Finding all saddle points in a matrix c++ - c++

Question

I'm working on a code that finds all saddle points in a matrix. Both smallest in their row and biggest in their column, and biggest in their row and smallest in their column fall under the definition (of my university) of a saddle point. Being a beginner I managed to get half of it done (finding saddle points which are smallest in their row and biggest in their column) by copying parts of what we've done in class and typing it myself. I have been stuck on it for quite some time and can't figure how to add the saddle points which are biggest in their row and smallest in their column to the program.
This is what I have so far:
#include <iostream>
#include <cstdlib>
using namespace std;
int a[10][10];
int x, y;
int pos_max(int j) //saddle points check
{
int max = 0;
for (int i = 1; i <= x - 1; i++) {
if (a[i][j] > a[max][j]) {
max = i;
}
}
return max;
}
int main() {
cout << "Enter the number of rows: ";
cin >> x;
cout << "Enter the number of columns: ";
cin >> y;
cout << "----------------------------" << endl;
for (int i = 0; i <= x - 1; i++) //input of the matrix
for (int j = 0; j <= y - 1; j++) {
cout << "a[" << i + 1 << ", " << j + 1 << "] = ";
cin >> a[i][j];
}
cout << "----------------------------\n";
for (int i = 0; i <= x - 1; i++) //visualization of the matrix
{
for (int j = 0; j <= y - 1; j++)
cout << a[i][j] << " ";
cout << endl;
}
cout << "----------------------------\n";
int r;
int flag = 0;
int i = y;
for (int j = 0; j <= y - 1; j++) {
r = pos_max(j);
for (i = 0; i <= y - 1; i++) {
if (a[r][i] < a[r][j]) {
break;
}
}
if (i == y) {
cout << "Saddle points are: ";
cout << "a[" << r + 1 << ", " << j + 1 << "] = " << a[r][j] << "\n";
flag = 1;
}
}
if (flag == 0) {
cout << "No saddle points\n";
}
cout << "----------------------------\n";
return 0;
}

Answer 1

First, there is a logical error with your code. In the pos_max function, it will return the index of the element which is maximum in the column. There can be a case when there are multiple maximum with the same value in the column, however, it returns the one which is not the minimum in the row, hence your program won't be able to print that saddle point.
To solve this, you can either return an array of all indices which are maximum in a column and then check for each of those points if it's minimum in their respective column, but I think it's not a very elegant solution. In any case, you will again have to write the entire code for the other condition for saddle points, minimum in column and maximum in row.
Hence, I would suggest a change in strategy. You create 4 arrays, max_row, max_col, min_row, min_col, where each array stores the minimum / maximum in that row / column respectively. Then you can traverse the array and check if that point satisfies saddle point condition.
Here is the code:
#include <iostream>
#include <cstdlib>
using namespace std;
int a[10][10];
int max_row[10], max_col[10], min_row[10], min_col[10];
int x, y;
bool is_saddle(int i, int j) {
int x = a[i][j];
return (max_row[i] == x && min_col[j] == x) || (min_row[i] == x && max_col[j] == x);
}
int main() {
/* code to input x, y and the matrix
...
*/
/* code to visualize the matrix
...
*/
/* populating max and min arrays */
for (int i = 0; i <= x-1; ++i) {
max_row[i] = a[i][0], min_row[i] = a[i][0];
for (int j = 0; j <= y-1; ++j) {
max_row[i] = max(max_row[i], a[i][j]);
min_row[i] = min(min_row[i], a[i][j]);
}
}
for (int j = 0; j <= y-1; ++j) {
max_col[j] = a[0][j], min_col[j] = a[0][j];
for (int i = 0; i <= x-1; ++i) {
max_col[j] = max(max_col[j], a[i][j]);
min_col[j] = min(min_col[j], a[i][j]);
}
}
/* Check for saddle point */
for (int i = 0; i <= x-1; ++i) {
for (int j = 0; j <= y-1; ++j) {
if (is_saddle(i, j)) {
cout << "Saddle points are: ";
cout << "a[" << i + 1 << ", " << j + 1 << "] = " << a[i][j] << "\n";
flag = 1;
}
}
}
if (flag == 0) {
cout << "No saddle points\n";
}
cout << "----------------------------\n";
return 0;
}

