# Find smallest and largest element from square matrix diagonals in C++

C++Server Side ProgrammingProgramming

In this problem, we are given a square matrix of size nXn. Our task is to Find smallest and largest element from square matrix diagonals. We need to find the smallest and largest elements of the primary and secondary diagonals of the matrox.

Let’s take an example to understand the problem,

## Input

mat[][] = {
{3, 4, 7},
{5, 2, 1},
{1, 8, 6}
}

## Output

Smallest element in Primary Diagonal = 2
Largest element in Primary Diagonal = 6
Smallest element in Secondary Diagonal = 1
Largest element in Secondary Diagonal = 7

## Solution Approach

A simple solution to solve the problem is using nested loops. For checking elements at primary diagonal we will consider i = j. And for the secondary diagonal, we will consider i + j = n -1. We will find the maximum and minimum elements of the matrix for both primary and secondary diagonal.

Program to illustrate the working of our solution,

## Example

Live Demo

#include<iostream>
using namespace std;
void findMaxAndMinOfDiagonals(int mat[3][3], int n){
if (n == 0)
return;
int pDiagMin = mat[0][0],
pDiagMax = mat[0][0];
int sDiagMin = mat[0][n - 1 ],
sDiagMax = mat[0][n - 1];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
if (i == j){
if (mat[i][j] < pDiagMin)
pDiagMin = mat[i][j];
if (mat[i][j] > pDiagMax)
pDiagMax = mat[i][j];
}
if ((i + j) == (n - 1)) {
if (mat[i][j] < sDiagMin){
sDiagMin = mat[i][j];
}
if (mat[i][j] > sDiagMax)
sDiagMax = mat[i][j];
}
}
}
cout<<("\nSmallest Element of Principal Diagonal : ")<<pDiagMin;
cout<<("\nGreatest Element of Principal Diagonal : ")<<pDiagMax;
cout<<("\nSmallest Element of Secondary Diagonal : ")<<sDiagMin;
cout<<("\nGreatest Element of Secondary Diagonal : ")<<sDiagMax;
}
int main(){
int mat[3][3] = {
{ 3, 4, 7 },
{ 0, 2, 1 },
{ 1, 7, 8 }
};
int n = sizeof(mat) / sizeof(mat[0]);
findMaxAndMinOfDiagonals(mat, n);
}

## Output

Smallest Element of Principal Diagonal : 2
Greatest Element of Principal Diagonal : 8
Smallest Element of Secondary Diagonal : 2
Greatest Element of Secondary Diagonal : 7

Another more effective solution is reducing the nested loop to single loop using the fact that for the primary diagonal both the indices of the mat are the same i.e.

primary diagonal elements = mat[i][j].
Similarly, for secondary diagonal elements = mat[i][n - i - 1]

Program to illustrate the working of our solution,

## Example

Live Demo

#include<iostream>
using namespace std;
void findMaxAndMinOfDiagonals(int mat[3][3], int n){
if (n == 0)
return;
int pDiagMin = mat[0][0],
pDiagMax = mat[0][0];
int sDiagMin = mat[0][n - 1 ],
sDiagMax = mat[0][n - 1];
for (int i = 1; i < n; i++) {
if (mat[i][i] < pDiagMin)
pDiagMin = mat[i][i];
if (mat[i][i] > pDiagMax)
pDiagMax = mat[i][i];
if (mat[i][n - 1 - i] < sDiagMin)
sDiagMin = mat[i][n - 1 - i];
if (mat[i][n - 1 - i] > sDiagMax)
sDiagMax = mat[i][n - 1 - i];
}
cout<<("\nSmallest Element of Principal Diagonal : ")<<pDiagMin;
cout<<("\nGreatest Element of Principal Diagonal : ")<<pDiagMax;
cout<<("\nSmallest Element of Secondary Diagonal : ")<<sDiagMin;
cout<<("\nGreatest Element of Secondary Diagonal : ")<<sDiagMax;
}
int main(){
int mat[3][3] = {
{ 3, 4, 7 },
{ 0, 2, 1 },
{ 1, 7, 8 }
};
int n = sizeof(mat) / sizeof(mat[0]);
findMaxAndMinOfDiagonals(mat, n);
}

## Output

Smallest Element of Principal Diagonal : 2
Greatest Element of Principal Diagonal : 8
Smallest Element of Secondary Diagonal : 1
Greatest Element of Secondary Diagonal : 7
Updated on 16-Mar-2021 06:33:02