Dado un entero N , la tarea es construir una array mat[][] de tamaño M x M (‘M’ es el número de dígitos en el entero dado) tal que cada diagonal de la array contenga el mismo dígito, colocado de acuerdo con la posición de los dígitos en el número entero dado y luego repita los pasos desde atrás.
Ejemplos:
Entrada: N = 123
Salida: {{1, 2, 3},
{2, 3, 2},
{3, 2, 1}}
Explicación: La array deseada debe tener un tamaño de 3*3. Los dígitos de N son 1, 2 y 3. Colocando 1, 2 y 3 a lo largo de las diagonales desde la celda superior izquierda hasta la diagonal N, y 2, 1 justo después de la diagonal N hasta la celda más inferior.Entrada: N = 3219
Salida: {{3, 2, 1, 9}, {2, 1, 9, 1}, {1, 9, 1, 2}, {9, 1, 2, 3}}
Enfoque: La tarea se puede resolver recorriendo la array en forma diagonal y asignando los valores de celda de acuerdo con el dígito correspondiente en el número dado.
- Extraiga y almacene los dígitos del entero dado en un vector, digamos v .
- Nuevamente almacene los dígitos en orden inverso para la segunda mitad de la diagonal de la array.
- Asigne los dígitos en el orden deseado.
- Imprime la array.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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.
Python3
# 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#
// 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.
Javascript
<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>
Producción:
3 2 1 9 2 1 9 1 1 9 1 2 9 1 2 3
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)