Dada una array de enteros positivos y negativos, la tarea es calcular la suma de los elementos mínimo y máximo de todos los subconjuntos de tamaño k.
Ejemplos:
Input : arr[] = {2, 5, -1, 7, -3, -1, -2}
K = 4
Output : 18
Explanation : Subarrays of size 4 are :
{2, 5, -1, 7}, min + max = -1 + 7 = 6
{5, -1, 7, -3}, min + max = -3 + 7 = 4
{-1, 7, -3, -1}, min + max = -3 + 7 = 4
{7, -3, -1, -2}, min + max = -3 + 7 = 4
Sum of all min & max = 6 + 4 + 4 + 4
= 18
Este problema es principalmente una extensión del siguiente problema.
Máximo de todos los subarreglos de tamaño k
Método 1 (Simple): Ejecute dos bucles para generar todos los subarreglos de tamaño k y encuentre los valores máximo y mínimo. Finalmente, devuelva la suma de todos los elementos máximos y mínimos.
El tiempo que tarda esta solución es O(n*k).
Método 2 (Eficiente usando Dequeue): La idea es usar la estructura de datos Dequeue y el concepto de ventana deslizante. Creamos dos colas vacías de doble extremo de tamaño k (‘S’, ‘G’) que solo almacenan índices de elementos de la ventana actual que no son inútiles. Un elemento es inútil si no puede ser máximo o mínimo de los siguientes subarreglos.
a) In deque 'G', we maintain decreasing order of
values from front to rear
b) In deque 'S', we maintain increasing order of
values from front to rear
1) First window size K
1.1) For deque 'G', if current element is greater
than rear end element, we remove rear while
current is greater.
1.2) For deque 'S', if current element is smaller
than rear end element, we just pop it while
current is smaller.
1.3) insert current element in both deque 'G' 'S'
2) After step 1, front of 'G' contains maximum element
of first window and front of 'S' contains minimum
element of first window. Remaining elements of G
and S may store maximum/minimum for subsequent
windows.
3) After that we do traversal for rest array elements.
3.1) Front element of deque 'G' is greatest and 'S'
is smallest element of previous window
3.2) Remove all elements which are out of this
window [remove element at front of queue ]
3.3) Repeat steps 1.1 , 1.2 ,1.3
4) Return sum of minimum and maximum element of all
sub-array size k.
A continuación se muestra la implementación de la idea anterior.
C++
// C++ program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
#include<bits/stdc++.h>
using namespace std;
// Returns sum of min and max element of all subarrays
// of size k
int SumOfKsubArray(int arr[] , int n , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
deque< int > S(k), G(k);
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( (!S.empty()) && arr[S.back()] >= arr[i])
S.pop_back(); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( (!G.empty()) && arr[G.back()] <= arr[i])
G.pop_back(); // Remove from rear
// Add current element at rear of both deque
G.push_back(i);
S.push_back(i);
}
// Process rest of the Array elements
for ( ; i < n; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S.front()] + arr[G.front()];
// Remove all elements which are out of this
// window
while ( !S.empty() && S.front() <= i - k)
S.pop_front();
while ( !G.empty() && G.front() <= i - k)
G.pop_front();
// remove all previous greater element that are
// useless
while ( (!S.empty()) && arr[S.back()] >= arr[i])
S.pop_back(); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( (!G.empty()) && arr[G.back()] <= arr[i])
G.pop_back(); // Remove from rear
// Add current element at rear of both deque
G.push_back(i);
S.push_back(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S.front()] + arr[G.front()];
return sum;
}
// Driver program to test above functions
int main()
{
int arr[] = {2, 5, -1, 7, -3, -1, -2} ;
int n = sizeof(arr)/sizeof(arr[0]);
int k = 3;
cout << SumOfKsubArray(arr, n, k) ;
return 0;
}
Java
// Java program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
import java.util.Deque;
import java.util.LinkedList;
public class Geeks {
// Returns sum of min and max element of all subarrays
// of size k
public static int SumOfKsubArray(int arr[] , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
Deque<Integer> S=new LinkedList<>(),G=new LinkedList<>();
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( !S.isEmpty() && arr[S.peekLast()] >= arr[i])
S.removeLast(); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( !G.isEmpty() && arr[G.peekLast()] <= arr[i])
G.removeLast(); // Remove from rear
// Add current element at rear of both deque
G.addLast(i);
S.addLast(i);
}
// Process rest of the Array elements
for ( ; i < arr.length; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S.peekFirst()] + arr[G.peekFirst()];
// Remove all elements which are out of this
// window
while ( !S.isEmpty() && S.peekFirst() <= i - k)
S.removeFirst();
while ( !G.isEmpty() && G.peekFirst() <= i - k)
G.removeFirst();
// remove all previous greater element that are
// useless
while ( !S.isEmpty() && arr[S.peekLast()] >= arr[i])
S.removeLast(); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( !G.isEmpty() && arr[G.peekLast()] <= arr[i])
G.removeLast(); // Remove from rear
// Add current element at rear of both deque
G.addLast(i);
S.addLast(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S.peekFirst()] + arr[G.peekFirst()];
return sum;
}
public static void main(String args[])
