通过交换和为奇数的元素来打印字典上最小的数组
原文:https://www . geeksforgeeks . org/print-按字典序最小数组交换元素-其和为奇数/
给定一组 N 个整数。任务是通过多次应用给定的操作来找到字典上最小的数组。操作是选取两个元素aIT5】和ajT9】(1<= I，j < =N)使得aI+ajT15】为奇数，然后互换aIT19】和ajT23】。 举例:**
输入: a[] = {1，5，4，3，2} 输出: 1 2 3 4 5 解释:先换(5，2)再换(4，3)。这是通过 交换满足给定条件 输入: a[] = {4，2} 输出: 4 2 解释:不可能交换任何元素可以得到的 字典上最小的可能数组。
接近:观察两个元素如果奇偶性不同，是否有可能互换。如果数组中的所有元素具有相同的奇偶校验(奇数+奇数和偶数+偶数不是奇数)，则不可能进行交换。因此，答案将只是输入数组。否则，您实际上可以交换任何一对元素。假设你想交换两个元素，a 和 b，它们具有相同的奇偶性。必须有第三个元素 c 具有不同的奇偶性。不失一般性，假设数组为【a，b，c】。让我们进行以下交换:
- 交换 a 和 c:
- 交换 b 和 c: [b，c，a]
- 交换 a 和 c: [b，a，c]
换句话说，用 c 作为一个中间元素来交换 a 和 b，之后它总是会回到原来的位置。因为任何一对元素之间都可以交换，所以我们总是可以对数组进行排序，这将是字典上最小的数组。 以下是上述办法的实施情况:
C++
// CPP program to find possible
// lexicographically smaller array
// by swapping only elements whose sum is odd
#include <bits/stdc++.h>
using namespace std;
// Function to find possible lexicographically smaller array
void lexicographically_smaller(int arr[], int n)
{
// To store number of even and odd integers
int odd = 0, even = 0;
// Find number of even and odd integers
for (int i = 0; i < n; i++) {
if (arr[i] % 2)
odd++;
else
even++;
}
// If it possible to make
// lexicographically smaller
if (odd && even)
sort(arr, arr + n);
// Print the array
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Driver code
int main()
{
int arr[] = { 1, 5, 4, 3, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
lexicographically_smaller(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find possible
// lexicographically smaller array
// by swapping only elements whose sum is odd
import java.util.*;
class GFG
{
// Function to find possible lexicographically smaller array
static void lexicographically_smaller(int arr[], int n)
{
// To store number of even and odd integers
int odd = 0, even = 0;
// Find number of even and odd integers
for (int i = 0; i < n; i++)
{
if (arr[i] % 2 == 1)
odd++;
else
even++;
}
// If it possible to make
// lexicographically smaller
if (odd > 0 && even > 0)
Arrays.sort(arr);
// Print the array
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 5, 4, 3, 2 };
int n = arr.length;
// Function call
lexicographically_smaller(arr, n);
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 program to find possible
# lexicographically smaller array
# by swapping only elements whose sum is odd
# Function to find possible
# lexicographically smaller array
def lexicographically_smaller(arr, n):
# To store number of even and odd integers
odd, even = 0, 0;
# Find number of even and odd integers
for i in range(n):
if (arr[i] % 2 == 1):
odd += 1;
else:
even += 1;
# If it possible to make
# lexicographically smaller
if (odd > 0 and even > 0):
arr.sort();
# Print the array
for i in range(n):
print(arr[i], end = " ");
# Driver code
if __name__ == '__main__':
arr = [ 1, 5, 4, 3, 2 ];
n = len(arr);
# Function call
lexicographically_smaller(arr, n);
# This code contributed by Rajput-Ji
C
// C# program to find possible
// lexicographically smaller array by
// swapping only elements whose sum is odd
using System;
class GFG
{
// Function to find possible
// lexicographically smaller array
static void lexicographically_smaller(int []arr, int n)
{
// To store number of even and odd integers
int odd = 0, even = 0;
// Find number of even and odd integers
for (int i = 0; i < n; i++)
{
if (arr[i] % 2 == 1)
odd++;
else
even++;
}
// If it possible to make
// lexicographically smaller
if (odd > 0 && even > 0)
Array.Sort(arr);
// Print the array
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 5, 4, 3, 2 };
int n = arr.Length;
// Function call
lexicographically_smaller(arr, n);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// javascript program to find possible
// lexicographically smaller array
// by swapping only elements whose sum is odd
// Function to find possible lexicographically smaller array
function lexicographically_smaller(arr , n)
{
// To store number of even and odd integers
var odd = 0, even = 0;
// Find number of even and odd integers
for (var i = 0; i < n; i++)
{
if (arr[i] % 2 == 1)
odd++;
else
even++;
}
// If it possible to make
// lexicographically smaller
if (odd > 0 && even > 0)
arr.sort((a,b)=>a-b);
// Print the array
for (i = 0; i < n; i++)
document.write(arr[i] + " ");
}
// Driver code
var arr = [ 1, 5, 4, 3, 2 ];
var n = arr.length;
// Function call
lexicographically_smaller(arr, n);
// This code is contributed by 29AjayKumar
</script>
Output:
1 2 3 4 5
时间复杂度 : O(N log N)
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