Dada una array ordenada de tamaño N , la tarea es reducir la array de modo que cada elemento pueda aparecer como máximo K veces.
Ejemplos:
Entrada: arr[] = {1, 2, 2, 2, 3}, K = 2
Salida: {1, 2, 2, 3}
Explicación:
elimine 2 una vez, ya que aparece más de 2 veces.
Entrada: arr[] = {3, 3, 3}, K = 1
Salida: {3}
Explicación:
elimine 3 dos veces, ya que aparece más de 1 vez.
Acercarse:
- Atraviesa la array dada arr
- Mantenga el conteo de cada elemento único en la array mientras atraviesa, usando un puntero i
- Si la frecuencia actual de arr[i] hasta el índice i es menor o igual a K, agregue el elemento arr[i] a la nueva array reducida e incremente la frecuencia en 1.
- Si la frecuencia actual de arr[i] hasta el índice i es mayor que K, omita hasta que encuentre el siguiente elemento único.
- Después de que termine el recorrido, imprima la array reducida.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to reduce the array
// such that each element appears
// at most K times
#include <bits/stdc++.h>
using namespace std;
// Function to reduce the array
void reduceArray(int arr[], int n, int K)
{
// Vector to store the reduced array
vector<int> vec;
int size = 0;
int curr_ele = arr[0], curr_freq = 1;
// Iterate over the array
for (int i = 0; i < n; i++) {
if (curr_ele == arr[i]
&& curr_freq <= K) {
vec.push_back(arr[i]);
size++;
}
else if (curr_ele != arr[i]) {
curr_ele = arr[i];
vec.push_back(arr[i]);
size++;
curr_freq = 1;
}
curr_freq++;
}
// Print the reduced array
cout << "{";
for (int i = 0; i < size; i++) {
cout << vec[i] << ", ";
}
cout << "}";
}
// Driver code
int main()
{
int arr[]
= { 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 };
int n = sizeof(arr)
/ sizeof(arr[0]);
int K = 2;
// Function call
reduceArray(arr, n, K);
return 0;
}
Java
// Java program to reduce the array
// such that each element appears
// at most K times
import java.util.*;
class GFG{
// Function to reduce the array
static void reduceArray(int arr[], int n, int K)
{
// Vector to store the reduced array
Vector<Integer> vec = new Vector<Integer>();
int size = 0;
int curr_ele = arr[0], curr_freq = 1;
// Iterate over the array
for (int i = 0; i < n; i++) {
if (curr_ele == arr[i]
&& curr_freq <= K) {
vec.add(arr[i]);
size++;
}
else if (curr_ele != arr[i]) {
curr_ele = arr[i];
vec.add(arr[i]);
size++;
curr_freq = 1;
}
curr_freq++;
}
// Print the reduced array
System.out.print("{");
for (int i = 0; i < size; i++) {
System.out.print(vec.get(i)+ ", ");
}
System.out.print("}");
}
// Driver code
public static void main(String[] args)
{
int arr[]
= { 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 };
int n = arr.length;
int K = 2;
// Function call
reduceArray(arr, n, K);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program to reduce the array
# such that each element appears
# at most K times
# Function to reduce the array
def reduceArray(arr, n, K) :
# List to store the reduced array
vec = [];
size = 0;
curr_ele = arr[0]; curr_freq = 1;
# Iterate over the array
for i in range(n) :
if (curr_ele == arr[i]
and curr_freq <= K) :
vec.append(arr[i]);
size += 1;
elif (curr_ele != arr[i]) :
curr_ele = arr[i];
vec.append(arr[i]);
size += 1;
curr_freq = 1;
curr_freq += 1;
# Print the reduced array
print("{",end= "");
for i in range(size) :
print(vec[i] ,end= ", ");
print("}",end="");
# Driver code
if __name__ == "__main__" :
arr = [ 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 ];
n = len(arr)
K = 2;
# Function call
reduceArray(arr, n, K);
# This code is contributed by AnkitRai01
C#
// C# program to reduce the array
// such that each element appears
// at most K times
using System;
using System.Collections.Generic;
class GFG{
// Function to reduce the array
static void reduceArray(int []arr, int n, int K)
{
// List to store the reduced array
List<int> vec = new List<int>();
int size = 0;
int curr_ele = arr[0], curr_freq = 1;
// Iterate over the array
for (int i = 0; i < n; i++) {
if (curr_ele == arr[i]
&& curr_freq <= K) {
vec.Add(arr[i]);
size++;
}
else if (curr_ele != arr[i]) {
curr_ele = arr[i];
vec.Add(arr[i]);
size++;
curr_freq = 1;
}
curr_freq++;
}
// Print the reduced array
Console.Write("{");
for (int i = 0; i < size; i++) {
Console.Write(vec[i]+ ", ");
}
Console.Write("}");
}
// Driver code
public static void Main(String[] args)
{
int []arr
= { 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 };
int n = arr.Length;
int K = 2;
// Function call
reduceArray(arr, n, K);
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// JavaScript program to reduce the array
// such that each element appears
// at most K times
// Function to reduce the array
function reduceArray( arr, n, K)
{
// Vector to store the reduced array
var vec=[];
var size = 0;
var curr_ele = arr[0], curr_freq = 1;
// Iterate over the array
for (var i = 0; i < n; i++) {
if (curr_ele == arr[i]
&& curr_freq <= K) {
vec.push(arr[i]);
size++;
}
else if (curr_ele != arr[i]) {
curr_ele = arr[i];
vec.push(arr[i]);
size++;
curr_freq = 1;
}
curr_freq++;
}
// Print the reduced array
document.write( "{");
for (var i = 0; i < size; i++) {
document.write( vec[i] +", ");
}
document.write("}");
}
var arr
= [ 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 ];
var n = 14;
var K = 2;
// Function call
reduceArray(arr, n, K);
</script>
Producción:
{1, 1, 2, 2, 3, 3, 4, 5, }
Complejidad temporal: O(N)
Complejidad espacial: O(1)