Dadas dos arrays A[] y B[] que constan de n y m elementos respectivamente. Encuentre la cantidad mínima de elementos para eliminar de cada array de modo que no exista ningún elemento común en ambos.
Ejemplos:
Input : A[] = { 1, 2, 3, 4}
B[] = { 2, 3, 4, 5, 8 }
Output : 3
We need to remove 2, 3 and 4 from any array.
Input : A[] = { 4, 2, 4, 4}
B[] = { 4, 3 }
Output : 1
We need to remove 4 from B[]
Input : A[] = { 1, 2, 3, 4 }
B[] = { 5, 6, 7 }
Output : 0
There is no common element in both.
Cuente la ocurrencia de cada número en ambas arrays. Si hay un número en ambas arrays, elimine el número de la array en la que aparece menos veces y agréguelo al resultado.
Implementación:
C++
// CPP program to find minimum element
// to remove so no common element
// exist in both array
#include <bits/stdc++.h>
using namespace std;
// To find no elements to remove
// so no common element exist
int minRemove(int a[], int b[], int n, int m)
{
// To store count of array element
unordered_map<int, int> countA, countB;
// Count elements of a
for (int i = 0; i < n; i++)
countA[a[i]]++;
// Count elements of b
for (int i = 0; i < m; i++)
countB[b[i]]++;
// Traverse through all common element, and
// pick minimum occurrence from two arrays
int res = 0;
for (auto x : countA)
if (countB.find(x.first) != countB.end())
res += min(x.second, countB[x.first]);
// To return count of minimum elements
return res;
}
// Driver program to test minRemove()
int main()
{
int a[] = { 1, 2, 3, 4 };
int b[] = { 2, 3, 4, 5, 8 };
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
cout << minRemove(a, b, n, m);
return 0;
}
Java
// JAVA Code to Remove minimum number of elements
// such that no common element exist in both array
import java.util.*;
class GFG {
// To find no elements to remove
// so no common element exist
public static int minRemove(int a[], int b[], int n,
int m)
{
// To store count of array element
HashMap<Integer, Integer> countA = new HashMap<
Integer, Integer>();
HashMap<Integer, Integer> countB = new HashMap<
Integer, Integer>();
// Count elements of a
for (int i = 0; i < n; i++){
if (countA.containsKey(a[i]))
countA.put(a[i], countA.get(a[i]) + 1);
else countA.put(a[i], 1);
}
// Count elements of b
for (int i = 0; i < m; i++){
if (countB.containsKey(b[i]))
countB.put(b[i], countB.get(b[i]) + 1);
else countB.put(b[i], 1);
}
// Traverse through all common element, and
// pick minimum occurrence from two arrays
int res = 0;
Set<Integer> s = countA.keySet();
for (int x : s)
if(countB.containsKey(x))
res += Math.min(countB.get(x),
countA.get(x));
// To return count of minimum elements
return res;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int a[] = { 1, 2, 3, 4 };
int b[] = { 2, 3, 4, 5, 8 };
int n = a.length;
int m = b.length;
System.out.println(minRemove(a, b, n, m));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find minimum # element to remove so no common # element exist in both array # To find no elements to remove # so no common element exist def minRemove(a, b, n, m): # To store count of array element countA = dict() countB = dict() # Count elements of a for i in range(n): countA[a[i]] = countA.get(a[i], 0) + 1 # Count elements of b for i in range(n): countB[b[i]] = countB.get(b[i], 0) + 1 # Traverse through all common # element, and pick minimum # occurrence from two arrays res = 0 for x in countA: if x in countB.keys(): res += min(countA[x],countB[x]) # To return count of # minimum elements return res # Driver Code a = [ 1, 2, 3, 4 ] b = [2, 3, 4, 5, 8 ] n = len(a) m = len(b) print(minRemove(a, b, n, m)) # This code is contributed # by mohit kumar
C#
// C# Code to Remove minimum number of elements
// such that no common element exist in both array
using System;
using System.Collections.Generic;
class GFG
{
// To find no elements to remove
// so no common element exist
public static int minRemove(int []a, int []b, int n,
int m)
{
// To store count of array element
Dictionary<int,int> countA = new Dictionary<int,int>();
Dictionary<int,int>countB = new Dictionary<int,int>();
// Count elements of a
for (int i = 0; i < n; i++)
{
if (countA.ContainsKey(a[i]))
{
var v = countA[a[i]];
countA.Remove(countA[a[i]]);
countA.Add(a[i], v + 1);
}
else countA.Add(a[i], 1);
}
// Count elements of b
for (int i = 0; i < m; i++)
{
if (countB.ContainsKey(b[i]))
{
var v = countB[b[i]];
countB.Remove(countB[b[i]]);
countB.Add(b[i], v + 1);
}
else countB.Add(b[i], 1);
}
// Traverse through all common element, and
// pick minimum occurrence from two arrays
int res = 0;
foreach (int x in countA.Keys)
if(countB.ContainsKey(x))
res += Math.Min(countB[x],
countA[x]);
// To return count of minimum elements
return res;
}
/* Driver code */
public static void Main(String[] args)
{
int []a = { 1, 2, 3, 4 };
int []b = { 2, 3, 4, 5, 8 };
int n = a.Length;
int m = b.Length;
Console.WriteLine(minRemove(a, b, n, m));
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// JavaScript Code to Remove
// minimum number of elements
// such that no common element
// exist in both array
// To find no elements to remove
// so no common element exist
function minRemove(a,b,n,m)
{
// To store count of array element
let countA = new Map();
let countB = new Map();
// Count elements of a
for (let i = 0; i < n; i++){
if (countA.has(a[i]))
countA.set(a[i],
countA.get(a[i]) + 1);
else
countA.set(a[i], 1);
}
// Count elements of b
for (let i = 0; i < m; i++){
if (countB.has(b[i]))
countB.set(b[i],
countB.get(b[i]) + 1);
else
countB.set(b[i], 1);
}
// Traverse through all
// common element, and
// pick minimum occurrence
// from two arrays
let res = 0;
for (let x of countA.keys())
if(countB.has(x))
res += Math.min(countB.get(x),
countA.get(x));
// To return count of minimum elements
return res;
}
/* Driver program to test above function */
let a=[1, 2, 3, 4 ];
let b=[2, 3, 4, 5, 8];
let n = a.length;
let m = b.length;
document.write(minRemove(a, b, n, m));
// This code is contributed by unknown2108
</script>
Producción
3
Complejidad temporal: O(n+m)
Espacio auxiliar: O(n+m)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA