Minimum number of steps to convert a given matrix into Upper Hessenberg matrix

Given a matrix of order NxN, Find the minimum number of steps to convert given matrix into Upper Hessenberg matrix. In each step, the only operation allowed is to decrease or increase any element value by 1.

Examples:

Input : N=3
1 2 8
1 3 4
2 3 4
Output :2
Decrease the element a[2][0] 2 times.
Now the matrix is upper hessenberg



Input : N=4
1 2 2 3
1 3 4 2
3 3 4 2
-1 0 1 4
Output :4

Approach:

  • For a matrix to be Upper Hessenberg matrix all of its elements below sub-diagonal must be equal zero, i.e Aij = 0 for all i > j+1..
  • The minimum number of steps required to convert a given matrix in the upper Hessenberg matrix is equal to the sum of the absolute values of all Aij for all i > j + 1.
  • The modulus value of the element is taken into account because both the increase and decrease of the element count as a single step.

Below is the implementation of the above approach:

C++

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// C++ implementation of above apporach
#include <bits/stdc++.h>
#define N 4
using namespace std;
  
// Function to count steps in
// conversion of matrix into upper
// Hessenberg matrix
int stepsRequired(int arr[][N])
{
    int result = 0;
    for (int i = 0; i < N; i++) {
  
        for (int j = 0; j < N; j++) {
  
            // if element is below sub-diagonal
            // add abs(element) into result
            if (i > j + 1)
                result += abs(arr[i][j]);
        }
    }
    return result;
}
  
// Driver code
int main()
{
    int arr[N][N] = { 1, 2, 3, 4,
                      3, 1, 0, 3,
                      3, 2, 1, 3,
                     -3, 4, 2, 1 };
  
    // Function call
    cout << stepsRequired(arr);
    return 0;
}

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Java














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// Java implementation of above apporach 


class GFG 


{ 


      


    static int N = 4; 


      


    // Function to count steps in 


    // conversion of matrix into upper 


    // Hessenberg matrix 


    static int stepsRequired(int arr[][]) 


    { 


        int result = 0; 


        for (int i = 0; i < N; i++) 


        { 


      


            for (int j = 0; j < N; j++) 


            { 


      


                // if element is below sub-diagonal 


                // add abs(element) into result 


                if (i > j + 1) 


                    result += Math.abs(arr[i][j]); 


            } 


        } 


        return result; 


    } 


      


    // Driver code 


    public static void main (String[] args) 


    { 


          


        int arr [][] = new int [][] {{1, 2, 3, 4}, 


                        {3, 1, 0, 3}, 


                        {3, 2, 1, 3}, 


                        {-3, 4, 2, 1 }}; 


      


        // Function call 


        System.out.println(stepsRequired(arr)); 


    } 


} 


  


// This code is contributed by ihritik 



















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Python3














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# Python3 implementation of above apporach 


N = 4; 


  


# Function to count steps in 


# conversion of matrix into upper 


# Hessenberg matrix 


def stepsRequired(arr): 


    result = 0; 


    for i in range(N): 


  


        for j in range(N): 


  


            # if element is below sub-diagonal 


            # add abs(element) into result 


            if (i > j + 1): 


                result += abs(arr[i][j]); 


  


    return result; 


  


# Driver code 


arr =   [[1, 2, 3, 4], 


         [3, 1, 0, 3], 


         [3, 2, 1, 3], 


         [-3, 4, 2, 1]]; 


  


# Function call 


print(stepsRequired(arr)); 


  


# This code is contributed by Rajput-Ji 



















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C#














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// C# implementation of above apporach 


using System; 


  


class GFG 


{ 


      


    static int N = 4; 


      


    // Function to count steps in 


    // conversion of matrix into upper 


    // Hessenberg matrix 


    static int stepsRequired(int [, ] arr) 


    { 


        int result = 0; 


        for (int i = 0; i < N; i++) 


        { 


      


            for (int j = 0; j < N; j++)  


            { 


      


                // if element is below sub-diagonal 


                // add abs(element) into result 


                if (i > j + 1) 


                    result += Math.Abs(arr[i, j]); 


            } 


        } 


        return result; 


    } 


      


    // Driver code 


    public static void Main ()  


    { 


          


        int [ , ] arr = new int [, ] { {1, 2, 3, 4}, 


                        {3, 1, 0, 3}, 


                        {3, 2, 1, 3}, 


                        {-3, 4, 2, 1}}; 


      


        // Function call 


        Console.WriteLine(stepsRequired(arr)); 


      


    } 


} 


  


// This code is contributed by ihritik 



















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Output:


10





 Time complexity :  O(N*N)



          
          
          

          

    

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Improved By :  ihritik, Rajput-Ji


					
		

    Article Tags : 
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Practice Tags : 
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