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📜  检查一个数组是否可以拆分为由单个不同元素组成的,长度为M的K个连续的不重叠子数组

📅  最后修改于: 2021-04-18 02:42:56             🧑  作者: Mango

给定两个整数MK,以及一个由N个正整数组成的数组arr [] ,任务是检查该数组是否可以拆分为长度为M的K个连续的不重叠子数组,以使每个子数组都由单个不同的元素组成。如果发现是真的,则打印“是” 。否则,打印“否”

例子:

方法:可以通过使用简单的数组遍历并检查当前索引i上的元素和索引(i + M)上的元素是否相同来解决给定的问题。请按照以下步骤解决问题:

  • 分别用10初始化两个变量countt,以分别存储匹配模式的总数和当前匹配模式的长度。
  • 使用变量i遍历[0,N – M – 1]范围内的给定数组,并执行以下操作:
    • 如果arr [i]arr [i + M]的值相同,则将t1 ;如果tm相同,则将t更新为0并增加count 。如果count的值为K,则打印“是”并退出循环。
    • 否则,如果tM,则将计数增加1
  • 完成上述步骤后,如果count的值与K不同,则打印“ No” ,因为不存在任何这种模式。

下面是上述方法的实现:

C++
// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if array can be split
// into K consecutive and non-overlapping
// subarrays of length M consisting of a
// single distinct element
string checkPattern(int arr[], int m,
                    int k, int n)
{
    int count = 1, t = 0;
 
    // Traverse over the range [0, N - M - 1]
    for (int i = 0; i < n - m; i++) {
 
        // Check if arr[i] is the
        // same as arr[i + m]
        if (arr[i] == arr[i + m]) {
 
            // Increment current length
            // t of pattern matched by 1
            t++;
 
            // Check if t is equal to m,
            // increment count of total
            // repeated pattern
            if (t == m) {
 
                t = 0;
                count++;
 
                // Return true if length of
                // total repeated pattern is k
                if (count == k) {
                    return "Yes";
                }
            }
        }
 
        else {
 
            // Update length of the
            // current pattern
            t = 0;
 
            // Update count to 1
            count = 1;
        }
    }
 
    // Finally return false if
    // no pattern found
    return "No";
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 1, 3, 3, 3, 3 };
    int M = 1, K = 3;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << checkPattern(arr, M, K, N);
 
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if array can be split
// into K consecutive and non-overlapping
// subarrays of length M consisting of a
// single distinct element
static String checkPattern(int arr[], int m,
                    int k, int n)
{
    int count = 1, t = 0;
 
    // Traverse over the range [0, N - M - 1]
    for (int i = 0; i < n - m; i++)
    {
 
        // Check if arr[i] is the
        // same as arr[i + m]
        if (arr[i] == arr[i + m])
        {
 
            // Increment current length
            // t of pattern matched by 1
            t++;
 
            // Check if t is equal to m,
            // increment count of total
            // repeated pattern
            if (t == m)
            {
                t = 0;
                count++;
 
                // Return true if length of
                // total repeated pattern is k
                if (count == k)
                {
                    return "Yes";
                }
            }
        }
        else
        {
 
            // Update length of the
            // current pattern
            t = 0;
 
            // Update count to 1
            count = 1;
        }
    }
 
    // Finally return false if
    // no pattern found
    return "No";
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 6, 1, 3, 3, 3, 3 };
    int M = 1, K = 3;
    int N = arr.length;
    System.out.print(checkPattern(arr, M, K, N));
}
}
 
// This code is contributed by 29AjayKumar


Python3
# Python 3 program for the above approach
 
# Function to check if array can be split
# into K consecutive and non-overlapping
# subarrays of length M consisting of a
# single distinct element
def checkPattern(arr, m, k, n):
    count = 1
    t = 0
 
    # Traverse over the range [0, N - M - 1]
    for i in range(n - m):
       
        # Check if arr[i] is the
        # same as arr[i + m]
        if (arr[i] == arr[i + m]):
           
            # Increment current length
            # t of pattern matched by 1
            t += 1
 
            # Check if t is equal to m,
            # increment count of total
            # repeated pattern
            if (t == m):
                t = 0
                count += 1
 
                # Return true if length of
                # total repeated pattern is k
                if (count == k):
                    return "Yes"
 
        else:
            # Update length of the
            # current pattern
            t = 0
 
            # Update count to 1
            count = 1
 
    # Finally return false if
    # no pattern found
    return "No"
 
# Driver Code
if __name__ == '__main__':
    arr  =  [6, 1, 3, 3, 3, 3]
    M = 1
    K = 3
    N = len(arr)
 
    print(checkPattern(arr, M, K, N))
     
    # This code is contributed by bgangwar59.


C#
// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to check if array can be split
  // into K consecutive and non-overlapping
  // subarrays of length M consisting of a
  // single distinct element
  static String checkPattern(int []arr, int m,
                             int k, int n)
  {
    int count = 1, t = 0;
 
    // Traverse over the range [0, N - M - 1]
    for (int i = 0; i < n - m; i++)
    {
 
      // Check if arr[i] is the
      // same as arr[i + m]
      if (arr[i] == arr[i + m])
      {
 
        // Increment current length
        // t of pattern matched by 1
        t++;
 
        // Check if t is equal to m,
        // increment count of total
        // repeated pattern
        if (t == m)
        {
          t = 0;
          count++;
 
          // Return true if length of
          // total repeated pattern is k
          if (count == k)
          {
            return "Yes";
          }
        }
      }
      else
      {
 
        // Update length of the
        // current pattern
        t = 0;
 
        // Update count to 1
        count = 1;
      }
    }
 
    // Finally return false if
    // no pattern found
    return "No";
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 6, 1, 3, 3, 3, 3 };
    int M = 1, K = 3;
    int N = arr.Length;
    Console.Write(checkPattern(arr, M, K, N));
  }
}
 
// This code is contributed by Rajput-Ji


输出:
Yes

时间复杂度: O(N)
辅助空间: O(1)