检查给定矩阵中的每一行是否包含从 1 到 N 的所有整数

给定一个大小为M*N的矩阵arr[][]只包含正整数，任务是检查每一行是否包含从1到N的所有整数。

例子：

Input: arr[][] = {{1, 4, 2, 3}, {2, 3, 4, 1}, {3, 4, 2, 1}}
Output: Yes
Explanation: Every row contains all the numbers from 1 to 4

Input: arr[][] = {{1, 2}, {2, 3}}
Output: No

朴素方法：最简单的方法是对矩阵的每一行进行排序，然后检查每一行是否包含从 1 到 N 的所有元素。

时间复杂度： O(M * N * logN)
辅助空间： O(M * N)

高效的方法：这个问题可以使用 Find the minimum positive number missing from an unsorted array 的方法来解决。这个想法是使用给定的矩阵作为结构来标记我们找到的数字。因此，如果找到1 到 N之间的数字，则通过反转该索引处的元素来标记与该数字对应的索引。

For example if 2 is found on row 1, change arr[1][2] to arr[1][2] = -arr[1][2]

对矩阵中的所有元素应用此操作后，矩阵应包含所有负数，否则所有行都不包含从 1 到N的数字。

下面是上述方法的实现：

C++

// C++ code for the above approach
 
#include 
using namespace std;
 
// Function to check if every row contains
// all the integers from 1 to N
bool checkMat(vector >& arr)
{
 
    int N = arr.size();
    int M = arr[0].size();
 
    for (auto& x : arr) {
        for (auto& y : x) {
            if (abs(y) >= 1 and abs(y) <= M) {
                x[abs(y) - 1] = -x[abs(y) - 1];
            }
        }
    }
 
    // Check if the matrix contains
    // all negatives or not
    for (auto x : arr) {
        for (auto y : x) {
            if (y > 0) {
                return 0;
            }
        }
    }
 
    return true;
}
 
// Driver Code
int main()
{
    vector > arr = { { 1, 4, 2, 3 },
                                 { 2, 3, 4, 1 },
                                 { 3, 4, 2, 1 } };
    if (checkMat(arr)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to check if every row contains
  // all the integers from 1 to N
  static Boolean checkMat(int[][] arr)
  {
 
    int N = arr.length;
    int M = arr[0].length;
 
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++) {
        if (Math.abs(arr[i][j]) >= 1 && Math.abs(arr[i][j]) <= M) {
          arr[i][Math.abs(arr[i][j]) - 1] = -arr[i][Math.abs(arr[i][j]) - 1];
        }
      }
    }
 
    // Check if the matrix contains
    // all negatives or not
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++) {
        if (arr[i][j] > 0) {
          return false;
        }
      }
    }
 
    return true;
  }
 
  // Driver Code
 
  public static void main (String[] args) {
    int[][] arr = { { 1, 4, 2, 3 },
                   { 2, 3, 4, 1 },
                   { 3, 4, 2, 1 } };
    if (checkMat(arr)) {
      System.out.print("Yes");
    }
    else {
      System.out.print("No");
    }
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach
 
# Function to check if every row contains
# all the integers from 1 to N
def checkMat(arr):
 
    N = len(arr)
    M = len(arr[0])
 
    for x in arr:
        for y in x:
            if (abs(y) >= 1 and abs(y) <= M):
                x[abs(y) - 1] = -x[abs(y) - 1]
 
    # Check if the matrix contains
    # all negatives or not
    for x in arr:
        for y in x:
            if (y > 0):
                return 0
 
    return True
 
# Driver Code
arr = [[1, 4, 2, 3],
       [2, 3, 4, 1],
       [3, 4, 2, 1]]
if (checkMat(arr)):
    print("Yes")
 
else:
    print("No")
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code to implement the above approach
using System;
class GFG {
 
  // Function to check if every row contains
  // all the integers from 1 to N
  static Boolean checkMat(int[,] arr)
  {
 
    int N = arr.GetLength(0);
    int M = arr.GetLength(1);
 
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++) {
        if (Math.Abs(arr[i, j]) >= 1 && Math.Abs(arr[i, j]) <= M) {
          arr[i, Math.Abs(arr[i, j]) - 1] = -arr[i, Math.Abs(arr[i, j]) - 1];
        }
      }
    }
 
    // Check if the matrix contains
    // all negatives or not
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++) {
        if (arr[i, j] > 0) {
          return false;
        }
      }
    }
 
    return true;
  }
 
  // Driver Code
  public static void Main () {
    int[,] arr = { { 1, 4, 2, 3 },
                  { 2, 3, 4, 1 },
                  { 3, 4, 2, 1 } };
    if (checkMat(arr)) {
      Console.Write("Yes");
    }
    else {
      Console.Write("No");
    }
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

     // JavaScript code for the above approach
 
     // Function to check if every row contains
     // all the integers from 1 to N
     function checkMat(arr)
     {
 
         let N = arr.length;
         let M = arr[0].length;
 
         for (let x of arr) {
             for (let y of x) {
                 if (Math.abs(y) >= 1 && Math.abs(y) <= M) {
                     x[Math.abs(y) - 1] = -x[Math.abs(y) - 1];
                 }
             }
         }
 
         // Check if the matrix contains
         // all negatives or not
         for (let x of arr) {
             for (let y of x) {
                 if (y > 0) {
                     return 0;
                 }
             }
         }
 
         return true;
     }
 
     // Driver Code
     let arr = [[1, 4, 2, 3],
     [2, 3, 4, 1],
     [3, 4, 2, 1]];
     if (checkMat(arr)) {
         document.write("Yes");
     }
     else {
         document.write("No");
     }
 
    // This code is contributed by Potta Lokesh
 

输出

Yes

时间复杂度： O(M * N)
辅助空间： O(1)