Answer 2

#include <iostream>
using namespace std;
int getMaxInRow(int[][5], int, int, int);
int getMinInColumn(int[][5], int, int, int);
void getSaddlePointCordinates(int [][5],int ,int );
void getInputOf2dArray(int a[][5], int, int);
int main()
{
int a[5][5] ;
int rows, columns;
cin >> rows >> columns;
getInputOf2dArray(a, 5, 5);
getSaddlePointCordinates(a,rows,columns);
}
void getInputOf2dArray(int a[][5], int rows, int columns)
{
for (int i = 0; i < rows; i = i + 1)
{
for (int j = 0; j < columns; j = j + 1)
{
cin >> a[i][j];
}
}
}
void getSaddlePointCordinates(int a[][5],int rows,int columns)
{
int flag = 0;
for (int rowNo = 0; rowNo < 5; rowNo++)
{
for (int columnNo = 0; columnNo < 5; columnNo++)
{
if (getMaxInRow(a, rows, columns, rowNo) == getMinInColumn(a, rows, columns, columnNo))
{
flag = 1;
cout << rowNo << columnNo;
}
}
}
if (flag == 0)
cout << "no saddle point";
cout << "\n";
}
int getMaxInRow(int a[][5], int row, int column, int rowNo)
{
int max = a[rowNo][0];
for (int i = 1; i < column; i = i + 1)
{
if (a[rowNo][i] > max)
max = a[rowNo][i];
}
return max;
}
int getMinInColumn(int a[][5], int row, int column, int columnNo)
{
int min = a[0][columnNo];
for (int i = 1; i < row; i = i + 1)
{
if (a[i][columnNo] < min)
min = a[i][columnNo];
}
return min;
}

Answer 3

just take the reference arr(ref[size]) // memorization method to check the minimum and maximum value in it.
Here is the Code Implementation with time complexity O(n *n) & space complexity O(n):
#include <bits/stdc++.h>
using namespace std;
#define size 5
void util(int arr[size][size], int *count)
{
int ref[size]; // array to hold all the max values of row's.
for(int r = 0; r < size; r++)
{
int max_row_val = arr[r][0];
for(int c = 1; c < size; c++)
{
if(max_row_val < arr[r][c])
max_row_val = arr[r][c];
}
ref[r] = max_row_val;
}
for(int c = 0; c < size; c++)
{
int min_col_val = arr[0][c];
for(int r = 1; r < size; r++) // min_val of the column
{
if(min_col_val > arr[r][c])
min_col_val = arr[r][c];
}
for(int r = 0; r < size; r++) // now search if the min_val of col and the ref[r] is same and the position is same, if both matches then print.
{
if(min_col_val == ref[r] && min_col_val == arr[r][c])
{
*count += 1;
if((*count) == 1)
cout << "The cordinate's are: \n";
cout << "(" << r << "," << c << ")" << endl;
}
}
}
}
// Driver function
int main()
{
int arr[size][size];
for(int i = 0; i < size; i++)
{
for(int j = 0; j < size; j++)
cin >> arr[i][j];
}
int count = 0;
util(arr, &count);
if(!count)
cout << "No saddle points" << endl;
}
// Test case -> Saddle Point
/*
Input1:
1 2 3 4 5
6 7 8 9 10
1 2 3 4 5
6 7 8 9 10
0 2 3 4 5
Output1:
The cordinate's are:
(0,4)
(2,4)
(4,4)
Input2:
1 2 3 4 5
6 7 8 9 1
10 11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
Output2:
No saddle points
*/