Somme du min et du max de tous les sous-tableaux de taille k.
Étant donné un tableau d'entiers positifs et négatifs, la tâche consiste à calculer la somme des éléments minimum et maximum de tous les sous-tableaux de taille k.
Exemples :
Saisir : arr[] = {2 5 -1 7 -3 -1 -2}
K = 4
Sortir : 18
Explication : Les sous-tableaux de taille 4 sont :
{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
Sous-tableaux manquants -
{2 -1 7 -3}
{2 7 -3 -1}
{2-3-1-2}
{5 7 -3 -1}
{5-3-1-2}
et quelques autres - pourquoi ceux-ci n'ont pas été pris en compte ??
Considérant que le résultat des tableaux manquants est 27
Somme de tous les min et max = 6 + 4 + 4 + 4 = 18
Ce problème est principalement une extension du problème ci-dessous.
Maximum de tous les sous-tableaux de taille k
Approche naïve :
C++Exécutez deux boucles pour générer tous les sous-tableaux, puis choisissez tous les sous-tableaux de taille k et recherchez les valeurs maximales et minimales. Enfin, renvoyez la somme de tous les éléments maximum et minimum.
// C++ program to find sum of all minimum and maximum // elements Of Sub-array Size k. #include using namespace std ; // Returns sum of min and max element of all subarrays // of size k int SumOfKsubArray ( int arr [] int N int k ) { // To store final answer int sum = 0 ; // Find all subarray for ( int i = 0 ; i < N ; i ++ ) { // To store length of subarray int length = 0 ; for ( int j = i ; j < N ; j ++ ) { // Increment the length length ++ ; // When there is subarray of size k if ( length == k ) { // To store maximum and minimum element int maxi = INT_MIN ; int mini = INT_MAX ; for ( int m = i ; m <= j ; m ++ ) { // Find maximum and minimum element maxi = max ( maxi arr [ m ]); mini = min ( mini arr [ m ]); } // Add maximum and minimum element in sum sum += maxi + mini ; } } } 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 = 4 ; 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.Arrays ; class GFG { // Returns sum of min and max element of all subarrays // of size k static int SumOfKsubArray ( int [] arr int N int k ) { // To store the final answer int sum = 0 ; // Find all subarrays for ( int i = 0 ; i < N ; i ++ ) { // To store the length of the subarray int length = 0 ; for ( int j = i ; j < N ; j ++ ) { // Increment the length length ++ ; // When there is a subarray of size k if ( length == k ) { // To store the maximum and minimum element int maxi = Integer . MIN_VALUE ; int mini = Integer . MAX_VALUE ; for ( int m = i ; m <= j ; m ++ ) { // Find the maximum and minimum element maxi = Math . max ( maxi arr [ m ] ); mini = Math . min ( mini arr [ m ] ); } // Add the maximum and minimum element to the sum sum += maxi + mini ; } } } return sum ; } // Driver program to test above functions public static void main ( String [] args ) { int [] arr = { 2 5 - 1 7 - 3 - 1 - 2 }; int N = arr . length ; int k = 4 ; System . out . println ( SumOfKsubArray ( arr N k )); } } //This code is contributed by Vishal Dhaygude
Python # Returns sum of min and max element of all subarrays # of size k def sum_of_k_subarray ( arr N k ): # To store final answer sum = 0 # Find all subarrays for i in range ( N ): # To store length of subarray length = 0 for j in range ( i N ): # Increment the length length += 1 # When there is a subarray of size k if length == k : # To store maximum and minimum element maxi = float ( '-inf' ) mini = float ( 'inf' ) for m in range ( i j + 1 ): # Find maximum and minimum element maxi = max ( maxi arr [ m ]) mini = min ( mini arr [ m ]) # Add maximum and minimum element to sum sum += maxi + mini return sum # Driver program to test above function def main (): arr = [ 2 5 - 1 7 - 3 - 1 - 2 ] N = len ( arr ) k = 4 print ( sum_of_k_subarray ( arr N k )) if __name__ == '__main__' : main ()
C# using System ; class Program { // Returns sum of min and max element of all subarrays // of size k static int SumOfKSubArray ( int [] arr int N int k ) { // To store the final answer int sum = 0 ; // Find all subarrays for ( int i = 0 ; i < N ; i ++ ) { // To store the length of subarray int length = 0 ; for ( int j = i ; j < N ; j ++ ) { // Increment the length length ++ ; // When there is a subarray of size k if ( length == k ) { // To store the maximum and minimum // element int maxi = int . MinValue ; int mini = int . MaxValue ; for ( int m = i ; m <= j ; m ++ ) { // Find maximum and minimum element maxi = Math . Max ( maxi arr [ m ]); mini = Math . Min ( mini arr [ m ]); } // Add maximum and minimum element in // sum sum += maxi + mini ; } } } return sum ; } // Driver program to test above functions static void Main () { int [] arr = { 2 5 - 1 7 - 3 - 1 - 2 }; int N = arr . Length ; int k = 4 ; Console . WriteLine ( SumOfKSubArray ( arr N k )); } }
