Find absolute difference between product of rowwise and columnwise min and max in given Matrix
Last Updated :
01 Nov, 2022
Given a square matrix M[][] of size N x N, the task is to find the maximums & minimums of each of the rows and columns, multiply the corresponding maximums & minimums of each row with those of each column, Finally return the absolute difference of both of these.
Examples:
Input: M[][] = [ [ 3, 2, 5 ], [ 7, 5, 4 ], [ 7, 2, 9 ] ]
Output: 11
Explanation:
Input: M = [ [1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
Output: 40
Naive Approach: The task can be solved by simulating the given operations. Follow the below steps to solve the problem:
- Take 2 arrays max1[ ] and min1[ ] for storing the maximum and minimum of each row or each column.
- First store the maximum of each row in max1[ ] and a minimum of each row in min1[ ].
- Multiple both matrix and return res, store it in R.
- Take transpose of matrix M[ ].
- Then store the maximum of each column in max1[ ] and the minimum of each column in min1[ ].
- Multiply both matrices and return res, store it in C.
- Print absolute difference of R and C.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the transpose matrix M[]
vector<vector<int>> transpose(vector<vector<int>>M, int N){
for(int i = 0; i < N; i++)
{
for(int j = 0; j < i + 1; j++)
{
int temp = M[i][j];
M[i][j] = M[j][i];
M[j][i] = temp;
}
}
return M;
}
// Function to convert matrix M[][] to size 1 * 1
int matOne(vector<vector<int>>M, int N)
{
// Max1 and min1 to store max and min
// of each row pr column
vector<int>max1;
vector<int>min1;
// Loop to store element in max1 and min1
for(int r = 0; r < N; r++)
{
max1.push_back(*max_element(M[r].begin(), M[r].end()));
min1.push_back(*min_element(M[r].begin(), M[r].end()));
}
// Initialise res to 0
int res = 0;
// Multiply and add both array
for(int i = 0; i < N; i++){
res += max1[i]*min1[i];
}
// Return res
return res;
}
// Driver code
int main()
{
// Original matrix
vector<vector<int>>M = {{3, 2, 5},{7, 5, 4},{7, 2, 9}};
// N size of matrix
int N = M.size();
// Call matOne function for rows
int R = matOne(M, N);
// Transpose the matrix
vector<vector<int>>T = transpose(M, N);
// Call matOne function for column
int C = matOne(T, N);
// Print the absolute difference
cout << abs(R - C) << endl;
}
// This code is contributed by shinjanpatra
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to find the transpose matrix M[]
static int[][] transpose(int[][] M, int N) {
int transpose[][] = new int[N][N];
// Code to transpose a matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
transpose[i][j] = M[j][i];
}
}
return transpose;
}
// Function to convert matrix M[][] to size 1 * 1
static int matOne(int[][] M, int N) {
// Max1 and min1 to store max and min
// of each row pr column
ArrayList<Integer> max1 = new ArrayList<Integer>();
ArrayList<Integer> min1 = new ArrayList<Integer>();
// Loop to calculate res.
for (int[] r : M)
{
max1.add(Arrays.stream(r).max().getAsInt());
min1.add(Arrays.stream(r).min().getAsInt());
}
// Initialise res to 0
int res = 0;
// Multiply and add both array
for(int i = 0; i < N; i++)
{
res += max1.get(i)*min1.get(i);
}
// Return res
return res;
}
// Driver code
public static void main(String[] args)
{
// Original matrix
int[][] M = { { 3, 2, 5 }, { 7, 5, 4 }, { 7, 2, 9 } };
// N size of matrix
int N = M.length;
// Call matOne function for rows
int R = matOne(M, N);
// Transpose the matrix
int[][] T = transpose(M, N);
// Call matOne function for column
int C = matOne(T, N);
// Print the absolute difference
System.out.println((Math.abs(R - C)));
}
}
// This code is contributed by aditya patil
# Python program for the above approach
# Function to find the transpose matrix M[]
def transpose(M, N):
for i in range(N):
for j in range(i + 1):
M[i][j], M[j][i] = M[j][i], M[i][j]
return M
# Function to convert matrix M[][] to size 1 * 1
def matOne(M, N):
# Max1 and min1 to store max and min
# of each row pr column
max1 = []
min1 = []
# Loop to store element in max1 and min1
for r in M:
max1.append(max(r))
min1.append(min(r))
# Initialise res to 0
res = 0
# Multiply and add both array
for i in range(N):
res += max1[i]*min1[i]
# Return res
return res
# Driver code
if __name__ == "__main__":
# Original matrix
M = [[3, 2, 5],
[7, 5, 4],
[7, 2, 9]]
# N size of matrix
N = len(M)
# Call matOne function for rows
R = matOne(M, N)
# Transpose the matrix
T = transpose(M, N)
# Call matOne function for column
C = matOne(T, N)
# Print the absolute difference
print(str(abs(R-C)))
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
public class GFG {
// Function to find the transpose matrix []M
static int[, ] transpose(int[, ] M, int N)
{
int[, ] transpose = new int[N, N];
// Code to transpose a matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
transpose[i, j] = M[j, i];
}
}
return transpose;
}
// Function to convert matrix M[][] to size 1 * 1
static int matOne(int[, ] M, int N)
{
// Max1 and min1 to store max and min
// of each row pr column
ArrayList max1 = new ArrayList();
ArrayList min1 = new ArrayList();
// Loop to calculate res.
