Dados dos arreglos A[] y B[] y un entero K , la tarea es maximizar el conteo de enteros del arreglo B[] que pueden igualarse con el arreglo A[] sumando o restando cualquier entero en el rango [0 , K] .
Ejemplos:
Entrada: K=5, A[] = [100, 65, 35, 85, 55], B[] = [30, 60, 75, 95]
Salida: 3
Explicación:
30 + 5, 60 + 5, 95 + 5 da los valores que son iguales con pocos elementos de la array A[].Entrada: K = 5, A[] = [10, 20, 30, 40, 50], B[] = [1, 20, 3]
Salida: 1
Explicación:
Solo se puede igualar el 2º valor, ya que su valor [20] se puede cambiar a [20] sumando o restando 0.
Enfoque: La idea es verificar si la diferencia absoluta entre los elementos del arreglo B[] con cualquier elemento del arreglo A[] es menor o igual a K . En caso afirmativo, inclúyalo en el recuento. Imprima el recuento de todos esos elementos después de verificar la condición anterior para todos los elementos en la array B[] .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
void countElement(int A[], int N,
int B[], int M, int K)
{
// To store the count of element
int cnt = 0;
// Traverse the array B[]
for (int i = 0; i < M; i++) {
int currentElement = B[i];
// Traverse the array A[]
for (int j = 0; j < N; j++) {
// Find the difference
int diff
= abs(currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K) {
cnt++;
break;
}
}
}
// Print the count
cout << cnt;
}
// Driver Code
int main()
{
// Given array A[] and B[]
int A[] = { 100, 65, 35, 85, 55 };
int B[] = { 30, 60, 75, 95 };
// Given K
int K = 5;
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
// Function Call
countElement(A, N, B, M, K);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
static void countElement(int A[], int N,
int B[], int M, int K)
{
// To store the count of element
int cnt = 0;
// Traverse the array B[]
for(int i = 0; i < M; i++)
{
int currentElement = B[i];
// Traverse the array A[]
for(int j = 0; j < N; j++)
{
// Find the difference
int diff = Math.abs(
currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K)
{
cnt++;
break;
}
}
}
// Print the count
System.out.print(cnt);
}
// Driver Code
public static void main(String[] args)
{
// Given array A[] and B[]
int A[] = { 100, 65, 35, 85, 55 };
int B[] = { 30, 60, 75, 95 };
// Given K
int K = 5;
int N = A.length;
int M = B.length;
// Function call
countElement(A, N, B, M, K);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program to implement # the above approach # Function that count the number of # integers from array B such that # subtracting element in the range # [0, K] given any element in A def countElement(A, N, B, M, K): # To store the count of element cnt = 0 # Traverse the array B for i in range(M): currentElement = B[i] # Traverse the array A for j in range(N): # Find the difference diff = abs(currentElement - A[j]) # If difference is atmost # K then increment the cnt if(diff <= K): cnt += 1 break # Print the count print(cnt) # Driver Code if __name__ == '__main__': # Given array A and B A = [ 100, 65, 35, 85, 55 ] B = [ 30, 60, 75, 95 ] N = len(A) M = len(B) # Given K K = 5 # Function call countElement(A, N, B, M, K) # This code is contributed by Shivam Singh
C#
// C# program for the above approach
using System;
class GFG{
// Function that count the number of
// integers from array []B such that
// subtracting element in the range
// [0, K] given any element in []A
static void countElement(int []A, int N,
int []B, int M, int K)
{
// To store the count of element
int cnt = 0;
// Traverse the array []B
for(int i = 0; i < M; i++)
{
int currentElement = B[i];
// Traverse the array []A
for(int j = 0; j < N; j++)
{
// Find the difference
int diff = Math.Abs(
currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K)
{
cnt++;
break;
}
}
}
// Print the count
Console.Write(cnt);
}
// Driver Code
public static void Main(String[] args)
{
// Given array []A and []B
int []A = { 100, 65, 35, 85, 55 };
int []B = { 30, 60, 75, 95 };
// Given K
int K = 5;
int N = A.Length;
int M = B.Length;
// Function call
countElement(A, N, B, M, K);
}
}
// This code is contributed by Rohit_ranjan
Javascript
<script>
// Javascript Script program to implement
// the above approach
// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
function countElement(A, N, B, M, K)
{
// To store the count of element
let cnt = 0;
// Traverse the array B[]
for(let i = 0; i < M; i++)
{
let currentElement = B[i];
// Traverse the array A[]
for(let j = 0; j < N; j++)
{
// Find the difference
let diff = Math.abs(
currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K)
{
cnt++;
break;
}
}
}
// Print the count
document.write(cnt);
}
// Driver Code
// Given array A[] and B[]
let A = [ 100, 65, 35, 85, 55 ];
let B = [ 30, 60, 75, 95 ];
// Given K
let K = 5;
let N = A.length;
let M = B.length;
// Function call
countElement(A, N, B, M, K);
</script>
3
Complejidad temporal: O(N*M), donde N y M son las longitudes de las arrays A[] y B[].
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por jatindscl015 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA