Dada una array 2D de orden M * N, la tarea es encontrar la suma de los elementos de la array.
Ejemplos:
Entrada: array[2][2] = {{1, 2}, {3, 4}};
Salida: 10Entrada: array[4][4] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16} };
Salida: 136
Acercarse:
La suma de cada elemento de la array 2D se puede calcular recorriendo la array y sumando los elementos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find sum of
// elements in 2D array
#include <iostream>
using namespace std;
// Get the size m and n
#define M 4
#define N 4
// Function to calculate sum
// of elements in 2d array
int sum(int arr[M][N])
{
int i, j;
int sum = 0;
// Finding the sum
for (i = 0; i < M; ++i) {
for (j = 0; j < N; ++j) {
// Add the element
sum = sum + arr[i][j];
}
}
return sum;
}
// Driver code
int main()
{
int i, j;
int arr[M][N];
// Get the matrix elements
int x = 1;
for (i = 0; i < M; i++)
for (j = 0; j < N; j++)
arr[i][j] = x++;
// Get sum
cout << sum(arr);
return 0;
}
Java
// Java code for the above approach
import java.io.*;
class GFG {
// Get the size m and n
static int M = 4;
static int N = 4;
// Function to calculate sum
// of elements in 2d array
static int sum(int arr[][])
{
int i, j;
int sum = 0;
// Finding the sum
for (i = 0; i < M; ++i) {
for (j = 0; j < N; ++j) {
// Add the element
sum = sum + arr[i][j];
}
}
return sum;
}
public static void main (String[] args)
{
int i, j;
int arr[][]= new int[M][N];
// Get the matrix elements
int x = 1;
for (i = 0; i < M; i++)
for (j = 0; j < N; j++)
arr[i][j] = x++;
// Get sum
System.out.println(sum(arr));
}
}
// This code is contributed by Potta Lokesh
Python3
# phython program to find sum of # elements in 2D array # Get the size m and n M = 4 N = 4 # Function to calculate sum # of elements in 2d array def sum(arr): sum = 0 # Finding the sum for i in range(M): for j in range(N): # Add the element sum = sum + arr[i][j] return sum # Driver code arr = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] # Get the matrix elements x = 1 for i in range(M): for j in range(N): arr[i][j] = x x += 1 # Get sum print(sum(arr)) # This code is contributed by ninja_hattori
C#
using System;
class GFG
{
// Get the size m and n
static int M = 4;
static int N = 4;
// Function to calculate sum
// of elements in 2d array
static int sum(int [,]arr)
{
int i, j;
int sum = 0;
// Finding the sum
for (i = 0; i < M; ++i)
{
for (j = 0; j < N; ++j)
{
// Add the element
sum = sum + arr[i,j];
}
}
return sum;
}
static public void Main (String[] args){
int i, j;
int [,]arr= new int[M,N];
// Get the matrix elements
int x = 1;
for (i = 0; i < M; i++)
for (j = 0; j < N; j++)
arr[i,j] = x++;
// Get sum
Console.WriteLine(sum(arr));
// Code
}
}
// This code is contributed by Kritima Gupta
Javascript
<script>
// JavaScript program to find sum of
// elements in 2D array
// Get the size m and n
const M = 4;
const N = 4;
// Function to calculate sum
// of elements in 2d array
const sum = (arr) => {
let i, j;
let sum = 0;
// Finding the sum
for (i = 0; i < M; ++i) {
for (j = 0; j < N; ++j) {
// Add the element
sum = sum + arr[i][j];
}
}
return sum;
}
// Driver code
let i, j;
let arr = new Array(M).fill(0).map(() => new Array(N).fill(0));
// Get the matrix elements
let x = 1;
for (i = 0; i < M; i++)
for (j = 0; j < N; j++)
arr[i][j] = x++;
// Get sum
document.write(sum(arr));
// This code is contributed by rakeshsahni.
</script>
Producción
136
Complejidad temporal: O(M*N)
Espacio auxiliar: O(1)
Otro método: usando STL. Llamar a la función incorporada para la suma de elementos de una array en STL. Usamos la función de acumulación (primero, último, suma) para devolver la suma de la array 1D.
A continuación se muestra la implementación de la idea.
C++
// C++ program to find sum of
// elements in 2D array
#include <iostream>
#include <numeric>
using namespace std;
// Get the size m and n
#define M 4
#define N 4
// Function to calculate sum
// of elements in 2d array
int sum(int arr[M][N])
{
int i, j;
int sum = 0;
// Finding the sum
for (i = 0; i < M; ++i) {
int ans = 0;
sum += accumulate(arr[i], arr[i] + N, ans);
}
return sum;
}
// Driver code
int main()
{
int i, j;
int arr[M][N];
// Get the matrix elements
int x = 1;
for (i = 0; i < M; i++)
for (j = 0; j < N; j++)
arr[i][j] = x++;
// Get sum
cout << sum(arr);
return 0;
}
Python
# phython program to find sum of # elements in 2D array # Get the size m and n M = 4 N = 4 # Function to calculate sum # of elements in 2d array def Findsum(arr): ans = 0 # Finding the sum for i in range(M): ans += int(sum(arr[i])) return ans # Driver code arr = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] # Get the matrix elements x = 1 for i in range(M): for j in range(N): arr[i][j] = x x += 1 # Get sum print(Findsum(arr)) # This code is contributed by Sam_snehil
Javascript
// JavaScript program to find sum of
// elements in 2D array
// Get the size m and n
const M = 4;
const N = 4;
// Function to calculate sum
// of elements in 2d array
const sum = (arr) => {
let i, j;
let sum = 0;
// Finding the sum
for (i = 0; i < M; ++i) {
sum = sum + arr[i].reduce(function(accumulator, currentValue){ return accumulator + currentValue;}, 0);
}
return sum;
}
// Driver code
let i, j;
let arr = new Array(M).fill(0).map(() => new Array(N).fill(0));
// Get the matrix elements
let x = 1;
for (i = 0; i < M; i++)
for (j = 0; j < N; j++)
arr[i][j] = x++;
// Get sum
document.write(sum(arr));
// This code is contributed by rakeshsahni.
Producción
136
Complejidad temporal: O(n*m) (donde n = nº de filas y m = nº de columna)
Espacio Auxiliar: O(1)