给定大小为N且整数K的数组arr [] ，任务是通过删除数组元素并增加所有其他数组元素的值以使数组元素的总和来检查是否可以使所有数组元素相等保持原样。如果发现是真的，则打印“是” 。否则，打印“否” 。
例子 ：
Input: arr[] = { 2, 2, 2, 3 }
Output: YES
Explanation:
Removing arr[2] ( 0-based indexing ) from the array and incrementing all other elements by 1 modifies arr[] to { 3, 3, 3 }
Since all array elements are equal. Therefore, the required output is “YES”.
Input: arr[] = { 0, 3, 0 }, K = 3
Output:: NO
Removing arr[1] ( 0-based indexing ) from the array and incrementing the value arr[0] by 1 and arr[2] by 2 modifies arr[] to { 1, 2 }
Since all array elements are not equal. Therefore, the required output is “NO”.
方法：可以使用贪婪技术解决问题。这个想法是删除最大的数组元素并增加其他元素的值，以使总和保持不变。请按照以下步骤解决问题：
- 初始化一个变量，例如secMax ，以存储数组的第二大元素。
- 初始化一个变量totalSum ，以存储数组元素的总和。
- 遍历数组并计算数组元素的总和，找到第二大的数组元素。
- 如果totalSum <(N – 1)* secMax或totalSum％(N – 1)！= 0 ，则打印“ NO” 。
- 否则，打印“是” 。
下面是上述方法的实现
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to check if an array of
// equal elements with sum equal to
// the given array can be obtained or not
bool CheckAllarrayEqual(int arr[], int N)
{
// Base case
if (N == 1) {
return true;
}
// Stores sum of
// array elements
int totalSum = arr[0];
// Stores second largest
// array element
int secMax = INT_MIN;
// Stores the largest
// array element
int Max = arr[0];
// Traverse the array
for (int i = 1; i < N; i++) {
if (arr[i] >= Max) {
// Update secMax
secMax = Max;
// Update Max
Max = arr[i];
}
else if (arr[i] > secMax) {
// Update secMax
secMax = arr[i];
}
// Update totalSum
totalSum += arr[i];
}
// If totalSum is less than
// secMax * (N - 1))
if ((secMax * (N - 1)) > totalSum) {
return false;
}
// If totalSum is not
// divisible by (N - 1)
if (totalSum % (N - 1)) {
return false;
}
return true;
}
// Driver Code
int main()
{
int arr[] = { 6, 2, 2, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
if (CheckAllarrayEqual(arr, N)) {
cout << "YES";
}
else {
cout << "NO";
}
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to check if an array of
// equal elements with sum equal to
// the given array can be obtained or not
static boolean CheckAllarrayEqual(int[] arr,
int N)
{
// Base case
if (N == 1)
{
return true;
}
// Stores sum of
// array elements
int totalSum = arr[0];
// Stores second largest
// array element
int secMax = Integer.MIN_VALUE;
// Stores the largest
// array element
int Max = arr[0];
// Traverse the array
for(int i = 1; i < N; i++)
{
if (arr[i] >= Max)
{
// Update secMax
secMax = Max;
// Update Max
Max = arr[i];
}
else if (arr[i] > secMax)
{
// Update secMax
secMax = arr[i];
}
// Update totalSum
totalSum += arr[i];
}
// If totalSum is less than
// secMax * (N - 1))
if ((secMax * (N - 1)) > totalSum)
{
return false;
}
// If totalSum is not
// divisible by (N - 1)
if (totalSum % (N - 1) != 0)
{
return false;
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 6, 2, 2, 2 };
int N = arr.length;
if (CheckAllarrayEqual(arr, N))
{
System.out.print("YES");
}
else
{
System.out.print("NO");
}
}
}
// This code is contributed by susmitakundugoaldangaPython3
# Python3 program to implement
# the above approach
# Function to check if an array of
# equal elements with sum equal to
# the given array can be obtained or not
def CheckAllarrayEqual(arr, N):
# Base case
if (N == 1):
return True
# Stores sum of
# array elements
totalSum = arr[0]
# Stores second largest
# array element
secMax = -10**19
# Stores the largest
# array element
Max = arr[0]
# Traverse the array
for i in range(1,N):
if (arr[i] >= Max):
# Update secMax
secMax = Max
# Update Max
Max = arr[i]
elif (arr[i] > secMax):
# Update secMax
secMax = arr[i]
# Update totalSum
totalSum += arr[i]
# If totalSum is less than
# secMax * (N - 1))
if ((secMax * (N - 1)) > totalSum):
return False
# If totalSum is not
# divisible by (N - 1)
if (totalSum % (N - 1)):
return False
return True
# Driver Code
if __name__ == '__main__':
arr=[6, 2, 2, 2]
N = len(arr)
if (CheckAllarrayEqual(arr, N)):
print("YES")
else:
print("NO")
# This code is contributd by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to check if an array of
// equal elements with sum equal to
// the given array can be obtained or not
static bool CheckAllarrayEqual(int[] arr, int N)
{
// Base case
if (N == 1) {
return true;
}
// Stores sum of
// array elements
int totalSum = arr[0];
// Stores second largest
// array element
int secMax = Int32.MinValue;
// Stores the largest
// array element
int Max = arr[0];
// Traverse the array
for (int i = 1; i < N; i++) {
if (arr[i] >= Max) {
// Update secMax
secMax = Max;
// Update Max
Max = arr[i];
}
else if (arr[i] > secMax) {
// Update secMax
secMax = arr[i];
}
// Update totalSum
totalSum += arr[i];
}
// If totalSum is less than
// secMax * (N - 1))
if ((secMax * (N - 1)) > totalSum) {
return false;
}
// If totalSum is not
// divisible by (N - 1)
if (totalSum % (N - 1) != 0) {
return false;
}
return true;
}
// Driver Code
public static void Main()
{
int[] arr = { 6, 2, 2, 2 };
int N = arr.Length;
if (CheckAllarrayEqual(arr, N)) {
Console.Write("YES");
}
else {
Console.Write("NO");
}
}
}
// This code is contributed by sanjoy_62输出：
YES时间复杂度： O(N)
辅助空间： O(1)