Dadas dos arrays (que pueden o no estar ordenadas). Estas arrays son tales que pueden tener algunos elementos comunes en ellas. Necesitamos encontrar elementos cuyos recuentos de ocurrencias estén más en la primera array que en la segunda.
Ejemplos:
Input : ar1[] = {1, 2, 2, 2, 3, 3, 4, 5}
ar2[] = {2, 2, 3, 3, 3, 4}
Output : 1 2 5
1 occurs one times in first and zero times in second
2 occurs three times in first and two times in second
............................
Input : ar1[] = {1, 3, 4, 2, 3}
ar2[] = {3, 4, 5}
Output : 3
La idea es usar hashing. Atravesamos el primer arreglo e insertamos todos los elementos y sus frecuencias en una tabla hash. Ahora recorremos la segunda array y reducimos las frecuencias en la tabla hash para los elementos comunes. Ahora recorremos la primera array nuevamente e imprimimos aquellos elementos cuyas frecuencias aún son mayores que 0. Para evitar la impresión repetida de los mismos elementos, establecemos la frecuencia en 0.
C++
// C++ program to print all those elements of
// first array that have more frequencies than
// second array.
#include <bits/stdc++.h>
using namespace std;
// Compares two intervals according to starting times.
void moreFreq(int ar1[], int ar2[], int m, int n)
{
// Traverse first array and store frequencies
// of all elements
unordered_map<int, int> mp;
for (int i = 0; i < m; i++)
mp[ar1[i]]++;
// Traverse second array and reduce frequencies
// of common elements.
for (int i = 0; i < n; i++)
if (mp.find(ar2[i]) != mp.end())
mp[ar2[i]]--;
// Now traverse first array again and print
// all those elements whose frequencies are
// more than 0. To avoid repeated printing,
// we set frequency as 0 after printing.
for (int i = 0; i < m; i++) {
if (mp[ar1[i]] > 0) {
cout << ar1[i] << " ";
mp[ar1[i]] = 0;
}
}
}
// Driver code
int main()
{
int ar1[] = { 1, 2, 2, 2, 3, 3, 4, 5 };
int ar2[] = { 2, 2, 3, 3, 3, 4 };
int m = sizeof(ar1) / sizeof(ar1[0]);
int n = sizeof(ar2) / sizeof(ar2[0]);
moreFreq(ar1, ar2, m, n);
return 0;
}
Java
// Java program to print all those elements of
// first array that have more frequencies than
// second array.
import java.util.*;
class GFG
{
// Compares two intervals according to starting times.
static void moreFreq(int ar1[], int ar2[], int m, int n)
{
// Traverse first array and store frequencies
// of all elements
Map<Integer,Integer> mp = new HashMap<>();
for (int i = 0 ; i < m; i++)
{
if(mp.containsKey(ar1[i]))
{
mp.put(ar1[i], mp.get(ar1[i])+1);
}
else
{
mp.put(ar1[i], 1);
}
}
// Traverse second array and reduce frequencies
// of common elements.
for (int i = 0; i < n; i++)
if (mp.containsKey(ar2[i]))
mp.put(ar2[i], mp.get(ar2[i])-1);
// Now traverse first array again and print
// all those elements whose frequencies are
// more than 0. To avoid repeated printing,
// we set frequency as 0 after printing.
