Summe aus Minimum und Maximum aller Subarrays der Größe k.
Bei einem Array aus positiven und negativen ganzen Zahlen besteht die Aufgabe darin, die Summe der minimalen und maximalen Elemente aller Unterarrays der Größe k zu berechnen.
Beispiele:
Eingang : arr[] = {2 5 -1 7 -3 -1 -2}
K = 4
Ausgabe : 18
Erläuterung : Subarrays der Größe 4 sind:
{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
Fehlende Unterarrays -
{2 -1 7 -3}
{2 7 -3 -1}
{2 -3 -1 -2}
{5 7 -3 -1}
{5 -3 -1 -2}
und noch ein paar mehr – warum wurden diese nicht berücksichtigt??
Unter Berücksichtigung fehlender Arrays ergibt sich ein Ergebnis von 27
Summe aller Min. und Max. = 6 + 4 + 4 + 4 = 18
Dieses Problem ist hauptsächlich eine Erweiterung des folgenden Problems.
Maximum aller Subarrays der Größe k
Naiver Ansatz:
C++Führen Sie zwei Schleifen aus, um alle Subarrays zu generieren, wählen Sie dann alle Subarrays der Größe k aus und ermitteln Sie die Maximal- und Minimalwerte. Geben Sie schließlich die Summe aller maximalen und minimalen Elemente zurück.
// 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 ));
Ausgabe
18
Zeitkomplexität: AN 2 *k), weil zwei Schleifen zum Finden aller Subarrays und eine Schleife zum Finden der maximalen und minimalen Elemente im Subarray der Größe k vorhanden sind
Hilfsraum: O(1), da kein zusätzlicher Speicherplatz verwendet wurde
Methode 2 (mit MultiSet):
Die Idee besteht darin, die Multiset-Datenstruktur und das Schiebefensterkonzept zu verwenden.
- Zuerst erstellen wir eine Multiset des Paares von {numberindex}, da der Index uns dabei helfen würde, das i-te Element zu entfernen und zum nächsten Fenster der Größe zu wechseln k .
- Zweitens haben wir ich Und J Dies sind hintere und vordere Zeiger, die zum Verwalten eines Fensters verwendet werden.
- Durchlaufen Sie das Array und fügen Sie es in das Multiset-Paar von {numberindex} ein und prüfen Sie auch die Fenstergröße, sobald sie gleich wird k Beginnen Sie mit Ihrem Hauptziel, d. h. die Summe der maximalen und minimalen Elemente zu ermitteln.
- Löschen Sie dann die i-te Indexnummer aus dem Satz und verschieben Sie den i-ten Zeiger zur nächsten Position, d. h. zum neuen Fenster mit der Größe k.
Durchführung:
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 ; }
Ausgabe
18
Zeitkomplexität: O(nlogk)
Hilfsraum: O(k)
Methode 3 (Effiziente Verwendung von Dequeue):
C++Die Idee besteht darin, die Dequeue-Datenstruktur und das Schiebefensterkonzept zu verwenden. Wir erstellen zwei leere doppelendige Warteschlangen der Größe k („S“, „G“), die nur Indizes von Elementen des aktuellen Fensters speichern, die nicht nutzlos sind. Ein Element ist nutzlos, wenn es nicht das Maximum oder Minimum der nächsten Subarrays sein kann.
// 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
Ausgabe
18
Zeitkomplexität: O(n)
Hilfsraum: O(k)
