给定大小为N的数组arr [] ，任务是找到最小的正整数K ，以使每个数组元素最多递增或递减K一次，使所有元素相等。如果不可能使所有数组元素相等，则打印-1 。
例子 ：
Input: arr[] = { 5, 7, 9 }
Output: 2
Explanation:
Incrementing the value of arr[0] by K(= 2) modifies arr[] to { 7, 7, 9 }.
Decrementing the value of arr[2] by K(= 2) modifies arr[] to { 7, 7, 7 }
Input: arr[] = {1, 3, 9}, N = 3
Output: -1
方法：请按照以下步骤解决问题：
- 初始化一个Set以存储数组中存在的所有不同元素。
- 计算数组中的不同元素，该元素等于Set的大小，例如M。
- 如果M> 3 ，则打印-1 。
- 如果M = 3 ，则检查集合最大和第二大元素之间的差是否等于集合最大第二和第三大元素之间的差。如果发现是正确的，则打印差异。否则，打印-1 。
- 如果M = 2 ，则检查集合中最大元素和第二大元素之间的差异是否相等。如果发现是正确的，则打印其差额的一半。否则，打印差异。
- 如果M <= 1 ，则打印0 。
下面是上述解决方案的实现：
C++
// C++ program for the above approach
#include
#include
using namespace std;
// Function to find smallest integer K such that
// incrementing or decrementing each element by
// K at most once makes all elements equal
void findMinKToMakeAllEqual(int N, int A[])
{
// Store distinct
// array elements
set B;
// Traverse the array, A[]
for (int i = 0; i < N; i++)
B.insert(A[i]);
// Count elements into the set
int M = B.size();
// Iterator to store first
// element of B
set::iterator itr = B.begin();
// If M is greater than 3
if (M > 3)
printf("-1");
// If M is equal to 3
else if (M == 3) {
// Stores the first
// smallest element
int B_1 = *itr;
// Stores the second
// smallest element
int B_2 = *(++itr);
// Stores the largest element
int B_3 = *(++itr);
// IF difference between B_2 and B_1
// is equal to B_3 and B_2
if (B_2 - B_1 == B_3 - B_2)
printf("%d", B_2 - B_1);
else
printf("-1");
}
// If M is equal to 2
else if (M == 2) {
// Stores the smallest element
int B_1 = *itr;
// Stores the largest element
int B_2 = *(++itr);
// If difference is an even
if ((B_2 - B_1) % 2 == 0)
printf("%d", (B_2 - B_1) / 2);
else
printf("%d", B_2 - B_1);
}
// If M is equal to 1
else
printf("%d", 0);
}
// Driver Code
int main()
{
// Given array
int A[] = { 1, 3, 5, 1 };
// Given size
int N = sizeof(A) / sizeof(A[0]);
// Print the required smallest integer
findMinKToMakeAllEqual(N, A);
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find smallest integer K such that
// incrementing or decrementing each element by
// K at most once makes all elements equal
static void findMinKToMakeAllEqual(int N, int A[])
{
// Store distinct
// array elements
Set B = new HashSet();
// Traverse the array, A[]
for (int i = 0; i < N; i++)
{
B.add(A[i]);
}
ArrayList b = new ArrayList(B);
// Count elements into the set
int M = b.size();
int i = 0;
// If M is greater than 3
if (M > 3)
{ System.out.print("-1");}
// If M is equal to 3
else if (M == 3)
{
// Stores the first
// smallest element
int B_1 = b.get(i++);
// Stores the second
// smallest element
int B_2 = b.get(i++);
// Stores the largest element
int B_3 = b.get(i++);
// IF difference between B_2 and B_1
// is equal to B_3 and B_2
if (B_2 - B_1 == B_3 - B_2)
{
System.out.print(B_2 - B_1);
}
else
{
System.out.print("-1");
}
}
// If M is equal to 2
else if (M == 2)
{
// Stores the smallest element
int B_1 = b.get(i++);
// Stores the largest element
int B_2 = b.get(i++);
// If difference is an even
if ((B_2 - B_1) % 2 == 0)
{
