Summan av min och max för alla undergrupper av storlek k.
Givet en matris av både positiva och negativa heltal är uppgiften att beräkna summan av minsta och maximala element för alla undermatriser av storlek k.
Exempel:
Input : arr[] = {2 5 -1 7 -3 -1 -2}
K = 4
Produktion : 18
Förklaring : Undergrupper av storlek 4 är:
{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
Saknade undermatriser -
{2 -1 7 -3}
{2 7 -3 -1}
{2 -3 -1 -2}
{5 7 -3 -1}
{5 -3 -1 -2}
och några fler -- varför övervägdes inte dessa??
Med tanke på att arrayer saknas kommer resultatet som 27
Summan av alla min & max = 6 + 4 + 4 + 4 = 18
Detta problem är huvudsakligen en förlängning av nedanstående problem.
Maximalt av alla undergrupper av storlek k
Naivt förhållningssätt:
C++Kör två slingor för att generera alla subarrayer och välj sedan alla subarrays av storlek k och hitta max- och minimumvärden. Returnera slutligen summan av alla max- och minimumelement.
// 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 ));
Produktion
18
Tidskomplexitet: PÅ 2 *k) eftersom två slingor för att hitta alla underarrayer och en slinga för att hitta de maximala och minimala elementen i underarrayen av storlek k
Hjälputrymme: O(1) eftersom inget extra utrymme har använts
Metod 2 (Använda MultiSet):
Tanken är att använda Multiset-datastruktur och skjutfönsterkoncept.
- Först skapar vi en multiset av par av {numberindex} eftersom index skulle hjälpa oss att ta bort elementet ith och gå till nästa storleksfönster k .
- För det andra har vi i och j som är bak- och frontpekare som används för att underhålla ett fönster.
- Gå igenom arrayen och infoga i multiset-paret av {numberindex} och kontrollera även för fönsterstorlek när den blir lika med k börja ditt primära mål, dvs att hitta summan av max och min element.
- Radera sedan det i:te indexnumret från uppsättningen och flytta i:t-pekaren till nästa plats, dvs. ett nytt fönster med storlek k.
Genomförande:
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 ; }
Produktion
18
Tidskomplexitet: O(nlogk)
Hjälputrymme: O(k)
Metod 3 (effektiv med Dequeue):
C++Tanken är att använda Dequeue-datastruktur och skjutfönsterkoncept. Vi skapar två tomma dubbelköer av storlek k ('S' 'G') som bara lagrar index för element i det aktuella fönstret som inte är värdelösa. Ett element är värdelöst om det inte kan vara maximalt eller minimum av nästa subarrayer.
// 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
Produktion
18
Tidskomplexitet: O(n)
Hjälputrymme: O(k)
