Minimale Anfangsenergie, die zum Überqueren der Straße erforderlich ist
Probieren Sie es bei GfG Practice aus 
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#practiceLinkDiv { display: none !important; } Gegeben sei ein Array mit positiven und negativen Zahlen. Das Array stellt Kontrollpunkte von einem Ende bis zum anderen Ende der Straße dar. Positive und negative Werte stellen die Energiemenge an diesem Kontrollpunkt dar. Positive Zahlen erhöhen die Energie und negative Zahlen verringern sie. Finden Sie die minimale Anfangsenergie, die zum Überqueren der Straße erforderlich ist, sodass das Energieniveau niemals 0 oder weniger als 0 wird.
Notiz : Der Wert der minimal erforderlichen Anfangsenergie beträgt 1, selbst wenn wir die Straße erfolgreich überqueren, ohne an einem Kontrollpunkt Energie auf weniger als oder gleich 0 zu verlieren. Die 1 ist für den ersten Kontrollpunkt erforderlich.
Beispiele:
Input : arr[] = {4 -10 4 4 4}
Output: 7
Suppose initially we have energy = 0 now at 1st
checkpoint we get 4. At 2nd checkpoint energy gets
reduced by -10 so we have 4 + (-10) = -6 but at any
checkpoint value of energy can not less than equals
to 0. So initial energy must be at least 7 because
having 7 as initial energy value at 1st checkpoint
our energy will be = 7+4 = 11 and then we can cross
2nd checkpoint successfully. Now after 2nd checkpoint
all checkpoint have positive value so we can cross
street successfully with 7 initial energy.
Input : arr[] = {3 5 2 6 1}
Output: 1
We need at least 1 initial energy to reach first
checkpoint
Input : arr[] = {-1 -5 -9}
Output: 16
Recommended Practice Minimale Energie Probieren Sie es aus!Brute-Force-Ansatz:
- Simulieren Sie für jedes mögliche anfängliche Energieniveau (beginnend mit 1) das Überqueren der Straße mit diesem Energieniveau und prüfen Sie, ob das Energieniveau jederzeit positiv bleibt.
- Gibt das minimale anfängliche Energieniveau zurück, das sicherstellt, dass das Energieniveau niemals Null oder negativ wird.
Unten ist der Code für den obigen Ansatz:
C++Java#includeusing namespace std ; // Function to check if energy level never becomes negative or zero bool check ( int arr [] int n int initEnergy ) { int energy = initEnergy ; for ( int i = 0 ; i < n ; i ++ ) { energy += arr [ i ]; if ( energy <= 0 ) { return false ; } } return true ; } // Function to calculate minimum initial energy // arr[] stores energy at each checkpoints on street int minInitialEnergy ( int arr [] int n ) { int minEnergy = 1 ; while ( ! check ( arr n minEnergy )) { minEnergy ++ ; } return minEnergy ; } // Driver code int main () { int arr [] = { 4 -10 4 4 4 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < minInitialEnergy ( arr n ); return 0 ; } Python3import java.util.* ; public class GFG { // Function to check if energy level never becomes // negative or zero static boolean check ( int [] arr int n int initEnergy ) { int energy = initEnergy ; for ( int i = 0 ; i < n ; i ++ ) { energy += arr [ i ] ; if ( energy <= 0 ) { return false ; } } return true ; } // Function to calculate minimum initial energy // arr[] stores energy at each checkpoints on the street static int minInitialEnergy ( int [] arr int n ) { int minEnergy = 1 ; while ( ! check ( arr n minEnergy )) { minEnergy ++ ; } return minEnergy ; } // Driver code public static void main ( String [] args ) { int [] arr = { 4 - 10 4 4 4 }; int n = arr . length ; System . out . println ( minInitialEnergy ( arr n )); } } // This code is contributed by akshitaguprzj3C## Function to check if energy level never becomes negative or zero def check ( arr n initEnergy ): energy = initEnergy for i in range ( n ): energy += arr [ i ] if energy <= 0 : return False return True # Function