for (int i = 0; i < m; i++)
{
if (mp.get(ar1[i]) > 0)
{
System.out.print(ar1[i] + " ");
mp.put(ar1[i], 0);
}
}
}
// Driver code
public static void main(String[] args)
{
int ar1[] = { 1, 2, 2, 2, 3, 3, 4, 5 };
int ar2[] = { 2, 2, 3, 3, 3, 4 };
int m = ar1.length;
int n = ar2.length;
moreFreq(ar1, ar2, m, n);
}
}
// This code has been contributed by 29AjayKumar
Python3
# Python3 program to print all those elements of # first array that have more frequencies than # second array. import math as mt # Compares two intervals according to # starting times. def moreFreq(ar1, ar2, m, n): # Traverse first array and store # frequencies of all elements mp = dict() for i in range(m): if ar1[i] in mp.keys(): mp[ar1[i]] += 1 else: mp[ar1[i]] = 1 # Traverse second array and reduce # frequencies of common elements. for i in range(n): if ar2[i] in mp.keys(): mp[ar2[i]] -= 1 # Now traverse first array again and print # all those elements whose frequencies are # more than 0. To avoid repeated printing, # we set frequency as 0 after printing. for i in range(m): if (mp[ar1[i]] > 0): print(ar1[i], end = " ") mp[ar1[i]] = 0 # Driver code ar1 = [ 1, 2, 2, 2, 3, 3, 4, 5 ] ar2 = [ 2, 2, 3, 3, 3, 4 ] m = len(ar1) n = len(ar2) moreFreq(ar1, ar2, m, n) # This code is contributed # by mohit kumar 29
C#
// C# program to print all those elements of
// first array that have more frequencies than
// second array.
using System;
using System.Collections.Generic;
class GFG
{
// Compares two intervals according to
// starting times.
static void moreFreq(int []ar1, int []ar2,
int m, int n)
{
// Traverse first array and store frequencies
// of all elements
Dictionary<int,
int> mp = new Dictionary<int,
int>();
for (int i = 0 ; i < m; i++)
{
if(mp.ContainsKey(ar1[i]))
{
mp[ar1[i]] = mp[ar1[i]] + 1;
}
else
{
mp.Add(ar1[i], 1);
}
}
// Traverse second array and reduce frequencies
// of common elements.
for (int i = 0; i < n; i++)
if (mp.ContainsKey(ar2[i]))
mp[ar2[i]] = mp[ar2[i]] - 1;
// Now traverse first array again and print
// all those elements whose frequencies are
// more than 0. To avoid repeated printing,
// we set frequency as 0 after printing.
for (int i = 0; i < m; i++)
{
if (mp[ar1[i]] > 0)
{
Console.Write(ar1[i] + " ");
mp[ar1[i]] = 0;
}
}
}
// Driver code
public static void Main(String[] args)
{
int []ar1 = { 1, 2, 2, 2, 3, 3, 4, 5 };
int []ar2 = { 2, 2, 3, 3, 3, 4 };
int m = ar1.Length;
int n = ar2.Length;
moreFreq(ar1, ar2, m, n);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript program to print all those elements of
// first array that have more frequencies than
// second array.
// Compares two intervals according to starting times.
function moreFreq(ar1, ar2, m, n)
{
// Traverse first array and store frequencies
// of all elements
var mp = new Map();
for (var i = 0; i < m; i++)
{
if(mp.has(ar1[i]))
mp.set(ar1[i], mp.get(ar1[i])+1)
else
mp.set(ar1[i], 1)
}
// Traverse second array and reduce frequencies
// of common elements.
for (var i = 0; i < n; i++)
if (mp.has(ar2[i]))
{
mp.set(ar2[i], mp.get(ar2[i])-1)
}
// Now traverse first array again and print
// all those elements whose frequencies are
// more than 0. To avoid repeated printing,
// we set frequency as 0 after printing.
for (var i = 0; i < m; i++) {
if (mp.get(ar1[i]) > 0) {
document.write( ar1[i] + " ");
mp.set(ar1[i], 0);
}
}
}
// Driver code
var ar1 = [1, 2, 2, 2, 3, 3, 4, 5];
var ar2 = [2, 2, 3, 3, 3, 4];
var m = ar1.length;
var n = ar2.length;
moreFreq(ar1, ar2, m, n);
</script>
Producción:
1 2 5
Complejidad de tiempo: O(m + n) bajo el supuesto de que unordered_map find() e insert() funcionan en tiempo O(1).