for (int r = 0; r < N; r++) {
int[] t = new int[N];
for (int j = 0; j < N; j++) {
t[j] = M[r, j];
}
min1.Add(t.Min());
max1.Add(t.Max());
}
// Initialise res to 0
int res = 0;
// Multiply and add both array
for (int i = 0; i < N; i++) {
res += (int)max1[i] * (int)min1[i];
}
// Return res
return res;
}
static public void Main()
{
// Code
// Original matrix
int[, ] M
= { { 3, 2, 5 }, { 7, 5, 4 }, { 7, 2, 9 } };
// N size of matrix
int N = M.GetLength(0);
// Call matOne function for rows
int R = matOne(M, N);
// Transpose the matrix
int[, ] T = transpose(M, N);
// Call matOne function for column
int C = matOne(T, N);
// Print the absolute difference
Console.WriteLine((Math.Abs(R - C)));
}
}
// This code is contributed by lokeshmvs21.
<script>
// Javascript program for the above approach
// Function to find the transpose matrix M[]
function transpose(M) {
return M[0].map((col, i) => M.map(row => row[i]));
}
// Function to convert matrix M[][] to size 1 * 1
function matOne(M, N) {
// Max1 and min1 to store max and min
// of each row pr column
let max1 = []
let min1 = []
// Loop to store element in max1 and min1
for (r of M){
max1.push([...r].sort((a, b) => b - a)[0])
min1.push([...r].sort((a, b) => a - b)[0])
}
// Initialise res to 0
let res = 0
// Multiply and add both array
for (let i = 0; i < N; i++)
res += max1[i] * min1[i]
// Return res
return res
}
// Driver code
// Original matrix
let M = [[3, 2, 5], [7, 5, 4], [7, 2, 9]];
// N size of matrix
let N = M.length;
// Call matOne function for rows
let R = matOne(M, N)
// Transpose the matrix
let T = transpose(M)
// Call matOne function for column
let C = matOne(T, N)
// Print the absolute difference
document.write(Math.abs(R - C))
// This code is contributed by saurabh_jaiswal.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
Space Optimized Approach: This approach is similar to approach 1, but in this, the auxiliary space will be optimized by not using max1 and min1 arrays. In this, direct multiple and add the element and return it.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the transpose matrix M[]
vector<vector<int>> transpose(vector<vector<int>>M, int N){
for(int i = 0; i < N; i++)
{
for(int j = 0; j < i + 1; j++)
{
int temp = M[i][j];
M[i][j] = M[j][i];
M[j][i] = temp;
}
}
return M;
}
// Function to convert matrix M[][] to size 1 * 1
int matOne(vector<vector<int>>M, int N)
{
// Max1 and min1 to store max and min
// of each row pr column
vector<int>arr;
int res = 0;
for (int r =0;r< N;r++){
arr.clear();
for(int j =0;j<N;j++){
arr.push_back(M[r][j]);
}
sort(arr.begin(),arr.end());
res += arr[arr.size()-1] * arr[0];
}
// Return res
return res;
}
// Driver code
int main()
{
// Original matrix
vector<vector<int>>M = {{3, 2, 5},{7, 5, 4},{7, 2, 9}};
// N size of matrix
int N = M.size();
// Call matOne function for rows
int R = matOne(M, N);
// Transpose the matrix
vector<vector<int>>T = transpose(M, N);
// Call matOne function for column
int C = matOne(T, N);
// Print the absolute difference
cout << abs(R - C) << endl;
}
// This code is contributed by akshitsaxenaa09.
