给定一个偶数长度的数组arr[] ,任务是检查是否可以重新排序数组元素,使得左半部分的总和不等于数组右半部分的总和。如果可能,则使用重新排序的序列打印“是” ,否则打印“否” 。
例子:
Input: arr[] = {1, 2, 2, 1, 3, 1}
Output:
Yes
1 1 1 2 2 3
Explanation:
sum of left half = 1 + 2 + 2 = 5
sum of right half = 2 + 2 + 3 = 7
both the sums are not equal.
Input: arr[] = {1, 1}
Output: No
Explanation:
There is no such arrangement possible.
方法:如果数组的所有元素都相等,那么我们不能重新排序数组元素以满足给定的条件。否则在给定数组的排序序列中,数组左半部分和右半部分的总和将始终不相等。
下面是上述方法的实现:
C
// C program for the above approach
#include
#include
// A comparator function used by qsort
int compare(const void * a, const void * b)
{
return (*(int*)a - *(int*)b);
}
// Function to print the required
// reordering of array if possible
void printArr(int arr[], int n)
{
// Sort the array in increasing order
qsort(arr, n, sizeof(int), compare);
// If all elements are equal, then
// it is not possible
if (arr[0] == arr[n - 1])
{
printf("No\n");
}
// Else print the sorted array arr[]
else
{
printf("Yes\n");
for(int i = 0; i < n; i++)
{
printf("%d ", arr[i]);
}
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 2, 2, 1, 3, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
printArr(arr, N);
return 0;
}
// This code is contributed by equbalzeeshan
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the required
// reordering of array if possible
void printArr(int arr[], int n)
{
// Sort the array in increasing order
sort(arr, arr + n);
// If all elements are equal, then
// it is not possible
if (arr[0] == arr[n - 1]) {
cout << "No" << endl;
}
// Else print the sorted array arr[]
else {
cout << "Yes" << endl;
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 2, 2, 1, 3, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
printArr(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the required
// reordering of array if possible
public static void printArr(int[] arr, int n)
{
// Sort the array in increasing order
Arrays.sort(arr);
// If all elements are equal, then
// it is not possible
if (arr[0] == arr[n - 1])
{
System.out.println("No");
}
// Else print the sorted array arr[]
else
{
System.out.println("Yes");
for(int i = 0; i < n; i++)
{
System.out.print(arr[i] + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 2, 2, 1, 3, 1 };
int N = arr.length;
// Function call
printArr(arr, N);
}
}
// This code is contributed by equbalzeeshan
Python3
# Python3 program for the above approach
# Function to print the required
# reordering of the array if possible
def printArr(arr, n):
# Sort the array in increasing
# order
arr.sort()
# If all elements are equal,
# then it is not possible
if(arr[0] == arr[n - 1]):
print("No")
# Else print the sorted
# array arr[]
else:
print("Yes")
for i in range(n):
print(arr[i], end = " ")
print()
# Driver Code
if __name__ == '__main__':
# Given array
arr = [ 1, 2, 2, 1, 3, 1 ]
N = len(arr)
# Function Call
printArr(arr, N)
# This code is contributed by Shivam Singh
C#
// C# code for the above approach
using System;
class GFG{
// Function to print the required
// reordering of array if possible
public static void printArr(int[] arr, int n)
{
// Sort the array in increasing order
Array.Sort(arr);
// If all elements are equal, then
// it is not possible
if (arr[0] == arr[n - 1])
{
Console.Write("No\n");
}
// Else print the sorted array arr[]
else
{
Console.Write("Yes\n");
for(int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
}
// Driver code
public static void Main()
{
// Given array
int[] arr = new int [6]{ 1, 2, 2, 1, 3, 1 };
int N = arr.Length;
// Function call
printArr(arr, N);
}
}
// This code is contributed by equbalzeeshan
Javascript
输出:
Yes
1 1 1 2 2 3
时间复杂度: O(N*log(N))
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