Dadas N filas con M elementos cada una y un arreglo arr[] de L números, la tarea es imprimir el conteo de elementos de ese arreglo presente en cada fila de la array.
Ejemplos:
Input: {8 27 39 589 23
23 34 589 12 45
939 32 27 12 78
23 349 48 21 32},
arr[] = {589, 39, 27}
Output: 1st row - 3
2nd row - 1
3rd row - 1
4th row - 0
In 1st row, all three elements in array z[] are present
In 2nd row, only 589 in array z[] are present
In 3rd row, only 27 in array z[] are present
In 4th row, none of the elements are present.
Input: {1, 2, 3
4, 5, 6},
arr[] = {2, 3, 4}
Output: 1st row - 2
2nd row - 1
Un enfoque ingenuo es iterar para cada elemento en el arreglo arr[] y para la i -ésima fila hacer una búsqueda lineal para cada elemento en el arreglo arr[]. Cuente el número de elementos e imprima el resultado para cada fila.
Complejidad de tiempo: O(N*M*L)
Un enfoque eficiente es iterar para todos los elementos en la i -ésima fila de la array. Marque todos los elementos usando una tabla hash . Iterar en la array de números en la array Z, verificar si el número está presente en la tabla hash. Aumente el recuento de cada elemento presente. Una vez comprobados todos los elementos, imprima el recuento.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print the count of
// elements present in the NxM matrix
#include <bits/stdc++.h>
using namespace std;
// Function to print the count of
// elements present in the NxM matrix
void printCount(int a[][5], int n, int m, int z[], int l)
{
// iterate in the n rows
for (int i = 0; i < n; i++) {
// map to mark elements in N-th row
unordered_map<int, int> mp;
// mark all elements in the n-th row
for (int j = 0; j < m; j++)
mp[a[i][j]] = 1;
int count = 0;
// check for occurrence of all elements
for (int j = 0; j < l; j++) {
if (mp[z[j]])
count += 1;
}
// print the occurrence of all elements
cout << "row" << i + 1 << " = " << count << endl;
}
}
// Driver Code
int main()
{
// NxM matrix
int a[][5] = { { 8, 27, 39, 589, 23 },
{ 23, 34, 589, 12, 45 },
{ 939, 32, 27, 12, 78 },
{ 23, 349, 48, 21, 32 } };
// elements array
int arr[] = { 589, 39, 27 };
int n = sizeof(a) / sizeof(a[0]);
int m = 5;
int l = sizeof(arr) / sizeof(arr[0]);
printCount(a, n, m, arr, l);
return 0;
}
Java
// Java program to print the count of
// elements present in the NxM matrix
import java.util.*;
class GFG
{
// Function to print the count of
// elements present in the NxM matrix
static void printCount(int a[][], int n, int m,
int z[], int l)
{
// iterate in the n rows
for (int i = 0; i < n; i++)
{
// map to mark elements in N-th row
Map<Integer,Integer> mp = new HashMap<>();
// mark all elements in the n-th row
for (int j = 0; j < m; j++)
mp.put(a[i][j], 1);
int count = 0;
// check for occurrence of all elements
for (int j = 0; j < l; j++)
{
if (mp.containsKey(z[j]))
count += 1;
}
// print the occurrence of all elements
System.out.println("row" +(i + 1) + " = " + count);
}
}
// Driver Code
public static void main(String[] args)
{
// NxM matrix
int a[][] = { { 8, 27, 39, 589, 23 },
{ 23, 34, 589, 12, 45 },
{ 939, 32, 27, 12, 78 },
{ 23, 349, 48, 21, 32 } };
// elements array
int arr[] = { 589, 39, 27 };
int n = a.length;
int m = 5;
int l = arr.length;
printCount(a, n, m, arr, l);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to print the count of
# elements present in the NxM matrix
# Function to print the count of
# elements present in the NxM matrix
def printCount(a, n, m, z, l):
# iterate in the n rows
for i in range(n):
# map to mark elements in N-th row
mp = dict()
# mark all elements in the n-th row
for j in range(m):
mp[a[i][j]] = 1
count = 0
# check for occurrence of all elements
for j in range(l):
if z[j] in mp.keys():
count += 1
# print the occurrence of all elements
print("row", i + 1, " = ", count )
# Driver Code
# NxM matrix
a = [[ 8, 27, 39, 589, 23 ],
[ 23, 34, 589, 12, 45 ],
[ 939, 32, 27, 12, 78 ],
[ 23, 349, 48, 21, 32 ]]
# elements array
arr = [ 589, 39, 27 ]
n = len(a)
m = 5
l = len(arr)
printCount(a, n, m, arr, l)
# This code is contributed by mohit kumar 29
C#
// C# program to print the count of
// elements present in the NxM matrix
using System;
using System.Collections.Generic;
class GFG
{
// Function to print the count of
// elements present in the NxM matrix
static void printCount(int [,]a, int n, int m,
int []z, int l)
{
// iterate in the n rows
for (int i = 0; i < n; i++)
{
// map to mark elements in N-th row
Dictionary<int,int> mp = new Dictionary<int,int>();
// mark all elements in the n-th row
for (int j = 0; j < m; j++)
mp.Add(a[i,j], 1);
int count = 0;
// check for occurrence of all elements
for (int j = 0; j < l; j++)
{
if (mp.ContainsKey(z[j]))
count += 1;
}
// print the occurrence of all elements
Console.WriteLine("row" +(i + 1) + " = " + count);
}
}
// Driver Code
public static void Main(String[] args)
{
// NxM matrix
int [,]a = { { 8, 27, 39, 589, 23 },
{ 23, 34, 589, 12, 45 },
{ 939, 32, 27, 12, 78 },
{ 23, 349, 48, 21, 32 } };
// elements array
int []arr = { 589, 39, 27 };
int n = a.GetLength(0);
int m = 5;
int l = arr.Length;
printCount(a, n, m, arr, l);
}
}
/* This code is contributed by PrinciRaj1992 */
Javascript
<script>
// JavaScript program to print the count of
// elements present in the NxM matrix
// Function to print the count of
// elements present in the NxM matrix
function printCount(a,n,m,z,l)
{
// iterate in the n rows
for (let i = 0; i < n; i++)
{
// map to mark elements in N-th row
let mp = new Map();
// mark all elements in the n-th row
for (let j = 0; j < m; j++)
mp.set(a[i][j], 1);
let count = 0;
// check for occurrence of all elements
for (let j = 0; j < l; j++)
{
if (mp.has(z[j]))
count += 1;
}
// print the occurrence of all elements
document.write("row" +(i + 1) +
" = " + count+"<br>");
}
}
// Driver Code
// NxM matrix
let a = [[ 8, 27, 39, 589, 23 ],
[ 23, 34, 589, 12, 45 ],
[ 939, 32, 27, 12, 78 ],
[ 23, 349, 48, 21, 32 ]];
// elements array
let arr=[ 589, 39, 27];
let n = a.length;
let m = 5;
let l = arr.length;
printCount(a, n, m, arr, l);
// This code is contributed by patel2127
</script>
Producción:
row1 = 3 row2 = 1 row3 = 1 row4 = 0
Complejidad de tiempo: O(N*M)