Som van min en max van alle subarrays van grootte k.
Gegeven een array van zowel positieve als negatieve gehele getallen, is het de taak om de som te berekenen van de minimum- en maximumelementen van alle sub-arrays met grootte k.
Voorbeelden:
Invoer : arr[] = {2 5 -1 7 -3 -1 -2}
K = 4
Uitvoer : 18
Uitleg : Subarrays van grootte 4 zijn:
{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
Ontbrekende subarrays -
{2 -1 7 -3}
{2 7 -3 -1}
{2 -3 -1 -2}
{5 7 -3 -1}
{5 -3 -1 -2}
en nog een paar -- waarom werden deze niet overwogen?
Gezien ontbrekende arrays komt het resultaat op 27
Som van alle min en max = 6 + 4 + 4 + 4 = 18
Dit probleem is voornamelijk een uitbreiding van het onderstaande probleem.
Maximaal van alle subarrays van grootte k
Naïeve aanpak:
C++Voer twee lussen uit om alle subarrays te genereren en kies vervolgens alle subarrays van grootte k en vind de maximale en minimale waarden. Geef ten slotte de som van alle maximale en minimale elementen terug.
// 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 ));
Uitvoer
18
Tijdcomplexiteit: OP 2 *k) omdat twee lussen om alle subarrays te vinden en één lus om de maximale en minimale elementen in de subarray van grootte k te vinden
Hulpruimte: O(1) omdat er geen extra ruimte is gebruikt
Methode 2 (MultiSet gebruiken):
Het idee is om Multiset-datastructuur en een schuifraamconcept te gebruiken.
- Eerst creëren we een meervoudig ingesteld van paar {numberindex} omdat index ons zou helpen bij het verwijderen van het i-de element en naar het volgende venster van grootte zou gaan k .
- Ten tweede hebben we dat gedaan i En J Dit zijn de voor- en achterwijzers die worden gebruikt om een venster te behouden.
- Doorloop de array en plaats deze in een multiset-paar van {numberindex} en controleer ook op windowSize zodra deze gelijk wordt aan k begin met uw primaire doel, d.w.z. het vinden van de som van max en min elementen.
- Wis vervolgens het i-de indexnummer uit de set en verplaats de i-de wijzer naar de volgende locatie, d.w.z. een nieuw venster van grootte k.
Uitvoering:
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 ; }
Uitvoer
18
Tijdcomplexiteit: O(nlogk)
Hulpruimte: O(k)
Methode 3 (Efficiënt met behulp van Dequeue):
C++Het idee is om de Dequeue-datastructuur en het schuifvensterconcept te gebruiken. We creëren twee lege dubbele wachtrijen van grootte k ('S' 'G') die alleen indices opslaan van elementen van het huidige venster die niet nutteloos zijn. Een element is nutteloos als het niet het maximum of minimum van volgende subarrays kan zijn.
// 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
Uitvoer
18
Tijdcomplexiteit: O(n)
Hulpruimte: O(k)
