd = gcdd(gcd, arr[i]);
}
return gcd;
}
// Function to find minimum count of elements
// inserted into the array such that absolute
// difference of pairs present in the array
static void findMax(int arr[], int N)
{
// Stores gcd of the array
int gcd = findGcd(arr, N);
// Stores the largest elemetn
// in the array
int Max = Integer.MIN_VALUE;
// Traverse the array
for (int i = 0; i < N; i++)
{
// Update Max
Max = Math.max(Max, arr[i]);
}
// Stores minimum count of elements inserted
// into the array such that absolute difference
// of pairs present in the array
int ans = (Max / gcd) - N;
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
// Given array
int arr[] = { 3, 5 };
// Size of the array
int N = arr.length;
// Function Call
findMax(arr, N);
}
}
// This code is contributed by susmitakundugoaldanga.Python3
# Python3 program for the above approach
import math
import sys
# Function to find the GCD of an array
def findGcd(arr, N):
# Stores GCD of an array
gcd = arr[0]
# Traverse the array
for i in range(1, N):
# Update gcd
gcd = math.gcd(gcd, arr[i])
return gcd
# Function to find minimum count of elements
# inserted into the array such that absolute
# difference of pairs present in the array
def findMax(arr, N):
# Stores gcd of the array
gcd = findGcd(arr, N)
# Stores the largest elemetn
# in the array
Max = -sys.maxsize - 1
# Traverse the array
for i in range(N):
# Update Max
Max = max(Max, arr[i])
# Stores minimum count of elements inserted
# into the array such that absolute difference
# of pairs present in the array
ans = (Max // gcd) - N
print(ans)
# Driver Code
if __name__ == "__main__":
# Given array
arr = [3, 5]
# Size of the array
N = len(arr)
# Function Call
findMax(arr, N)
# This code is contributed by chitranayalC#
// C# program for the above approach
using System;
class GFG
{
// Recursive function to return gcd of a and b
static int gcdd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcdd(a - b, b);
return gcdd(a, b - a);
}
// Function to find the GCD of an array
static int findGcd(int[] arr, int N)
{
// Stores GCD of an array
int gcd = arr[0];
// Traverse the array
for (int i = 1; i < N; i++)
{
// Update gcd
gcd = gcdd(gcd, arr[i]);
}
return gcd;
}
// Function to find minimum count of elements
// inserted into the array such that absolute
// difference of pairs present in the array
static void findMax(int[] arr, int N)
{
// Stores gcd of the array
int gcd = findGcd(arr, N);
// Stores the largest elemetn
// in the array
int Max = Int32.MinValue;
// Traverse the array
for (int i = 0; i < N; i++)
{
// Update Max
Max = Math.Max(Max, arr[i]);
}
// Stores minimum count of elements inserted
// into the array such that absolute difference
// of pairs present in the array
int ans = (Max / gcd) - N;
Console.WriteLine(ans);
}
// Driver code
static public void Main()
{
// Given array
int[] arr = new int[] { 3, 5 };
// Size of the array
int N = arr.Length;
// Function Call
findMax(arr, N);
}
}
// This code is contributed by Dharanendra L V.Javascript
输出：
3时间复杂度： O(N * Log(X))，其中X是数组中最大的元素。
辅助空间： O(1)