Construct a square Matrix using digits of given number N based on given pattern
Last Updated :
23 Jul, 2025
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Given an integer N, The task is to construct a matrix mat[][] of size M x M ('M' is the number of digits in the given integer) such that each diagonal of the matrix contains the same digit, placed according to the position of the digits in the given integer and then again repeat the steps from the back.
Examples:
Input: N = 123
Output: {{1, 2, 3},
{2, 3, 2},
{3, 2, 1}}
Explanation: The desired matrix must be of size 3*3. The digits of N are 1, 2, and 3. Placing 1, 2 and 3 along the diagonals from the top left cell till the Nth diagonal, and 2, 1 just after the Nth diagonal till the bottom-most cell.Input: N = 3219
Output: {{3, 2, 1, 9}, {2, 1, 9, 1}, {1, 9, 1, 2}, {9, 1, 2, 3}}
Approach: The task can be solved by traversing the matrix in a diagonal fashion and assigning the cell values according to the corresponding digit in the given number.
- Extract and store the digits of the given integer in a vector say v.
- Again store the digits in reverse order for 2nd half diagonal of the matrix.
- Assign the digits in the desired order.
- Print the matrix.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct the matrix
void constructMatrix(int n)
{
// Vector to store the
// digits of the integer
vector<int> v;
// Extracting the digits
// from the integer
while (n > 0) {
v.push_back(n % 10);
n = n / 10;
}
// Reverse the vector
reverse(v.begin(), v.end());
// Size of the vector
int N = v.size();
// Loop to store the digits in
// reverse order in the same vector
for (int i = N - 2; i >= 0; i--) {
v.push_back(v[i]);
}
// Matrix to be constructed
int mat[N][N];
// Assign the digits and
// print the desired matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
mat[i][j] = v[i + j];
cout << mat[i][j] << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
int n = 3219;
// Passing n to constructMatrix function
constructMatrix(n);
return 0;
}
// Java program for the above approach
import java.util.ArrayList;
import java.util.Collections;
class GFG {
// Function to construct the matrix
public static void constructMatrix(int n)
{
// Vector to store the
// digits of the integer
ArrayList<Integer> v = new ArrayList<Integer>();
// Extracting the digits
// from the integer
while (n > 0) {
v.add(n % 10);
n = n / 10;
}
// Reverse the vector
Collections.reverse(v);
// Size of the vector
int N = v.size();
// Loop to store the digits in
// reverse order in the same vector
for (int i = N - 2; i >= 0; i--) {
v.add(v.get(i));
}
// Matrix to be constructed
int[][] mat = new int[N][N];
// Assign the digits and
// print the desired matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
mat[i][j] = v.get(i + j);
System.out.print(mat[i][j] + " ");
}
System.out.println("");
}
}
// Driver Code
public static void main(String args[]) {
int n = 3219;
// Passing n to constructMatrix function
constructMatrix(n);
}
}
// This code is contributed by gfgking.
# python program for the above approach
# Function to construct the matrix
def constructMatrix(n):
# Vector to store the
# digits of the integer
v = []
# Extracting the digits
# from the integer
while (n > 0):
v.append(n % 10)
n = n // 10
# Reverse the vector
v.reverse()
# Size of the vector
N = len(v)
# Loop to store the digits in
# reverse order in the same vector
for i in range(N-2, -1, -1):
v.append(v[i])
# Matrix to be constructed
mat = [[0 for _ in range(N)] for _ in range(N)]
# Assign the digits and
# print the desired matrix
for i in range(0, N):
for j in range(0, N):
mat[i][j] = v[i + j]
print(mat[i][j], end=" ")
print()
# Driver Code
if __name__ == "__main__":
n = 3219
# Passing n to constructMatrix function
constructMatrix(n)
# This code is contributed by rakeshsahni
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to construct the matrix
static void constructMatrix(int n)
{
// Vector to store the
// digits of the integer
List<int> v = new List<int>();
// Extracting the digits
// from the integer
while (n > 0) {
v.Add(n % 10);
n = n / 10;
}
// Reverse the vector
v.Reverse();
// Size of the vector
int N = v.Count;
// Loop to store the digits in
// reverse order in the same vector
for (int i = N - 2; i >= 0; i--) {
v.Add(v[i]);
}
// Matrix to be constructed
int[,] mat = new int[N, N];
// Assign the digits and
// print the desired matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
mat[i, j] = v[i + j];
Console.Write(mat[i, j] + " ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main()
{
int n = 3219;
// Passing n to constructMatrix function
constructMatrix(n);
}
}
// This code is contributed by sanjoy_62.
<script>
// JavaScript Program to implement
// the above approach
// Function to construct the matrix
function constructMatrix(n)
{
// Vector to store the
// digits of the integer
let v = [];
// Extracting the digits
// from the integer
while (n > 0) {
v.push(n % 10);
n = Math.floor(n / 10);
}
// Reverse the vector
v.reverse();
// Size of the vector
let N = v.length;
// Loop to store the digits in
// reverse order in the same vector
for (let i = N - 2; i >= 0; i--) {
v.push(v[i]);
}
// Matrix to be constructed
let mat = new Array(N);
for (let i = 0; i < mat.length; i++) {
mat[i] = new Array(N).fill(0);
}
// Assign the digits and
// print the desired matrix
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
mat[i][j] = v[i + j];
document.write(mat[i][j] + " ");
}
document.write("<br>")
}
}
// Driver Code
let n = 3219;
// Passing n to constructMatrix function
constructMatrix(n);
// This code is contributed by Potta Lokesh
</script>
Output:
3 2 1 9 2 1 9 1 1 9 1 2 9 1 2 3
Time Complexity: O(N2)
Auxiliary Space: O(1)