System.out.print((B_2 - B_1) / 2);
}
else
{
System.out.print(B_2 - B_1);
}
}
// If M is equal to 1
else
{
System.out.print(0);
}
}
// Driver code
public static void main (String[] args)
{
// Given array
int A[] = { 1, 3, 5, 1 };
// Given size
int N = A.length;
// Print the required smallest integer
findMinKToMakeAllEqual(N, A);
}
}
// This code is contributed by rag2127 Python3
# Python3 program for the above approach
# Function to find smallest integer K such
# that incrementing or decrementing each
# element by K at most once makes all
# elements equal
def findMinKToMakeAllEqual(N, A):
# Store distinct
# array elements
B = {}
# Traverse the array, A[]
for i in range(N):
B[A[i]] = 1
# Count elements into the set
M = len(B)
# Iterator to store first
# element of B
itr, i = list(B.keys()), 0
# If M is greater than 3
if (M > 3):
print("-1")
# If M is equal to 3
elif (M == 3):
# Stores the first
# smallest element
B_1, i = itr[i], i + 1
# Stores the second
# smallest element
B_2, i = itr[i], i + 1
# Stores the largest element
B_3, i = itr[i], i + 1
# IF difference between B_2 and B_1
# is equal to B_3 and B_2
if (B_2 - B_1 == B_3 - B_2):
print(B_2 - B_1)
else:
print("-1")
# If M is equal to 2
elif (M == 2):
# Stores the smallest element
B_1, i = itr[i], i + 1
# Stores the largest element
B_2, i = itr[i], i + 1
# If difference is an even
if ((B_2 - B_1) % 2 == 0):
print((B_2 - B_1) // 2)
else:
print(B_2 - B_1)
# If M is equal to 1
else:
print(0)
# Driver Code
if __name__ == '__main__':
# Given array
A = [ 1, 3, 5, 1 ]
# Given size
N = len(A)
# Print the required smallest integer
findMinKToMakeAllEqual(N, A)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find smallest integer K such that
// incrementing or decrementing each element by
// K at most once makes all elements equal
static void findMinKToMakeAllEqual(int N, int[] A)
{
// Store distinct
// array elements
var B = new HashSet();
// Traverse the array, A[]
for (int i = 0; i < N; i++) {
B.Add(A[i]);
}
List b = new List(B);
// Count elements into the set
int M = b.Count;
int j = 0;
// If M is greater than 3
if (M > 3) {
Console.Write("-1");
}
// If M is equal to 3
else if (M == 3) {
// Stores the first
// smallest element
int B_1 = b[j++];
// Stores the second
// smallest element
int B_2 = b[j++];
// Stores the largest element
int B_3 = b[j++];
// IF difference between B_2 and B_1
// is equal to B_3 and B_2
if (B_2 - B_1 == B_3 - B_2) {
Console.Write(B_2 - B_1);
}
else {
Console.Write("-1");
}
}
// If M is equal to 2
else if (M == 2) {
// Stores the smallest element
int B_1 = b[j++];
// Stores the largest element
int B_2 = b[j++];
// If difference is an even
if ((B_2 - B_1) % 2 == 0) {
Console.Write((B_2 - B_1) / 2);
}
else {
Console.Write(B_2 - B_1);
}
}
// If M is equal to 1
else {
Console.Write(0);
}
}
// Driver code
public static void Main(string[] args)
{
// Given array
int[] A = { 1, 3, 5, 1 };
// Given size
int N = A.Length;
// Print the required smallest integer
findMinKToMakeAllEqual(N, A);
}
}
// This code is contributed by chitranayal. 输出：
2时间复杂度： O(N * log(N))
辅助空间： O(N)