Dada una array de N enteros donde N es par. Hay dos tipos de operaciones permitidas en la array.
- Aumenta el valor de cualquier elemento A[i] en 1.
- Si dos elementos adyacentes en la array son números primos consecutivos, elimine ambos elementos. Es decir, A[i] es un número primo y A[i+1] es el siguiente número primo.
La tarea es encontrar el número mínimo de operaciones requeridas para eliminar todos los elementos de la array.
Ejemplos:
Input : arr[] = { 1, 2, 4, 3 }
Output : 5
Minimum 5 operation are required.
1. Increase the 2nd element by 1
{ 1, 2, 4, 3 } -> { 1, 3, 4, 3 }
2. Increase the 3rd element by 1
{ 1, 3, 4, 3 } -> { 1, 3, 5, 3 }
3. Delete the 2nd and 3rd element
{ 1, 3, 5, 3 } -> { 1, 3 }
4. Increase the 1st element by 1
{ 1, 3 } -> { 2, 3 }
5. Delete the 1st and 2nd element
{ 2, 3 } -> { }
Input : arr[] = {10, 12}
Output : 3
Para eliminar números, debemos transformar dos números en dos números primos consecutivos. ¿Cómo calcular el costo mínimo de transformar dos números a, b en dos números primos consecutivos, donde el costo es el número de incremento de ambos números?
Usamos criba para precalcular números primos y luego encontramos el primer primo p que no es mayor que a y el primero mayor que p usando una array.
Una vez que hemos calculado dos números primos más cercanos, usamos Programación Dinámica para resolver el problema. Sea dp[i][j] el costo mínimo de limpiar el subarreglo A[i, j]. Si hay dos números en la array, la respuesta es fácil de encontrar. Ahora, para N > 2, pruebe cualquier elemento en la posición k > i como un par para A[i], tal que haya un número par de elementos de A[i, j] entre A[i] y A[k]. Para un k fijo, el costo mínimo de compensación A[i, j], es decir, dp[i][j], es igual a dp[i + 1][k – 1] + dp[k + 1][j] + ( costo de transformar A[i] y A[k] en números primos consecutivos). Podemos calcular la respuesta iterando sobre todos los k posibles.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to count minimum operations
// required to remove an array
#include<bits/stdc++.h>
#define MAX 100005
using namespace std;
// Return the cost to convert two numbers into
// consecutive prime number.
int cost(int a, int b, int prev[], int nxt[])
{
int sub = a + b;
if (a <= b && prev[b-1] >= a)
return nxt[b] + prev[b-1] - a - b;
a = max(a, b);
a = nxt[a];
b = nxt[a + 1];
return a + b - sub;
}
// Sieve to store next and previous prime
// to a number.
void sieve(int prev[], int nxt[])
{
int pr[MAX] = { 0 };
pr[1] = 1;
for (int i = 2; i < MAX; i++)
{
if (pr[i])
continue;
for (int j = i*2; j < MAX; j += i)
pr[j] = 1;
}
// Computing next prime each number.
for (int i = MAX - 1; i; i--)
{
if (pr[i] == 0)
nxt[i] = i;
else
nxt[i] = nxt[i+1];
}
// Computing previous prime each number.
for (int i = 1; i < MAX; i++)
{
if (pr[i] == 0)
prev[i] = i;
else
prev[i] = prev[i-1];
}
}
// Return the minimum number of operation required.
int minOperation(int arr[], int nxt[], int prev[], int n)
{
int dp[n + 5][n + 5] = { 0 };
// For each index.
for (int r = 0; r < n; r++)
{
// Each subarray.
for (int l = r-1; l >= 0; l -= 2)
{
dp[l][r] = INT_MAX;
for (int ad = l; ad < r; ad += 2)
dp[l][r] = min(dp[l][r], dp[l][ad] +
dp[ad+1][r-1] +
cost(arr[ad], arr[r], prev, nxt));
}
}
return dp[0][n - 1] + n/2;
}
// Driven Program
int main()
{
int arr[] = { 1, 2, 4, 3 };
int n = sizeof(arr)/sizeof(arr[0]);
int nxt[MAX], prev[MAX];
sieve(prev, nxt);
cout << minOperation(arr, nxt, prev, n);
return 0;
}
Java
// Java program to count minimum operations
// required to remove an array
class GFG
{
static final int MAX = 100005;
// Return the cost to convert two
// numbers into consecutive prime number.
static int cost(int a, int b,
int prev[], int nxt[])
{
int sub = a + b;
if (a <= b && prev[b - 1] >= a)
{
return nxt[b] + prev[b - 1] - a - b;
}
a = Math.max(a, b);
a = nxt[a];
b = nxt[a + 1];
return a + b - sub;
}
// Sieve to store next and previous
// prime to a number.
static void sieve(int prev[], int nxt[])
{
int pr[] = new int[MAX];
pr[1] = 1;
for (int i = 2; i < MAX; i++)
{
if (pr[i] == 1)
{
continue;
}
for (int j = i * 2; j < MAX; j += i)
{
pr[j] = 1;
}
}
// Computing next prime each number.
