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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 the given matrix into the 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 :
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:  

Below is the implementation of the above approach: 
 




// C++ implementation of above approach
#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;
}




// Java implementation of above approach
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




# Python3 implementation of above approach
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




// C# implementation of above approach
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




<script>
// Java script implementation of above approach
let N = 4;
     
    // Function to count steps in
    // conversion of matrix into upper
    // Hessenberg matrix
    function stepsRequired(arr)
    {
        let result = 0;
        for (let i = 0; i < N; i++)
        {
     
            for (let 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   
        let arr =[[1, 2, 3, 4],
                        [3, 1, 0, 3],
                        [3, 2, 1, 3],
                        [-3, 4, 2, 1 ]];
     
        // Function call
        document.write(stepsRequired(arr));
 
// This code is contributed by mohan pavan
 
</script>

Output: 
10

 

Time complexity : O(N*N)
Auxiliary space: O(1) because it is using constant space for variables


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