{
int arr[] = {2, 5, -1, 7, -3, -1, -2} ;
int k = 3;
System.out.println(SumOfKsubArray(arr, k));
}
}
//This code is contributed by Gaurav Tiwari
Python
# Python3 program to find Sum of all minimum and maximum # elements Of Sub-array Size k. from collections import deque # Returns Sum of min and max element of all subarrays # of size k def SumOfKsubArray(arr, n , k): Sum = 0 # Initialize result # The queue will store indexes of useful elements # in every window # In deque 'G' we maintain decreasing order of # values from front to rear # In deque 'S' we maintain increasing order of # values from front to rear S = deque() G = deque() # Process first window of size K for i in range(k): # Remove all previous greater elements # that are useless. while ( len(S) > 0 and arr[S[-1]] >= arr[i]): S.pop() # Remove from rear # Remove all previous smaller that are elements # are useless. while ( len(G) > 0 and arr[G[-1]] <= arr[i]): G.pop() # Remove from rear # Add current element at rear of both deque G.append(i) S.append(i) # Process rest of the Array elements for i in range(k, n): # Element at the front of the deque 'G' & 'S' # is the largest and smallest # element of previous window respectively Sum += arr[S[0]] + arr[G[0]] # Remove all elements which are out of this # window while ( len(S) > 0 and S[0] <= i - k): S.popleft() while ( len(G) > 0 and G[0] <= i - k): G.popleft() # remove all previous greater element that are # useless while ( len(S) > 0 and arr[S[-1]] >= arr[i]): S.pop() # Remove from rear # remove all previous smaller that are elements # are useless while ( len(G) > 0 and arr[G[-1]] <= arr[i]): G.pop() # Remove from rear # Add current element at rear of both deque G.append(i) S.append(i) # Sum of minimum and maximum element of last window Sum += arr[S[0]] + arr[G[0]] return Sum # Driver program to test above functions arr=[2, 5, -1, 7, -3, -1, -2] n = len(arr) k = 3 print(SumOfKsubArray(arr, n, k)) # This code is contributed by mohit kumar
C#
// C# program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
using System;
using System.Collections.Generic;
class Geeks
{
// Returns sum of min and max element of all subarrays
// of size k
public static int SumOfKsubArray(int []arr , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
List<int> S = new List<int>();
List<int> G = new List<int>();
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( S.Count != 0 && arr[S[S.Count - 1]] >= arr[i])
S.RemoveAt(S.Count - 1); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( G.Count != 0 && arr[G[G.Count - 1]] <= arr[i])
G.RemoveAt(G.Count - 1); // Remove from rear
// Add current element at rear of both deque
G.Add(i);
S.Add(i);
}
// Process rest of the Array elements
for ( ; i < arr.Length; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S[0]] + arr[G[0]];
// Remove all elements which are out of this
// window
while ( S.Count != 0 && S[0] <= i - k)
S.RemoveAt(0);
while ( G.Count != 0 && G[0] <= i - k)
G.RemoveAt(0);
// remove all previous greater element that are
// useless
while ( S.Count != 0 && arr[S[S.Count-1]] >= arr[i])
S.RemoveAt(S.Count - 1 ); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( G.Count != 0 && arr[G[G.Count - 1]] <= arr[i])
G.RemoveAt(G.Count - 1); // Remove from rear
// Add current element at rear of both deque
G.Add(i);
S.Add(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S[0]] + arr[G[0]];
return sum;
}
// Driver code
public static void Main(String []args)
{
int []arr = {2, 5, -1, 7, -3, -1, -2} ;
int k = 3;
Console.WriteLine(SumOfKsubArray(arr, k));
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// JavaScript program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
// Returns sum of min and max element of all subarrays
// of size k
function SumOfKsubArray(arr , k)
{
let sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
let S = [];
let G = [];
// Process first window of size K
let i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( S.length != 0 && arr[S[S.length - 1]] >= arr[i])
S.pop(); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( G.length != 0 && arr[G[G.length - 1]] <= arr[i])
G.pop(); // Remove from rear
// Add current element at rear of both deque
G.push(i);
S.push(i);
}
// Process rest of the Array elements
for ( ; i < arr.length; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S[0]] + arr[G[0]];
// Remove all elements which are out of this
// window
while ( S.length != 0 && S[0] <= i - k)
S.shift(0);
while ( G.length != 0 && G[0] <= i - k)
G.shift(0);
// remove all previous greater element that are
// useless
while ( S.length != 0 && arr[S[S.length-1]] >= arr[i])
S.pop(); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( G.length != 0 && arr[G[G.length - 1]] <= arr[i])
G.pop(); // Remove from rear
// Add current element at rear of both deque
G.push(i);
S.push(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S[0]] + arr[G[0]];
return sum;
}
// Driver code
let arr = [2, 5, -1, 7, -3, -1, -2];
let k = 3;
document.write(SumOfKsubArray(arr, k));
// This code is contributed by _saurabh_jaiswal
</script>
Producción
14
Complejidad temporal: O(n)
Espacio auxiliar: O(k)
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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