for (int i = MAX - 2; i > 0; i--)
{
if (pr[i] == 0)
{
nxt[i] = i;
}
else
{
nxt[i] = nxt[i + 1];
}
}
// Computing previous prime each number.
for (int i = 1; i < MAX; i++)
{
if (pr[i] == 0)
{
prev[i] = i;
}
else
{
prev[i] = prev[i - 1];
}
}
}
// Return the minimum number
// of operation required.
static int minOperation(int arr[], int nxt[],
int prev[], int n)
{
int dp[][] = new int[n + 5][n + 5];
// For each index.
for (int r = 0; r < n; r++)
{
// Each subarray.
for (int l = r - 1; l >= 0; l -= 2)
{
dp[l][r] = Integer.MAX_VALUE;
for (int ad = l; ad < r; ad += 2)
{
dp[l][r] = Math.min(dp[l][r], dp[l][ad] +
dp[ad + 1][r - 1] +
cost(arr[ad], arr[r],
prev, nxt));
}
}
}
return dp[0][n - 1] + n / 2;
}
// Driver Code
public static void main(String args[])
{
int arr[] = {1, 2, 4, 3};
int n = arr.length;
int nxt[] = new int[MAX], prev[] = new int[MAX];
sieve(prev, nxt);
System.out.println(minOperation(arr, nxt, prev, n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python 3 program to count minimum # operations required to remove an array MAX = 100005 INT_MAX = 10000000 # Return the cost to convert two numbers # into consecutive prime number. def cost(a, b, prev, nxt): sub = a + b if (a <= b and prev[b - 1] >= a): return nxt[b] + prev[b - 1] - a - b a = max(a, b) a = nxt[a] b = nxt[a + 1] return a + b - sub # Sieve to store next and previous # prime to a number. def sieve(prev, nxt): pr = [0 for i in range(MAX)] pr[1] = 1 for i in range(1, MAX): if (pr[i]): continue for j in range(i * 2, MAX, i): pr[j] = 1 # Computing next prime each number. for i in range(MAX - 2, -1, -1): if (pr[i] == 0): nxt[i] = i else: nxt[i] = nxt[i + 1] # Computing previous prime each number. for i in range(1, MAX): if (pr[i] == 0): prev[i] = i else: prev[i] = prev[i - 1] # Return the minimum number of # operation required. def minOperation(arr, nxt, prev, n): dp = [[0 for i in range(n + 5)] for i in range(n + 5)] # For each index. for r in range(n): # Each subarray. for l in range(r - 1, -1, -2): dp[l][r] = INT_MAX; for ad in range(l, r, 2): dp[l][r] = min(dp[l][r], dp[l][ad] + dp[ad + 1][r - 1] + cost(arr[ad], arr[r], prev, nxt)) return dp[0][n - 1] + n // 2 # Driver Code arr = [1, 2, 4, 3] n = len(arr) nxt = [0 for i in range(MAX)] prev = [0 for i in range(MAX)] sieve(prev, nxt) print(minOperation(arr, nxt, prev, n)) # This code is contributed by sahishelangia
C#
// C# program to count minimum operations
// required to remove an array
using System;
class GFG
{
static int MAX = 100005;
// Return the cost to convert two
// numbers into consecutive prime number.
static int cost(int a, int b,
int[] prev, int[] nxt)
{
int sub = a + b;
if (a <= b && prev[b - 1] >= a)
{
return nxt[b] + prev[b - 1] - a - b;
}
a = Math.Max(a, b);
a = nxt[a];
b = nxt[a + 1];
return a + b - sub;
}
// Sieve to store next and previous
// prime to a number.
static void sieve(int[] prev, int[] nxt)
{
int[] pr = new int[MAX];
pr[1] = 1;
for (int i = 2; i < MAX; i++)
{
if (pr[i] == 1)
{
continue;
}
for (int j = i * 2; j < MAX; j += i)
{
pr[j] = 1;
}
}
// Computing next prime each number.
for (int i = MAX - 2; i > 0; i--)
{
if (pr[i] == 0)
{
nxt[i] = i;
}
else
{
nxt[i] = nxt[i + 1];
}
}
// Computing previous prime each number.
for (int i = 1; i < MAX; i++)
{
if (pr[i] == 0)
{
prev[i] = i;
}
else
{
prev[i] = prev[i - 1];
}
}
}
// Return the minimum number
// of operation required.
static int minOperation(int[] arr, int[] nxt,
int[] prev, int n)
{
int[,] dp = new int[n + 5, n + 5];
// For each index.
for (int r = 0; r < n; r++)
{
// Each subarray.
for (int l = r - 1; l >= 0; l -= 2)
{
dp[l, r] = Int32.MaxValue;
for (int ad = l; ad < r; ad += 2)
{
dp[l, r] = Math.Min(dp[l, r], dp[l, ad] +
dp[ad + 1, r - 1] +
cost(arr[ad], arr[r],
prev, nxt));
}
}
}
return dp[0, n - 1] + n / 2;
}
// Driver Code
public static void Main()
{
int[] arr = {1, 2, 4, 3};
int n = arr.Length;
int[] nxt = new int[MAX];
int[] prev = new int[MAX];
sieve(prev, nxt);
Console.WriteLine(minOperation(arr, nxt, prev, n));
}
}
// This code is contributed by Mukul Singh
PHP
<?php
// PHP program to count minimum operations
// required to remove an array
$MAX = 100005;
// Return the cost to convert two numbers into
// consecutive prime number.