to calculate minimum initial energy # arr stores energy at each checkpoints on street def minInitialEnergy ( arr n ): minEnergy = 1 while not check ( arr n minEnergy ): minEnergy += 1 return minEnergy # Driver code arr = [ 4 - 10 4 4 4 ] n = len ( arr ) print ( minInitialEnergy ( arr n )) # THIS CODE IS CONTRIBUTED BY CHANDAN AGARWALJavaScriptusing System ; namespace EnergyCheck { class GFG { // Function to check if energy level never becomes negative or zero static bool Check ( int [] arr int n int initEnergy ) { int energy = initEnergy ; for ( int i = 0 ; i < n ; i ++ ) { energy += arr [ i ]; if ( energy <= 0 ) { return false ; } } return true ; } // Function to calculate minimum initial energy // arr[] stores energy at each checkpoints on street static int MinInitialEnergy ( int [] arr int n ) { int minEnergy = 1 ; while ( ! Check ( arr n minEnergy )) { minEnergy ++ ; } return minEnergy ; } // Driver code static void Main ( string [] args ) { int [] arr = { 4 - 10 4 4 4 }; int n = arr . Length ; Console . WriteLine ( MinInitialEnergy ( arr n )); } } }Ausgabe :
7
Zeitkomplexität: O(2^n)
Hilfsraum: An)
Wir nehmen die anfängliche Mindestenergie 0 an, d. h. initMinEnergy = 0 und Energie an jedem Kontrollpunkt als currEnergy = 0. Durchqueren Sie nun jeden Kontrollpunkt linear und fügen Sie das Energieniveau an jedem i-ten Kontrollpunkt hinzu, d. h. currEnergy = currEnergy + arr[i]. Wenn currEnergy nicht positiv wird, benötigen wir mindestens 'abs(currEnergy) + 1' zusätzliche Anfangsenergie, um diesen Punkt zu überschreiten. Daher aktualisieren wir initMinEnergy = (initMinEnergy + abs(currEnergy) + 1). Wir aktualisieren auch currEnergy = 1, da wir jetzt über die erforderliche zusätzliche minimale Anfangsenergie für den nächsten Punkt verfügen.
Nachfolgend finden Sie die Umsetzung der obigen Idee.
C++Java// C++ program to find minimum initial energy to // reach end #includeusing namespace std ; // Function to calculate minimum initial energy // arr[] stores energy at each checkpoints on street int minInitialEnergy ( int arr [] int n ) { // initMinEnergy is variable to store minimum initial // energy required. int initMinEnergy = 0 ; // currEnergy is variable to store current value of // energy at i'th checkpoint on street int currEnergy = 0 ; // flag to check if we have successfully crossed the // street without any energy loss <= o at any checkpoint bool flag = 0 ; // Traverse each check point linearly for ( int i = 0 ; i < n ; i ++ ) { currEnergy += arr [ i ]; // If current energy becomes negative or 0 increment // initial minimum energy by the negative value plus 1. // to keep current energy positive (at least 1). Also // update current energy and flag. if ( currEnergy <= 0 ) { initMinEnergy += abs ( currEnergy ) + 1 ; currEnergy = 1 ; flag = 1 ; } } // If energy never became negative or 0 then // return 1. Else return computed initMinEnergy return ( flag == 0 ) ? 1 : initMinEnergy ; } // Driver Program to test the case int main () { int arr [] = { 4 -10 4 4 4 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < minInitialEnergy ( arr n ); return 0 ; } Python3// Java program to find minimum // initial energy to reach end class GFG { // Function to calculate minimum // initial energy arr[] stores energy // at each checkpoints on street static int minInitialEnergy ( int arr [] int n ) { // initMinEnergy is variable to store // minimum initial energy required. int initMinEnergy = 0 ; // currEnergy is variable to store // current value of energy at // i'th checkpoint on street int currEnergy = 0 ; // flag to check if we have successfully // crossed the street without any energy // loss <= o at any checkpoint boolean flag = false ; // Traverse each check point linearly for ( int i = 0 ; i < n ; i ++ ) { currEnergy += arr [ i ] ; // If current energy becomes negative or 0 // increment initial minimum energy by the negative // value plus 1. to keep current energy // positive (at least 1). Also // update current energy and flag. if ( currEnergy <= 0 ) { initMinEnergy += Math . abs ( currEnergy ) + 1 ; currEnergy = 1 ; flag = true ; } } // If energy never became negative or 0 then // return 1. Else return computed initMinEnergy return ( flag == false ) ? 