Dada una array arr[] que consta de N enteros y un entero K, la tarea es encontrar el máximo número posible de elementos únicos aumentando cualquier elemento de la array en K solo una vez.
Ejemplos:
Entrada: arr[] = {0, 2, 4, 3, 4}, K = 1
Salida: 5
Explicación:
Incrementar arr[2] ( = 4) por K ( = 1). Por lo tanto, la nueva array es {0, 2, 4, 3, 5} que tiene 5 elementos únicos.Entrada: arr[] = {2, 3, 2, 4, 5, 5, 7, 4}, K = 2
Salida: 7
Explicación:
Aumentar 4 en 2 = 6.
Aumentar elemento 7 en 2 = 9
Aumentar 5 en 2 = 7.
La nueva array es {2, 3, 2, 4, 5, 7, 9, 6} que contiene 7 elementos únicos.
Enfoque: la idea para resolver este problema es almacenar la frecuencia de los elementos de la array en un mapa y cambiar los elementos de la array en consecuencia para obtener elementos únicos en la array después de incrementar cualquier valor en K . Siga los pasos a continuación para resolver el problema:
- Atraviese la array y almacene la frecuencia de cada elemento de la array en un mapa .
- Recorra el Mapa y realice los siguientes pasos:
- Si el recuento del elemento actual es 1 , no cambie el elemento de la array.
- Si el recuento del elemento actual es superior a 1 , entonces aumente el elemento actual en K y aumente el recuento de (elemento_actual + K) en 1 en el Mapa .
- Después de completar los pasos anteriores, imprima el tamaño del Mapa como el número máximo de elementos únicos en la array.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
void maxDifferent(int arr[], int N, int K)
{
// Stores the count of element
// in array
map<int, int> M;
// Traverse the array
for (int i = 0; i < N; i++) {
// Increase the counter of
// the array element by 1
M[arr[i]]++;
}
// Traverse the map
for (auto it = M.begin();
it != M.end(); it++) {
// Extract the current element
int current_element = it->first;
// Number of times the current
// element is present in array
int count = it->second;
// If element is present only
// once, then do not change it
if (count == 1)
continue;
// If the count > 1 then change
// one of the same current
// elements to (current_element + K)
// and increase its count by 1
M[current_element + K]++;
}
// The size of the map is the
// required answer
cout << M.size();
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 2, 4, 5, 5, 7, 4 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
maxDifferent(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
static void maxDifferent(int arr[], int N, int K)
{
// Stores the count of element
// in array
HashMap<Integer,
Integer> M = new HashMap<Integer,
Integer>();
// Traverse the array
for(int i = 0; i < N; i++)
{
// Increase the counter of
// the array element by 1
Integer count = M.get(arr[i]);
if (count == null)
{
M.put(arr[i], 1);
}
else
{
M.put(arr[i], count + 1);
}
}
// Iterator itr = M.entrySet().iterator();
Iterator<Map.Entry<Integer,
Integer>> itr = M.entrySet().iterator();
int[] ar1 = new int[N];
// Traverse the map
while (itr.hasNext())
{
Map.Entry<Integer, Integer> Element = itr.next();
// Extract the current element
int current_element = (int)Element.getKey();
// Number of times the current
// element is present in array
int count = (int)Element.getValue();
// If element is present only
// once, then do not change it
if (count == 1)
continue;
// If the count > 1 then change
// one of the same current
// elements to (current_element + K)
// and increase its count by 1
ar1[current_element + K]++;
}
for(int i = 0; i < N; i++)
{
if (ar1[i] >= 0)
{
Integer count = M.get(ar1[i]);
if (count == null)
{
M.put(ar1[i], 1);
}
else
{
M.put(ar1[i], count + 1);
}
}
}
// The size of the map is the
// required answer
System.out.println(M.size());
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 3, 2, 4, 5, 5, 7, 4 };
int K = 2;
int N = arr.length;
// Function Call
maxDifferent(arr, N, K);
}
}
// This code is contributed by Dharanendra L V
Python3
# Python3 program for the above approach
# Function to find the maximum unique
# elements in array after incrementing
# any element by K
def maxDifferent(arr, N, K):
# Stores the count of element
# in array
M = {}
# Traverse the array
for i in range(N):
# Increase the counter of
# the array element by 1
M[arr[i]] = M.get(arr[i], 0) + 1
# Traverse the map
for it in list(M.keys()):
# Extract the current element
current_element = it
# Number of times the current
# element is present in array
count = M[it]
# If element is present only
# once, then do not change it
if (count == 1):
continue
# If the count > 1 then change
# one of the same current
# elements to (current_element + K)
# and increase its count by 1
M[current_element + K] = M.get(current_element, 0) + 1
# The size of the map is the
# required answer
print(len(M))
# Driver Code
if __name__ == '__main__':
arr=[2, 3, 2, 4, 5, 5, 7, 4]
K = 2
N = len(arr)
# Function Call
maxDifferent(arr, N, K)
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
static void maxDifferent(int []arr, int N, int K)
{
// Stores the count of element
// in array
Dictionary<int,
int> M = new Dictionary<int,
int>();
// Traverse the array
for(int i = 0; i < N; i++)
{
// Increase the counter of
// the array element by 1
int count = M.ContainsKey(arr[i]) ? M[arr[i]] : 0;
if (count == 0)
{
M.Add(arr[i], 1);
}
else
{
M[arr[i]] = count + 1;
}
}
int[] ar1 = new int[N];
// Traverse the map
foreach(KeyValuePair<int, int> Element in M)
{
// Extract the current element
int current_element = (int)Element.Key;
// Number of times the current
// element is present in array
int count = (int)Element.Value;
// If element is present only
// once, then do not change it
if (count == 1)
continue;
// If the count > 1 then change
// one of the same current
// elements to (current_element + K)
// and increase its count by 1
ar1[current_element + K]++;
}
for(int i = 0; i < N; i++)
{
if (ar1[i] >= 0)
{
int count = M.ContainsKey(ar1[i]) ? M[ar1[i]] : 0;
if (count == 0)
{
M.Add(ar1[i], 1);
}
else
{
M[ar1[i]] = count + 1;
}
}
}
// The size of the map is the
// required answer
Console.WriteLine(M.Count);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 3, 2, 4, 5, 5, 7, 4 };
int K = 2;
int N = arr.Length;
// Function Call
maxDifferent(arr, N, K);
}
}
// This code contributed by shikhasingrajput
Javascript
<script>
// JavaScript program for the above approach
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
function maxDifferent(arr, N, K)
{
// Stores the count of element
// in array
var M = new Map();
// Traverse the array
for (var i = 0; i < N; i++) {
// Increase the counter of
// the array element by 1
if(M.has(arr[i]))
{
M.set(arr[i], M.get(arr[i])+1);
}
else
{
M.set(arr[i],1);
}
}
M.forEach((value, key) => {
// Extract the current element
var current_element = key;
// Number of times the current
// element is present in array
var count = value;
// If element is present only
// once, then do not change it
if (count != 1)
{
// If the count > 1 then change
// one of the same current
// elements to (current_element + K)
// and increase its count by 1
if(M.has(current_element+K))
{
M.set(current_element+K,
M.get(current_element+K)+1)
}
else
{
M.set(current_element+K,1);
}
}
});
// The size of the map is the
// required answer
document.write( M.size);
}
// Driver Code
var arr = [2, 3, 2, 4, 5, 5, 7, 4];
var K = 2;
var N = arr.length;
// Function Call
maxDifferent(arr, N, K);
</script>
Producción:
7
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por adarsh_sinhg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA