Neka svi elementi niza budu jednaki uz minimalne troškove
S obzirom na niz veličine n zadatak je da se vrijednost svih elemenata izjednači sa minimalni trošak . Trošak promjene vrijednosti iz x u y je abs(x - y).
Primjeri:
Ulazni: arr[] = [1 100 101]
Izlaz : 100
Obrazloženje: Sve njegove vrijednosti možemo promijeniti na 100 uz minimalne troškove
|1 - 100| + |100 - 100| + |101 - 100| = 100Ulazni : arr[] = [4 6]
Izlaz : 2
Obrazloženje: Sve njegove vrijednosti možemo promijeniti na 5 uz minimalne troškove
|4 - 5| + |5 - 6| = 2Ulazni: arr[] = [5 5 5 5]
Izlaz:
Obrazloženje: Sve su vrijednosti već jednake.
[Naivni pristup] Korištenje 2 ugniježđene petlje - O(n^2) vremena i O(1) prostora
C++Imajte na umu da naš odgovor uvijek može biti jedna od vrijednosti polja. Čak iu drugom primjeru iznad možemo alternativno napraviti oba kao 4 ili oba kao 6 po istoj cijeni.
Ideja je razmotriti svaku vrijednost u nizu kao potencijalnu ciljanu vrijednost, a zatim izračunati ukupnu cijenu pretvaranja svih ostalih elemenata u tu ciljanu vrijednost. Provjerom svih mogućih ciljanih vrijednosti možemo pronaći onu koja rezultira minimalnim ukupnim troškom konverzije.
// C++ program to Make all array // elements equal with minimum cost #include using namespace std ; // Function which finds the minimum // cost to make array elements equal int minCost ( vector < int > & arr ) { int n = arr . size (); int ans = INT_MAX ; // Try each element as the target value for ( int i = 0 ; i < n ; i ++ ) { int currentCost = 0 ; // Calculate cost of making all // elements equal to arr[i] for ( int j = 0 ; j < n ; j ++ ) { currentCost += abs ( arr [ j ] - arr [ i ]); } // Update minimum cost if current cost is lower ans = min ( ans currentCost ); } return ans ; } int main () { vector < int > arr = { 1 100 101 }; cout < < minCost ( arr ) < < endl ; return 0 ; }
Java // Java program to Make all array // elements equal with minimum cost import java.util.* ; class GfG { // Function which finds the minimum // cost to make array elements equal static int minCost ( int [] arr ) { int n = arr . length ; int ans = Integer . MAX_VALUE ; // Try each element as the target value for ( int i = 0 ; i < n ; i ++ ) { int currentCost = 0 ; // Calculate cost of making all // elements equal to arr[i] for ( int j = 0 ; j < n ; j ++ ) { currentCost += Math . abs ( arr [ j ] - arr [ i ] ); } // Update minimum cost if current cost is lower ans = Math . min ( ans currentCost ); } return ans ; } public static void main ( String [] args ) { int [] arr = { 1 100 101 }; System . out . println ( minCost ( arr )); } }
Python # Python program to Make all array # elements equal with minimum cost # Function which finds the minimum # cost to make array elements equal def minCost ( arr ): n = len ( arr ) ans = float ( 'inf' ) # Try each element as the target value for i in range ( n ): currentCost = 0 # Calculate cost of making all # elements equal to arr[i] for j in range ( n ): currentCost += abs ( arr [ j ] - arr [ i ]) # Update minimum cost if current cost is lower ans = min ( ans currentCost ) return ans if __name__ == '__main__' : arr = [ 1 100 101 ] print ( minCost ( arr ))
C# // C# program to Make all array // elements equal with minimum cost using System ; class GfG { // Function which finds the minimum // cost to make array elements equal static int minCost ( int [] arr ) { int n = arr . Length ; int ans = int . MaxValue ; // Try each element as the target value for ( int i = 0 ; i < n ; i ++ ) { int currentCost = 0 ; // Calculate cost of making all // elements equal to arr[i] for ( int j = 0 ; j < n ; j ++ ) { currentCost += Math . Abs ( arr [ j ] - arr [ i ]); } // Update minimum cost if current cost is lower ans = Math . Min ( ans currentCost ); } return ans ; } static void Main () { int [] arr = { 1 100 101 }; Console . WriteLine ( minCost ( arr )); } }
JavaScript // JavaScript program to Make all array // elements equal with minimum cost // Function which finds the minimum // cost to make array elements equal function minCost ( arr ) { let n = arr . length ; let ans = Number . MAX_SAFE_INTEGER ; // Try each element as the target value for ( let i = 0 ; i < n ; i ++ ) { let currentCost = 0 ; // Calculate cost of making all // elements equal to arr[i] for ( let j = 0 ; j < n ; j ++ ) { currentCost += Math . abs ( arr [ j ] - arr [ i ]); } // Update minimum cost if current cost is lower ans = Math . min ( ans currentCost ); } return ans ; } let arr = [ 1 100 101 ]; console . log ( minCost ( arr ));
Izlaz
100
[Očekivani pristup - 1] Korištenje binarnog pretraživanja - O(n Log (Raspon)) vrijeme i O(1) razmak
Ideja je koristiti binarno pretraživanje za učinkovito pronalaženje optimalne vrijednosti u koju bi se svi elementi niza trebali pretvoriti. Budući da funkcija ukupnog troška tvori konveksnu krivulju (prvo opadajuću, a zatim rastuću) u rasponu mogućih vrijednosti, možemo koristiti binarno pretraživanje da lociramo minimalnu točku ove krivulje uspoređujući trošak na srednjoj točki s troškom na srednjoj točki minus jedan koji nam govori u kojem smjeru dalje tražiti.