function cost($a, $b, &$prev, &$nxt)
{
$sub = $a + $b;
if ($a <= $b && $prev[$b-1] >= $a)
return $nxt[$b] + $prev[$b-1] - $a - $b;
$a = max($a, $b);
$a = $nxt[$a];
$b = $nxt[$a + 1];
return $a + $b - $sub;
}
// Sieve to store next and previous prime
// to a number.
function sieve(&$prev, &$nxt)
{
global $MAX;
$pr = array_fill(0,$MAX,NULL);
$pr[1] = 1;
for ($i = 2; $i < $MAX; $i++)
{
if ($pr[$i])
continue;
for ($j = $i*2; $j < $MAX; $j += $i)
$pr[$j] = 1;
}
// Computing next prime each number.
for ($i = $MAX - 1; $i; $i--)
{
if ($pr[$i] == 0)
$nxt[$i] = $i;
else
$nxt[$i] = $nxt[$i+1];
}
// Computing previous prime each number.
for ($i = 1; $i < $MAX; $i++)
{
if ($pr[$i] == 0)
$prev[$i] = $i;
else
$prev[$i] = $prev[$i-1];
}
}
// Return the minimum number of operation required.
function minOperation(&$arr, &$nxt, &$prev, $n)
{
global $MAX;
$dp = array_fill(0,($n + 5),array_fill(0,($n + 5),NULL));
// For each index.
for ($r = 0; $r < $n; $r++)
{
// Each subarray.
for ($l = $r-1; $l >= 0; $l -= 2)
{
$dp[$l][$r] = PHP_INT_MAX;
for ($ad = $l; $ad < $r; $ad += 2)
$dp[$l][$r] = min($dp[$l][$r], $dp[$l][$ad] +
$dp[$ad+1][$r-1] +
cost($arr[$ad], $arr[$r], $prev, $nxt));
}
}
return $dp[0][$n - 1] + $n/2;
}
// Driven Program
$arr = array( 1, 2, 4, 3 );
$n = sizeof($arr)/sizeof($arr[0]);
$nxt = array_fill(0,$MAX,NULL);
$prev = array_fill(0,$MAX,NULL);
sieve($prev, $nxt);
echo minOperation($arr, $nxt, $prev, $n);
return 0;
?>
Javascript
<script>
// Javascript program to count minimum operations
// required to remove an array
let MAX = 100005;
// Return the cost to convert two
// numbers into consecutive prime number.
function cost(a,b,prev,nxt)
{
let sub = a + b;
if (a <= b && prev[b - 1] >= a)
{
return nxt[b] + prev[b - 1] - a - b;
}
a = Math.max(a, b);
a = nxt[a];
b = nxt[a + 1];
return a + b - sub;
}
// Sieve to store next and previous
// prime to a number.
function sieve(prev,nxt)
{
let pr = new Array(MAX);
for(let i=0;i<MAX;i++)
{
pr[i]=0;
}
pr[1] = 1;
for (let i = 2; i < MAX; i++)
{
if (pr[i] == 1)
{
continue;
}
for (let j = i * 2; j < MAX; j += i)
{
pr[j] = 1;
}
}
// Computing next prime each number.
for (let i = MAX - 2; i > 0; i--)
{
if (pr[i] == 0)
{
nxt[i] = i;
}
else
{
nxt[i] = nxt[i + 1];
}
}
// Computing previous prime each number.
for (let i = 1; i < MAX; i++)
{
if (pr[i] == 0)
{
prev[i] = i;
}
else
{
prev[i] = prev[i - 1];
}
}
}
// Return the minimum number
// of operation required.
function minOperation(arr,nxt,prev,n)
{
let dp = new Array(n + 5);
for(let i=0;i<n+5;i++)
{
dp[i]=new Array(n+5);
for(let j=0;j<n+5;j++)
{
dp[i][j]=0;
}
}
// For each index.
for (let r = 0; r < n; r++)
{
// Each subarray.
for (let l = r - 1; l >= 0; l -= 2)
{
dp[l][r] = Number.MAX_VALUE;
for (let ad = l; ad < r; ad += 2)
{
dp[l][r] = Math.min(dp[l][r], dp[l][ad] +
dp[ad + 1][r - 1] +
cost(arr[ad], arr[r],
prev, nxt));
}
}
}
return dp[0][n - 1] + n / 2;
}
// Driver Code
let arr=[1, 2, 4, 3];
let n = arr.length;
let nxt=new Array(MAX);
let prev=new Array(MAX);
sieve(prev, nxt);
document.write(minOperation(arr, nxt, prev, n));
// This code is contributed by rag2127
</script>
Producción
5
Complejidad Temporal: O(N 3 ).
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA