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 hessenbergInput : 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++
// 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
// 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
# 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#
// 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 |
Output:
10
Time complexity : O(N*N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.