import java.util.Arrays;
// Java program for the above approach
class GFG
{
// Function to find the transpose matrix M[]
static int[][] transpose(int[][] M, int N) {
int transpose[][] = new int[N][N];
// Code to transpose a matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
transpose[i][j] = M[j][i];
}
}
return transpose;
}
// Function to convert matrix M[][] to size 1 * 1
static int matOne(int[][] M, int N) {
// Loop to calculate res.
int res = 0;
for (int[] r : M)
res += Arrays.stream(r).max().getAsInt() * Arrays.stream(r).min().getAsInt();
// Return res
return res;
}
// Driver code
public static void main(String[] args)
{
// Original matrix
int[][] M = { { 3, 2, 5 }, { 7, 5, 4 }, { 7, 2, 9 } };
// N size of matrix
int N = M.length;
// Call matOne function for rows
int R = matOne(M, N);
// Transpose the matrix
int[][] T = transpose(M, N);
// Call matOne function for column
int C = matOne(T, N);
// Print the absolute difference
System.out.println((Math.abs(R - C)));
}
}
// This code is contributed by 29AjayKumar
# Python program for the above approach
# Function to find the transpose matrix M[]
def transpose(M, N):
for i in range(N):
for j in range(i + 1):
M[i][j], M[j][i] = M[j][i], M[i][j]
return M
# Function to convert matrix M[][] to size 1 * 1
def matOne(M, N):
# Loop to calculate res.
res = 0
for r in M:
res += max(r)*min(r)
# Return res
return res
# Driver code
if __name__ == "__main__":
# Original matrix
M = [[3, 2, 5],
[7, 5, 4],
[7, 2, 9]]
# N size of matrix
N = len(M)
# Call matOne function for rows
R = matOne(M, N)
# Transpose the matrix
T = transpose(M, N)
# Call matOne function for column
C = matOne(T, N)
# Print the absolute difference
print(str(abs(R-C)))
// C# program for the above approach
using System;
using System.Linq;
public class GFG
{
// Function to find the transpose matrix []M
static int[,] transpose(int[,] M, int N)
{
int [,]transpose = new int[N,N];
// Code to transpose a matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
transpose[i,j] = M[j,i];
}
}
return transpose;
}
// Function to convert matrix [,]M to size 1 * 1
static int matOne(int[,] M, int N) {
// Loop to calculate res.
int res = 0;
for (int r =0;r< N;r++){
int[] t = new int[N];
for(int j =0;j<N;j++){
t[j] = M[r,j];
}
res += t.Max() * t.Min();
}
// Return res
return res;
}
// Driver code
public static void Main(String[] args)
{
// Original matrix
int[,] M = { { 3, 2, 5 }, { 7, 5, 4 }, { 7, 2, 9 } };
// N size of matrix
int N = M.GetLength(0);
// Call matOne function for rows
int R = matOne(M, N);
// Transpose the matrix
int[,] T = transpose(M, N);
// Call matOne function for column
int C = matOne(T, N);
// Print the absolute difference
Console.WriteLine((Math.Abs(R - C)));
}
}
// This code is contributed by umadevi9616
<script>
// JavaScript code for the above approach
// Function to find the transpose matrix M[]
function transpose(M, N)
{
// This function stores transpose
// of A in B
let B = Array(M).fill(0);
for (let i = 0; i < N; i++) {
B[i] = new Array(N);
}
var i, j;
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
B[i][j] = M[j][i];
return B
}
// Function to convert matrix M[][] to size 1 * 1
function matOne(M, N) {
// Max1 and min1 to store max and min
// of each row pr column
max1 = []
min1 = []
// Loop to store element in max1 and min1
for (let i = 0; i < M.length; i++) {
r = M[i]
max1.push(Math.max(...r))
min1.push(Math.min(...r))
}
// Initialise res to 0
res = 0
// Multiply and add both array
for (let i = 0; i < N; i++)
res += (max1[i] * min1[i])
// Return res
return res
}
// Driver code
// Original matrix
let M = [[3, 2, 5],
[7, 5, 4],
[7, 2, 9]]
// N size of matrix
let N = M.length
// Call matOne function for rows
let R = matOne(M, N)
// Transpose the matrix
T = transpose(M, N)
// Call matOne function for column
C = matOne(T, N)
// Print the absolute difference
document.write(Math.abs(R - C))
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N2log(N))
Auxiliary Space: O(N) because using extra space for array
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