给定一个表示前N个自然数的排列的数组arr [] ，任务是通过交换最多一对数组元素来找到给定数组arr []的词典最小排列。如果无法按字典顺序缩小数组，请打印“ -1” 。
例子：
Input: arr[] = {3, 2, 1, 4}
Output: 1 2 3 4
Explanation: Swapping elements at index 2 and 0, the modified array is {1, 2, 3, 4}, which is lexicographically the smallest permutation of the given array arr[].
Input: arr[] = {1, 2, 3, 4}
Output: -1
方法：想法是找到不在正确位置的第一个数组元素，即arr [i]与索引(i + 1)不同，然后将其与正确位置的元素交换。请按照以下步骤解决此问题：
- 遍历数组arr []并找到索引i ，以使arr [i]不等于(i + 1) (例如idx) ，并将(i + 1)存储在变量中，例如ele 。
- 现在，找到ele的索引，例如newIdx 。
- 完成上述步骤后，存在两个索引idx和newIdx 。通过在索引idx和newIdx处交换元素，可以形成数组在字典上最小的排列。现在，打印数组arr [] 。否则，打印“ -1” 。
下面是上述方法的实现：
C++14
// C++ implementation of the above approach
#include
#include
using namespace std;
// Function to print the
// elements of the array arr[]
void print(int arr[], int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
// Function to convert given array to
// lexicographically smallest permutation
// possible by swapping at most one pair
void makeLexicographically(int arr[], int N)
{
// Stores the index of the first
// element which is not at its
// correct position
int index = 0;
int temp = 0;
// Checks if any such array
// element exists or not
int check = 0;
int condition = 0;
int element = 0;
// Traverse the given array
for (int i = 0; i < N; ++i) {
// If element is found at i
if (element == arr[i]) {
check = i;
break;
}
// If the first array is
// not in correct position
else if (arr[i] != i + 1 && check == 0) {
// Store the index of
// the first elements
index = i;
check = 1;
condition = -1;
// Store the index of
// the first element
element = i + 1;
}
}
// Swap the pairs
if (condition == -1) {
temp = arr[index];
arr[index] = arr[check];
arr[check] = temp;
}
// Print the array
print(arr, N);
}
// Driver code
int main()
{
// Given array
int arr[] = { 1, 2, 3, 4 };
// Store the size of the array
int N = sizeof(arr) / sizeof(arr[0]);
makeLexicographically(arr, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to print the
// elements of the array arr[]
static void print(int arr[])
{
// Traverse the array arr[]
for (int element : arr) {
System.out.print(element + " ");
}
}
// Function to convert given array to
// lexicographically smallest permutation
// possible by swapping at most one pair
static void makeLexicographically(
int arr[], int length)
{
// Stores the index of the first
// element which is not at its
// correct position
int index = 0;
int temp = 0;
// Checks if any such array
// element exists or not
int check = 0;
int condition = 0;
int element = 0;
// Traverse the given array
for (int i = 0; i < length; ++i) {
// If element is found at i
if (element == arr[i]) {
check = i;
break;
}
// If the first array is
// not in correct position
else if (arr[i] != i + 1
&& check == 0) {
// Store the index of
// the first elements
index = i;
check = 1;
condition = -1;
// Store the index of
// the first element
element = i + 1;
}
}
// Swap the pairs
if (condition == -1) {
temp = arr[index];
arr[index] = arr[check];
arr[check] = temp;
}
// Print the array
print(arr);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4 };
int N = arr.length;
makeLexicographically(arr, N);
}
}Python3
# Python3 program for the above approach
# Function to prthe
# elements of the array arr[]
def printt(arr, N) :
# Traverse the array arr[]
for i in range(N):
print(arr[i], end = " ")
# Function to convert given array to
# lexicographically smallest permutation
# possible by swapping at most one pair
def makeLexicographically(arr, N) :
# Stores the index of the first
# element which is not at its
# correct position
index = 0
temp = 0
# Checks if any such array
# element exists or not
check = 0
condition = 0
element = 0
# Traverse the given array
for i in range(N):
# If element is found at i
if (element == arr[i]) :
check = i
break
# If the first array is
# not in correct position
elif (arr[i] != i + 1 and check == 0) :
# Store the index of
# the first elements
index = i
check = 1
condition = -1
# Store the index of
# the first element
element = i + 1
# Swap the pairs
if (condition == -1) :
temp = arr[index]
arr[index] = arr[check]
arr[check] = temp
# Prthe array
printt(arr, N)
# Driver code
# Given array
arr = [ 1, 2, 3, 4 ]
# Store the size of the array
N = len(arr)
makeLexicographically(arr, N)
# This code is contributed by code_hunt.C#
// C# program for the above approach
using System;
class GFG {
// Function to print the
// elements of the array arr[]
static void print(int[] arr)
{
// Traverse the array arr[]
foreach (int element in arr) {
Console.Write(element + " ");
}
}
// Function to convert given array to
// lexicographically smallest permutation
// possible by swapping at most one pair
static void makeLexicographically(
int []arr, int length)
{
// Stores the index of the first
// element which is not at its
// correct position
int index = 0;
int temp = 0;
// Checks if any such array
// element exists or not
int check = 0;
int condition = 0;
int element = 0;
// Traverse the given array
for (int i = 0; i < length; ++i) {
// If element is found at i
if (element == arr[i]) {
check = i;
break;
}
// If the first array is
// not in correct position
else if (arr[i] != i + 1
&& check == 0) {
// Store the index of
// the first elements
index = i;
check = 1;
condition = -1;
// Store the index of
// the first element
element = i + 1;
}
}
// Swap the pairs
if (condition == -1) {
temp = arr[index];
arr[index] = arr[check];
arr[check] = temp;
}
// Print the array
print(arr);
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 1, 2, 3, 4 };
int N = arr.Length;
makeLexicographically(arr, N);
}
}
// This code is contributed by ukasp.输出：
1 2 3 4时间复杂度： O(N)
辅助空间： O(1)