Minimum steps required to convert the matrix into lower hessenberg matrix
Last Updated :
22 Mar, 2022
Given a matrix of order NxN, Find the minimum number of steps to convert given matrix into Lower Hessenberg matrix. In each step, the only operation allowed is to decrease or increase any element value by 1.
Examples:
Input: mat[][] = {
{1, 2, 8},
{1, 3, 4},
{2, 3, 4}}
Output: 8
Decrease the element a[0][2] 8 times.
Now the matrix is lower Hessenberg.
Input: mat[][] = {
{9, 2, 5, 5},
{12, 3, 4, 5},
{13, -3, 4, 2},
{-1, 10, 1, 4}}
Output: 15
Approach:
- For a matrix to be Lower Hessenberg matrix all of its elements above super-diagonal must be equal zero, i.e Aij = 0 for all j > i+1..
- The minimum number of steps required to convert a given matrix in the lower Hessenberg matrix is equal to the sum of the absolute values of all Aij for all j > i + 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++
#include <bits/stdc++.h>
#define N 4
using namespace std;
int stepsRequired( int arr[][N])
{
int result = 0;
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++) {
if (j > i + 1)
result += abs (arr[i][j]);
}
}
return result;
}
int main()
{
int arr[N][N] = { 1, 2, 3, 2,
3, 1, 0, 3,
3, 2, 1, 3,
-3, 4, 2, 1 };
cout << stepsRequired(arr);
return 1;
}
|
Java
import java.io.*;
class GFG
{
static int N = 4 ;
static int stepsRequired( int arr[][])
{
int result = 0 ;
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < N; j++)
{
if (j > i + 1 )
result += Math.abs(arr[i][j]);
}
}
return result;
}
public static void main (String[] args)
{
int [][]arr = { { 1 , 2 , 3 , 2 },
{ 3 , 1 , 0 , 3 },
{ 3 , 2 , 1 , 3 },
{- 3 , 4 , 2 , 1 }
};
System.out.println (stepsRequired(arr));
}
}
|
Python3
N = 4
def stepsRequired(arr):
result = 0
for i in range (N):
for j in range (N):
if (j > i + 1 ):
result + = abs (arr[i][j])
return result
arr = [[ 1 , 2 , 3 , 2 ],
[ 3 , 1 , 0 , 3 ],
[ 3 , 2 , 1 , 3 ],
[ - 3 , 4 , 2 , 1 ]]
print (stepsRequired(arr))
|
C#
using System;
class GFG
{
static int N = 4;
static int stepsRequired( int [,]arr)
{
int result = 0;
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
if (j > i + 1)
result += Math.Abs(arr[i,j]);
}
}
return result;
}
static public void Main ()
{
int [,]arr = { {1, 2, 3, 2},
{3, 1, 0, 3},
{3, 2, 1, 3},
{-3, 4, 2, 1}
};
Console.Write(stepsRequired(arr));
}
}
|
Javascript
<script>
let N = 4;
function stepsRequired(arr)
{
let result = 0;
for (let i = 0; i < N; i++)
{
for (let j = 0; j < N; j++)
{
if (j > i + 1)
result += Math.abs(arr[i][j]);
}
}
return result;
}
let arr= [[1, 2, 3, 2],
[3, 1, 0, 3],
[3, 2, 1, 3],
[-3, 4, 2, 1]
];
document.write (stepsRequired(arr));
</script>
|
Time complexity: O(N*N)
Auxiliary Space: O(1)
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