Pristup korak po korak:
- Pronađite minimalne i maksimalne vrijednosti u nizu kako biste uspostavili raspon pretraživanja
- Koristite binarno pretraživanje između minimalnih i maksimalnih vrijednosti kako biste pronašli optimalnu ciljnu vrijednost
- Za svaku probnu vrijednost izračunajte ukupnu cijenu pretvaranja svih elemenata niza u tu vrijednost
- Usporedite cijenu na trenutnoj srednjoj točki s cijenom na srednjoj točki minus jedan kako biste odredili smjer pretraživanja
- Nastavite sa sužavanjem raspona pretraživanja dok ne pronađete konfiguraciju minimalnog troška
// C++ program to Make all array // elements equal with minimum cost #include using namespace std ; // Function to find the cost of changing // array values to mid. int findCost ( vector < int > & arr int mid ) { int n = arr . size (); int ans = 0 ; for ( int i = 0 ; i < n ; i ++ ) { ans += abs ( arr [ i ] - mid ); } return ans ; } // Function which finds the minimum cost // to make array elements equal. int minCost ( vector < int > & arr ) { int n = arr . size (); int mini = INT_MAX maxi = INT_MIN ; // Find the minimum and maximum value. for ( int i = 0 ; i < n ; i ++ ) { mini = min ( mini arr [ i ]); maxi = max ( maxi arr [ i ]); } int s = mini e = maxi ; int ans = INT_MAX ; while ( s <= e ) { int mid = s + ( e - s ) / 2 ; int cost1 = findCost ( arr mid ); int cost2 = findCost ( arr mid -1 ); if ( cost1 < cost2 ) { ans = cost1 ; s = mid + 1 ; } else { e = mid - 1 ; } } return ans ; } int main () { vector < int > arr = { 1 100 101 }; cout < < minCost ( arr ); return 0 ; }
Java // Java program to Make all array // elements equal with minimum cost import java.util.* ; class GfG { // Function to find the cost of changing // array values to mid. static int findCost ( int [] arr int mid ) { int n = arr . length ; int ans = 0 ; for ( int i = 0 ; i < n ; i ++ ) { ans += Math . abs ( arr [ i ] - mid ); } return ans ; } // Function which finds the minimum cost // to make array elements equal. static int minCost ( int [] arr ) { int n = arr . length ; int mini = Integer . MAX_VALUE maxi = Integer . MIN_VALUE ; // Find the minimum and maximum value. for ( int i = 0 ; i < n ; i ++ ) { mini = Math . min ( mini arr [ i ] ); maxi = Math . max ( maxi arr [ i ] ); } int s = mini e = maxi ; int ans = Integer . MAX_VALUE ; while ( s <= e ) { int mid = s + ( e - s ) / 2 ; int cost1 = findCost ( arr mid ); int cost2 = findCost ( arr mid - 1 ); if ( cost1 < cost2 ) { ans = cost1 ; s = mid + 1 ; } else { e = mid - 1 ; } } return ans ; } public static void main ( String [] args ) { int [] arr = { 1 100 101 }; System . out . println ( minCost ( arr )); } }
Python # Python program to Make all array # elements equal with minimum cost # Function to find the cost of changing # array values to mid. def findCost ( arr mid ): n = len ( arr ) ans = 0 for i in range ( n ): ans += abs ( arr [ i ] - mid ) return ans # Function which finds the minimum cost # to make array elements equal. def minCost ( arr ): n = len ( arr ) mini = float ( 'inf' ) maxi = float ( '-inf' ) # Find the minimum and maximum value. for i in range ( n ): mini = min ( mini arr [ i ]) maxi = max ( maxi arr [ i ]) s = mini e = maxi ans = float ( 'inf' ) while s <= e : mid = s + ( e - s ) // 2 cost1 = findCost ( arr mid ) cost2 = findCost ( arr mid - 1 ) if cost1 < cost2 : ans = cost1 s = mid + 1 else : e = mid - 1 return ans if __name__ == '__main__' : arr = [ 1 100 101 ] print ( minCost ( arr ))
