Grootste product van een subarray van grootte k
Probeer het eens op GfG Practice 
#practiceLinkDiv {weergave: geen! belangrijk; }
#practiceLinkDiv {weergave: geen! belangrijk; } Gegeven een array bestaande uit n positieve gehele getallen en een geheel getal k. Vind de grootste productsubarray van grootte k, dat wil zeggen vind de maximale productie van k aaneengesloten elementen in de array waar k <= n.
Voorbeelden:
Input: arr[] = {1 5 9 8 2 4
1 8 1 2}
k = 6
Output: 4608
The subarray is {9 8 2 4 1 8}
Input: arr[] = {1 5 9 8 2 4 1 8 1 2}
k = 4
Output: 720
The subarray is {5 9 8 2}
Input: arr[] = {2 5 8 1 1 3};
k = 3
Output: 80
The subarray is {2 5 8} Recommended Practice Grootste product Probeer het!Brute Force-aanpak:
We herhalen alle subarrays van grootte k door twee geneste lussen te gebruiken. De buitenste lus loopt van 0 tot n-k en de binnenste lus loopt van i tot i+k-1. We berekenen het product van elke subarray en werken het maximale product dat tot nu toe is gevonden bij. Tenslotte retourneren wij het maximale product.
Hier zijn de stappen voor de bovenstaande aanpak:
- Initialiseer een variabele maxProduct naar INT_MIN, die de kleinst mogelijke gehele waarde vertegenwoordigt.
- Herhaal alle subarrays van grootte k met behulp van twee geneste lussen.
- De buitenste lus loopt van 0 tot n-k.
- De binnenste lus loopt van i tot i+k-1, waarbij i de startindex van de subarray is.
- Bereken het product van de huidige subarray met behulp van de binnenste lus.
- Als het product groter is dan maxProduct, update maxProduct dan naar het huidige product.
- Retourneer maxProduct als resultaat.
Hieronder vindt u de code van de bovenstaande aanpak:
C++Java// C++ program to find the maximum product of a subarray // of size k. #includeusing namespace std ; // This function returns maximum product of a subarray // of size k in given array arr[0..n-1]. This function // assumes that k is smaller than or equal to n. int findMaxProduct ( int arr [] int n int k ) { int maxProduct = INT_MIN ; for ( int i = 0 ; i <= n - k ; i ++ ) { int product = 1 ; for ( int j = i ; j < i + k ; j ++ ) { product *= arr [ j ]; } maxProduct = max ( maxProduct product ); } return maxProduct ; } // Driver code int main () { int arr1 [] = { 1 5 9 8 2 4 1 8 1 2 }; int k = 6 ; int n = sizeof ( arr1 ) / sizeof ( arr1 [ 0 ]); cout < < findMaxProduct ( arr1 n k ) < < endl ; k = 4 ; cout < < findMaxProduct ( arr1 n k ) < < endl ; int arr2 [] = { 2 5 8 1 1 3 }; k = 3 ; n = sizeof ( arr2 ) / sizeof ( arr2 [ 0 ]); cout < < findMaxProduct ( arr2 n k ); return 0 ; } Python3import java.util.Arrays ; public class Main { // This function returns the maximum product of a subarray of size k in the given array // It assumes that k is smaller than or equal to the length of the array. static int findMaxProduct ( int [] arr int n int k ) { int maxProduct = Integer . MIN_VALUE ; for ( int i = 0 ; i <= n - k ; i ++ ) { int product = 1 ; for ( int j = i ; j < i + k ; j ++ ) { product *= arr [ j ] ; } maxProduct = Math . max ( maxProduct product ); } return maxProduct ; } // Driver code public static void main ( String [] args ) { int [] arr1 = { 1 5 9 8 2 4 1 8 1 2 }; int k = 6 ; int n = arr1 . length ; System . out . println ( findMaxProduct ( arr1 n k )); k = 4 ; System . out . println ( findMaxProduct ( arr1 n k )); int [] arr2 = { 2 5 8 1 1 3 }; k = 3 ; n = arr2 . length ; System . out . println ( findMaxProduct ( arr2 n k )); } }C## Python Code def find_max_product ( arr k ): max_product = float ( '-inf' ) # Initialize max_product to negative infinity n = len ( arr ) # Get the length of the input array # Iterate through the array with a window of size k for i in range ( n - k + 1 ): product = 1 # Initialize product to 1 for each subarray for j in range ( i i + k ): product *= arr [ j ] # Calculate the product of the subarray max_product = max ( max_product product ) # Update max_product if necessary return max_product # Return the maximum product of a subarray of size k # Driver code if __name__ == '__main__' : arr1 = [ 1 5 9 8 2 4 1 8 1 2 ] k = 6 print ( find_max_product ( arr1 k )) # Output 25920 k = 4 print ( find_max_product ( arr1 k )) # Output 1728 arr2 = [ 2 5 8 1 1 3 ] k = 3 print ( find_max_product ( arr2 k )) # Output 80 # This code is contributed by guptapratikJavaScriptusing System ; public class GFG { // This function returns the maximum product of a subarray of size k in the given array // It assumes that k is smaller than or equal to the length of the array. static int FindMaxProduct ( int [] arr int n int k ) { int maxProduct = int . MinValue ; for ( int i = 0 ; i <= n - k ; i ++ ) { int product = 1 ; for ( int j = i ; j < i + k ; j ++ ) { product *= arr [ j ]; } maxProduct = Math . Max ( maxProduct product ); } return maxProduct ; } // Driver code public static void Main ( string [] args ) { int [] arr1 = { 1 5 9 8 2 4 1 8 1 2 }; int k = 6 ; int n = arr1 . Length ; Console . WriteLine ( FindMaxProduct ( arr1 n k )); k = 4 ; Console . WriteLine ( FindMaxProduct ( arr1 n k )); int [] arr2 = { 2 5 8 1 1 3 }; k = 3 ; n = arr2 . Length ; Console . WriteLine ( FindMaxProduct ( arr2 n k )); } }
Uitvoer4608 720 80Tijdcomplexiteit: O(n*k) waarbij n de lengte is van de invoerarray en k de grootte is van de subarray waarvoor we het maximale product vinden.
Hulpruimte: O(1) omdat we alleen een constante hoeveelheid extra ruimte gebruiken om het maximale product en het product van de huidige subarray op te slaan.Methode 2 (Efficiënt: O(n))
We kunnen het oplossen in O(n) door gebruik te maken van het feit dat het product van een subarray met grootte k in O(1) tijd kan worden berekend als we het product van een eerdere subarray bij ons hebben.
curr_product = (prev_product / arr[i-1]) * arr[i + k -1]
prev_product : Product of subarray of size k beginning
with arr[i-1]
curr_product : Product of subarray of size k beginning
with arr[i]C++
Op deze manier kunnen we het subarrayproduct met maximale k-grootte in slechts één doorgang berekenen. Hieronder vindt u de C++-implementatie van het idee.Java// C++ program to find the maximum product of a subarray // of size k. #includeusing namespace std ; // This function returns maximum product of a subarray // of size k in given array arr[0..n-1]. This function // assumes that k is smaller than or equal to n. int findMaxProduct ( int arr [] int n int k ) { // Initialize the MaxProduct to 1 as all elements // in the array are positive int MaxProduct = 1 ; for ( int i = 0 ; i < k ; i ++ ) MaxProduct *= arr [ i ]; int prev_product = MaxProduct ; // Consider every product beginning with arr[i] // where i varies from 1 to n-k-1 for ( int i = 1 ; i <= n - k ; i ++ ) { int curr_product = ( prev_product / arr [ i -1 ]) * arr [ i + k -1 ]; MaxProduct = max ( MaxProduct curr_product ); prev_product = curr_product ; } // Return the maximum product found return MaxProduct ; } // Driver code int main () { int arr1 [] = { 1 5 9 8 2 4 1 8 1 2 }; int k = 6 ; int n = sizeof ( arr1 ) / sizeof ( arr1 [ 0 ]); cout < < findMaxProduct ( arr1 n k ) < < endl ; k = 4 ; cout < < findMaxProduct ( arr1 n k ) < < endl ; int arr2 [] = { 2 5 8 1 1 3 }; k = 3 ; n = sizeof ( arr2 ) / sizeof ( arr2 [ 0 ]); cout < < findMaxProduct ( arr2 n k ); return 0 ; } Python3// Java program to find the maximum product of a subarray // of size k import java.io.* ; import java.util.* ; class GFG { // Function returns maximum product of a subarray // of size k in given array arr[0..n-1]. This function // assumes that k is smaller than or equal to n. static int findMaxProduct ( int arr [] int n int k ) { // Initialize the MaxProduct to 1 as all elements // in the array are positive int MaxProduct = 1 ; for ( int i = 0 ; i < k ; i ++ ) MaxProduct *= arr [ i ] ; int prev_product = MaxProduct ; // Consider every product beginning with arr[i] // where i varies from 1 to n-k-1 for ( int i = 1 ; i <= n - k ; i ++ ) { int curr_product = ( prev_product / arr [ i - 1 ] ) * arr [ i + k - 1 ] ; MaxProduct = Math . max ( MaxProduct curr_product ); prev_product = curr_product ; } // Return the maximum product found return MaxProduct ; } // driver program public static void main ( String [] args ) { int arr1 [] = { 1 5 9 8 2 4 1 8 1 2 }; int k = 6 ; int n = arr1 . length ; System . out . println ( findMaxProduct ( arr1 n k )); k = 4 ; System . out . println ( findMaxProduct ( arr1 n k )); int arr2 [] = { 2 5 8 1 1 3 }; k = 3 ; n = arr2 . length ; System . out . println ( findMaxProduct ( arr2 n k )); } } // This code is contributed by Pramod KumarC## Python 3 program to find the maximum # product of a subarray of size k. # This function returns maximum product # of a subarray of size k in given array # arr[0..n-1]. This function assumes # that k is smaller than or equal to n. def findMaxProduct ( arr n k ) : # Initialize the MaxProduct to 1 # as all elements in the array # are positive MaxProduct = 1 for i in range ( 0 k ) : MaxProduct = MaxProduct * arr [ i ] prev_product = MaxProduct # Consider every product beginning # with arr[i] where i varies from # 1 to n-k-1 for i in range ( 1 n - k + 1 ) : curr_product = ( prev_product // arr [ i - 1 ]) * arr [ i + k - 1 ] MaxProduct = max ( MaxProduct curr_product ) prev_product = curr_product # Return the maximum product found return MaxProduct # Driver code arr1 = [ 1 5 9 8 2 4 1 8 1 2 ] k = 6 n = len ( arr1 ) print ( findMaxProduct ( arr1 n k ) ) k = 4 print ( findMaxProduct ( arr1 n k )) arr2 = [ 2 5 8 1 1 3 ] k = 3 n = len ( arr2 ) print ( findMaxProduct ( arr2 n k )) # This code is contributed by Nikita Tiwari.JavaScript// C# program to find the maximum // product of a subarray of size k using System ; class GFG { // Function returns maximum // product of a subarray of // size k in given array // arr[0..n-1]. This function // assumes that k is smaller // than or equal to n. static int findMaxProduct ( int [] arr int n int k ) { // Initialize the MaxProduct // to 1 as all elements // in the array are positive int MaxProduct = 1 ; for ( int i = 0 ; i < k ; i ++ ) MaxProduct *= arr [ i ]; int prev_product = MaxProduct ; // Consider every product beginning // with arr[i] where i varies from // 1 to n-k-1 for ( int i = 1 ; i <= n - k ; i ++ ) { int curr_product = ( prev_product / arr [ i - 1 ]) * arr [ i + k - 1 ]; MaxProduct = Math . Max ( MaxProduct curr_product ); prev_product = curr_product ; } // Return the maximum // product found return MaxProduct ; } // Driver Code public static void Main () { int [] arr1 = { 1 5 9 8 2 4 1 8 1 2 }; int k = 6 ; int n = arr1 . Length ; Console . WriteLine ( findMaxProduct ( arr1 n k )); k = 4 ; Console . WriteLine ( findMaxProduct ( arr1 n k )); int [] arr2 = { 2 5 8 1 1 3 }; k = 3 ; n = arr2 . Length ; Console . WriteLine ( findMaxProduct ( arr2 n k )); } } // This code is contributed by anuj_67.PHP< script > // JavaScript program to find the maximum // product of a subarray of size k // Function returns maximum // product of a subarray of // size k in given array // arr[0..n-1]. This function // assumes that k is smaller // than or equal to n. function findMaxProduct ( arr n k ) { // Initialize the MaxProduct // to 1 as all elements // in the array are positive let MaxProduct = 1 ; for ( let i = 0 ; i < k ; i ++ ) MaxProduct *= arr [ i ]; let prev_product = MaxProduct ; // Consider every product beginning // with arr[i] where i varies from // 1 to n-k-1 for ( let i = 1 ; i <= n - k ; i ++ ) { let curr_product = ( prev_product / arr [ i - 1 ]) * arr [ i + k - 1 ]; MaxProduct = Math . max ( MaxProduct curr_product ); prev_product = curr_product ; } // Return the maximum // product found return MaxProduct ; } let arr1 = [ 1 5 9 8 2 4 1 8 1 2 ]; let k = 6 ; let n = arr1 . length ; document . write ( findMaxProduct ( arr1 n k ) + ' ' ); k = 4 ; document . write ( findMaxProduct ( arr1 n k ) + ' ' ); let arr2 = [ 2 5 8 1 1 3 ]; k = 3 ; n = arr2 . length ; document . write ( findMaxProduct ( arr2 n k ) + ' ' ); < /script>
Uitvoer4608 720 80Hulpruimte: O(1) omdat er geen extra ruimte wordt gebruikt.
Dit artikel is bijgedragen door Ashutosh Kumar .
