给定一个大小为N的圆形数组arr[]和两个整数K和M ,任务是从索引K开始对M数组元素进行排序。
例子:
Input: arr[] = {4, 1, 6, 5, 3}, K = 2, M = 3
Output: 4 1 3 5 6
Explanation: After sorting 3 array elements starting from index 2 modifies arr[] to {4, 1, 3, 5, 6}.
Input: arr[] = {67, 2, 9, 7, 1}, K = 4, M = 3
Output: 2 67 9 7 1
Explanation: After sorting 3 array elements starting from index 4 modifies arr[] to {2, 67, 9, 7, 1}.
方法:这个想法是如果它们的元素顺序不正确,则交换圆形数组中的相邻元素。请按照以下步骤解决给定的问题:
- 使用变量i在[0, M – 1]范围内遍历给定数组arr[] 。
- 使用变量j在[K, K + M – 1]范围内再次遍历数组arr[] ,如果arr[j%n]大于arr[(j + 1)%n],则交换当前相邻对值。
- 完成上述步骤后,打印数组arr[] 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the circular array
void printCircularArray(int arr[], int n)
{
// Print the array
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
// Function to sort m elements of diven
// circular array starting from index k
void sortCircularArray(int arr[], int n,
int k, int m)
{
// Traverse M elements
for (int i = 0; i < m; i++) {
// Iterate from index k to k + m - 1
for (int j = k; j < k + m - 1; j++) {
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
swap(arr[j % n], arr[(j + 1) % n]);
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
int main()
{
int arr[] = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
sortCircularArray(arr, N, K, M);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to print the circular array
static void printCircularArray(int arr[], int n)
{
// Print the array
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
// Function to sort m elements of diven
// circular array starting from index k
static void sortCircularArray(int arr[], int n,
int k, int m)
{
// Traverse M elements
for (int i = 0; i < m; i++) {
// Iterate from index k to k + m - 1
for (int j = k; j < k + m - 1; j++) {
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
int t = arr[j % n];
arr[j % n] = arr[(j + 1) % n];
arr[(j + 1) % n] = t;
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
public static void main (String[] args)
{
int[] arr = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = arr.length;
// Function Call
sortCircularArray(arr, N, K, M);
}
}
// This code is contributed by susmitakundugoaldanga
Python3
# Python3 program for the above approach
# Function to print the circular array
def printCircularArray(arr, n):
# Print the array
for i in range(n):
print(arr[i], end = " ")
# Function to sort m elements of diven
# circular array starting from index k
def sortCircularArray(arr, n, k, m):
# Traverse M elements
for i in range(m):
# Iterate from index k to k + m - 1
for j in range(k, k + m - 1):
# Check if the next element
# in the circular array is
# less than the current element
if (arr[j % n] > arr[(j + 1) % n]):
# Swap current element
# with the next element
arr[j % n], arr[(j + 1) % n] = (arr[(j + 1) % n],
arr[j % n])
# Print the circular array
printCircularArray(arr, n)
# Driver Code
if __name__ == "__main__" :
arr = [ 4, 1, 6, 5, 3 ]
K = 2
M = 3
N = len(arr)
# Function Call
sortCircularArray(arr, N, K, M)
# This code is contributed by AnkThon
C#
// C# program for the above approach
using System;
class GFG
{
// Function to print the circular array
static void printCircularArray(int []arr, int n)
{
// Print the array
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
// Function to sort m elements of diven
// circular array starting from index k
static void sortCircularArray(int []arr, int n,
int k, int m)
{
// Traverse M elements
for (int i = 0; i < m; i++)
{
// Iterate from index k to k + m - 1
for (int j = k; j < k + m - 1; j++)
{
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
int t = arr[j % n];
arr[j % n] = arr[(j + 1) % n];
arr[(j + 1) % n] = t;
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
public static void Main (string[] args)
{
int[] arr = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = arr.Length;
// Function Call
sortCircularArray(arr, N, K, M);
}
}
// This code is contributed by AnkThon
Javascript
输出:
4 1 3 5 6
时间复杂度: O(N 2 )
辅助空间: O(1)
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