C# // C# program to Make all array // elements equal with minimum cost using System ; class GfG { // Function to find the cost of changing // array values to mid. static int findCost ( int [] arr int mid ) { int n = arr . Length ; int ans = 0 ; for ( int i = 0 ; i < n ; i ++ ) { ans += Math . Abs ( arr [ i ] - mid ); } return ans ; } // Function which finds the minimum cost // to make array elements equal. static int minCost ( int [] arr ) { int n = arr . Length ; int mini = int . MaxValue maxi = int . MinValue ; // Find the minimum and maximum value. for ( int i = 0 ; i < n ; i ++ ) { mini = Math . Min ( mini arr [ i ]); maxi = Math . Max ( maxi arr [ i ]); } int s = mini e = maxi ; int ans = int . MaxValue ; while ( s <= e ) { int mid = s + ( e - s ) / 2 ; int cost1 = findCost ( arr mid ); int cost2 = findCost ( arr mid - 1 ); if ( cost1 < cost2 ) { ans = cost1 ; s = mid + 1 ; } else { e = mid - 1 ; } } return ans ; } static void Main () { int [] arr = { 1 100 101 }; Console . WriteLine ( minCost ( arr )); } }
JavaScript // JavaScript program to Make all array // elements equal with minimum cost // Function to find the cost of changing // array values to mid. function findCost ( arr mid ) { let n = arr . length ; let ans = 0 ; for ( let i = 0 ; i < n ; i ++ ) { ans += Math . abs ( arr [ i ] - mid ); } return ans ; } // Function which finds the minimum cost // to make array elements equal. function minCost ( arr ) { let n = arr . length ; let mini = Number . MAX_SAFE_INTEGER maxi = Number . MIN_SAFE_INTEGER ; // Find the minimum and maximum value. for ( let i = 0 ; i < n ; i ++ ) { mini = Math . min ( mini arr [ i ]); maxi = Math . max ( maxi arr [ i ]); } let s = mini e = maxi ; let ans = Number . MAX_SAFE_INTEGER ; while ( s <= e ) { let mid = Math . floor ( s + ( e - s ) / 2 ); let cost1 = findCost ( arr mid ); let cost2 = findCost ( arr mid - 1 ); if ( cost1 < cost2 ) { ans = cost1 ; s = mid + 1 ; } else { e = mid - 1 ; } } return ans ; } let arr = [ 1 100 101 ]; console . log ( minCost ( arr ));
Izlaz
100
[Očekivani pristup - 2] Korištenje sortiranja - O(n Log n) vremena i O(1) prostora
Ideja je pronaći optimalnu vrijednost na koju treba izjednačiti sve elemente koji mora biti jedan od postojećih elemenata niza. Prvo sortiranjem niza, a zatim ponavljanjem kroz svaki element kao potencijalnu ciljanu vrijednost, izračunavamo trošak transformacije svih ostalih elemenata u tu vrijednost učinkovitim praćenjem zbroja elemenata lijevo i desno od trenutne pozicije.
Pristup korak po korak:
- Sortirajte polje za obradu elemenata uzlaznim redoslijedom.
- Za svaki element kao potencijalnu ciljanu vrijednost izračunajte dva troška: podizanje manjih elemenata i smanjenje većih elemenata.
- Pratite lijeve i desne zbrojeve kako biste učinkovito izračunali te troškove u konstantnom vremenu po iteraciji.
- Dovođenje manjih elemenata košta: (trenutna vrijednost × broj manjih elemenata) - (zbroj manjih elemenata)
- Troškovi smanjenja većih elemenata: (zbroj većih elemenata) - (trenutna vrijednost × broj većih elemenata)
- Usporedite trenutni trošak s minimalnim troškom.