JavaScript // 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 N k ) { // To store final answer let sum = 0 ; // Find all subarray for ( let i = 0 ; i < N ; i ++ ) { // To store length of subarray let length = 0 ; for ( let j = i ; j < N ; j ++ ) { // Increment the length length ++ ; // When there is subarray of size k if ( length === k ) { // To store maximum and minimum element let maxi = Number . MIN_SAFE_INTEGER ; let mini = Number . MAX_SAFE_INTEGER ; for ( let m = i ; m <= j ; m ++ ) { // Find maximum and minimum element maxi = Math . max ( maxi arr [ m ]); mini = Math . min ( mini arr [ m ]); } // Add maximum and minimum element in sum sum += maxi + mini ; } } } return sum ; } // Driver program to test above function const arr = [ 2 5 - 1 7 - 3 - 1 - 2 ]; const N = arr . length ; const k = 4 ; console . log ( SumOfKsubArray ( arr N k ));
Sortir
18
Complexité temporelle : SUR 2 *k) car deux boucles pour trouver tous les sous-tableaux et une boucle pour trouver les éléments maximum et minimum dans le sous-tableau de taille k
Espace auxiliaire : O(1) car aucun espace supplémentaire n'a été utilisé
Méthode 2 (à l'aide de MultiSet) :
L'idée est d'utiliser la structure de données Multiset et le concept de fenêtre coulissante.
- Tout d'abord, nous créons un multiensemble de paire de {numberindex} car l'index nous aiderait à supprimer le ième élément et à passer à la fenêtre de taille suivante k .
- Deuxièmement, nous avons je et j qui sont des pointeurs arrière et avant utilisés pour maintenir une fenêtre.
- Parcourez le tableau et insérez-le dans une paire multiensemble de {numberindex} et vérifiez également windowSize une fois qu'il devient égal à k commencez votre objectif principal, c'est-à-dire trouver la somme des éléments max et min.
- Effacez ensuite le ième numéro d'index de l'ensemble et déplacez le ième pointeur vers l'emplacement suivant, c'est-à-dire une nouvelle fenêtre de taille k.
Mise en œuvre:
C++ // C++ program to find sum of all minimum and maximum // elements Of Sub-array Size k. #include 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 ; // to store our final sum // multiset because nos. could be repeated // multiset pair is {numberindex} multiset < pair < int int > > ms ; int i = 0 ; // back pointer int j = 0 ; // front pointer while ( j < n && i < n ) { ms . insert ( { arr [ j ] j }); // inserting {numberindex} // front pointer - back pointer + 1 is for checking // window size int windowSize = j - i + 1 ; // Once they become equal start what we need to do if ( windowSize == k ) { // extracting first since set is always in // sorted ascending order int mini = ms . begin () -> first ; // extracting last element aka beginning from // last (numbers extraction) int maxi = ms . rbegin () -> first ; // adding summation of maximum & minimum element // of each subarray of k into final sum sum += ( maxi + mini ); // erasing the ith index element from set as it // won't appaer in next window of size k ms . erase ({ arr [ i ] i }); // increasing back pointer for next window of // size k; i ++ ; } j ++ ; // always increments front pointer } 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 = 4 ; cout < < SumOfKsubArray ( arr n k ); return 0 ; }
Sortir
18
Complexité temporelle : O(nlogk)
Espace auxiliaire : O(k)
Méthode 3 (efficace en utilisant Dequeue) :
C++L'idée est d'utiliser la structure de données Dequeue et le concept de fenêtre coulissante. Nous créons deux files d'attente vides à double extrémité de taille k (« S » « G ») qui stockent uniquement les indices des éléments de la fenêtre courante qui ne sont pas inutiles. Un élément est inutile s'il ne peut pas être le maximum ou le minimum des sous-tableaux suivants.
// C++ program to find sum of all minimum and maximum // elements Of Sub-array Size k. #include 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 if ( ! S . empty () && S . front () == i - k ) S . pop_front (); if ( ! 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 = 4 ; 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 = 4 ; 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 = 4 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 = 4 ; Console . WriteLine ( SumOfKsubArray ( arr k )); } } // This code is contributed by gauravrajput1
JavaScript // 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 = 4 ; console . log ( SumOfKsubArray ( arr k )); // This code is contributed by _saurabh_jaiswal
Sortir
18
Complexité temporelle : O(n)
Espace auxiliaire : O(k)
