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Check if every row in given Matrix contains all the integers from 1 to N

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  • Difficulty Level : Medium
  • Last Updated : 31 Jan, 2022

Given a matrix arr[][] of size M*N containing only positive integers, the task is to check if every row contains all the integers from 1 to N

Examples: 

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

 

Naive Approach: The simplest approach is to sort each row of the matrix and then check each row if it contains all elements from 1 to N or not.

Time Complexity: O(M * N * logN)
Auxiliary Space: O(M * N)

Efficient approach: This problem can be solved using the approach of Find the smallest positive number missing from an unsorted array. The idea is to use the given matrix to use as a structure to mark the numbers that we found. So, if a number between 1 to N is found, mark the index corresponding to that number by inverting the element present at that index. 

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

After applying this operation on all elements in the matrix, the matrix should contain all negatives otherwise all rows do not contain numbers from 1 to N.

Below is the implementation of the above approach:

C++




// C++ code for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if every row contains
// all the integers from 1 to N
bool checkMat(vector<vector<int> >& 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<vector<int> > 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




<script>
     // 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
 </script>

 
 

Output

Yes

 

Time Complexity: O(M * N)
Auxiliary Space: O(1)

 


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