// C++ program to Make all array // elements equal with minimum cost #include using namespace std ; // Function which finds the minimum cost // to make array elements equal. int minCost ( vector < int > & arr ) { int n = arr . size (); // Sort the array sort ( arr . begin () arr . end ()); // Variable to store sum of elements // to the right side. int right = 0 ; for ( int i = 0 ; i < n ; i ++ ) { right += arr [ i ]; } int ans = INT_MAX ; int left = 0 ; for ( int i = 0 ; i < n ; i ++ ) { // Remove the current element from right sum. right -= arr [ i ]; // Find cost of incrementing left side elements int leftCost = i * arr [ i ] - left ; // Find cost of decrementing right side elements. int rightCost = right - ( n -1 - i ) * arr [ i ]; ans = min ( ans leftCost + rightCost ); // Add current value to left sum left += arr [ i ]; } return ans ; } int main () { vector < int > arr = { 1 100 101 }; cout < < minCost ( arr ); return 0 ; }
Java // Java program to Make all array // elements equal with minimum cost import java.util.* ; class GfG { // Function which finds the minimum cost // to make array elements equal. static int minCost ( int [] arr ) { int n = arr . length ; // Sort the array Arrays . sort ( arr ); // Variable to store sum of elements // to the right side. int right = 0 ; for ( int i = 0 ; i < n ; i ++ ) { right += arr [ i ] ; } int ans = Integer . MAX_VALUE ; int left = 0 ; for ( int i = 0 ; i < n ; i ++ ) { // Remove the current element from right sum. right -= arr [ i ] ; // Find cost of incrementing left side elements int leftCost = i * arr [ i ] - left ; // Find cost of decrementing right side elements. int rightCost = right - ( n - 1 - i ) * arr [ i ] ; ans = Math . min ( ans leftCost + rightCost ); // Add current value to left sum left += arr [ i ] ; } return ans ; } public static void main ( String [] args ) { int [] arr = { 1 100 101 }; System . out . println ( minCost ( arr )); } }
Python # Python program to Make all array # elements equal with minimum cost # Function which finds the minimum cost # to make array elements equal. def minCost ( arr ): n = len ( arr ) # Sort the array arr . sort () # Variable to store sum of elements # to the right side. right = sum ( arr ) ans = float ( 'inf' ) left = 0 for i in range ( n ): # Remove the current element from right sum. right -= arr [ i ] # Find cost of incrementing left side elements leftCost = i * arr [ i ] - left # Find cost of decrementing right side elements. rightCost = right - ( n - 1 - i ) * arr [ i ] ans = min ( ans leftCost + rightCost ) # Add current value to left sum left += arr [ i ] return ans if __name__ == '__main__' : arr = [ 1 100 101 ] print ( minCost ( arr ))
C# // C# program to Make all array // elements equal with minimum cost using System ; class GfG { // Function which finds the minimum cost // to make array elements equal. static int minCost ( int [] arr ) { int n = arr . Length ; // Sort the array Array . Sort ( arr ); // Variable to store sum of elements // to the right side. int right = 0 ; for ( int i = 0 ; i < n ; i ++ ) { right += arr [ i ]; } int ans = int . MaxValue ; int left = 0 ; for ( int i = 0 ; i < n ; i ++ ) { // Remove the current element from right sum. right -= arr [ i ]; // Find cost of incrementing left side elements int leftCost = i * arr [ i ] - left ; // Find cost of decrementing right side elements. int rightCost = right - ( n - 1 - i ) * arr [ i ]; ans = Math . Min ( ans leftCost + rightCost ); // Add current value to left sum left += arr [ i ]; } return ans ; } static void Main () { int [] arr = { 1 100 101 }; Console . WriteLine ( minCost ( arr )); } }
JavaScript // JavaScript program to Make all array // elements equal with minimum cost // Function which finds the minimum cost // to make array elements equal. function minCost ( arr ) { let n = arr . length ; // Sort the array arr . sort (( a b ) => a - b ); // Variable to store sum of elements // to the right side. let right = 0 ; for ( let i = 0 ; i < n ; i ++ ) { right += arr [ i ]; } let ans = Number . MAX_SAFE_INTEGER ; let left = 0 ; for ( let i = 0 ; i < n ; i ++ ) { // Remove the current element from right sum. right -= arr [ i ]; // Find cost of incrementing left side elements let leftCost = i * arr [ i ] - left ; // Find cost of decrementing right side elements. let rightCost = right - ( n - 1 - i ) * arr [ i ]; ans = Math . min ( ans leftCost + rightCost ); // Add current value to left sum left += arr [ i ]; } return ans ; } let arr = [ 1 100 101 ]; console . log ( minCost ( arr ));
Izlaz
100Napravi kviz