1 : initMinEnergy ; } // Driver code public static void main ( String [] args ) { int arr [] = { 4 - 10 4 4 4 }; int n = arr . length ; System . out . print ( minInitialEnergy ( arr n )); } } // This code is contributed by Anant Agarwal.C## Python program to find minimum initial energy to # reach end # Function to calculate minimum initial energy # arr[] stores energy at each checkpoints on street def minInitialEnergy ( arr ): n = len ( arr ) # initMinEnergy is variable to store minimum initial # energy required initMinEnergy = 0 ; # currEnergy is variable to store current value of # energy at i'th checkpoint on street currEnergy = 0 # flag to check if we have successfully crossed the # street without any energy loss <= 0 at any checkpoint flag = 0 # Traverse each check point linearly for i in range ( n ): currEnergy += arr [ i ] # If current energy becomes negative or 0 increment # initial minimum energy by the negative value plus 1. # to keep current energy positive (at least 1). Also # update current energy and flag. if currEnergy <= 0 : initMinEnergy += ( abs ( currEnergy ) + 1 ) currEnergy = 1 flag = 1 # If energy never became negative or 0 then # return 1. Else return computed initMinEnergy return 1 if flag == 0 else initMinEnergy # Driver program to test above function arr = [ 4 - 10 4 4 4 ] print ( minInitialEnergy ( arr )) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)JavaScript// C# program to find minimum // C# program to find minimum // initial energy to reach end using System ; class GFG { // Function to calculate minimum // initial energy arr[] stores energy // at each checkpoints on street static int minInitialEnergy ( int [] arr int n ) { // initMinEnergy is variable to store // minimum initial energy required. int initMinEnergy = 0 ; // currEnergy is variable to store // current value of energy at // i'th checkpoint on street int currEnergy = 0 ; // flag to check if we have successfully // crossed the street without any energy // loss <= o at any checkpoint bool flag = false ; // Traverse each check point linearly for ( int i = 0 ; i < n ; i ++ ) { currEnergy += arr [ i ]; // If current energy becomes negative or 0 // negativeincrement initial minimum energy // by the value plus 1. to keep current // energy positive (at least 1). Also // update current energy and flag. if ( currEnergy <= 0 ) { initMinEnergy += Math . Abs ( currEnergy ) + 1 ; currEnergy = 1 ; flag = true ; } } // If energy never became negative // or 0 then return 1. Else return // computed initMinEnergy return ( flag == false ) ? 1 : initMinEnergy ; } // Driver code public static void Main () { int [] arr = { 4 - 10 4 4 4 }; int n = arr . Length ; Console . Write ( minInitialEnergy ( arr n )); } } // This code is contributed by Nitin Mittal.PHP< script > // Javascript program to find minimum // initial energy to reach end // Function to calculate minimum // initial energy arr[] stores // energy at each checkpoints on street function minInitialEnergy ( arr n ) { // initMinEnergy is variable // to store minimum initial // energy required. let initMinEnergy = 0 ; // currEnergy is variable to // store current value of energy // at i'th checkpoint on street let currEnergy = 0 ; // flag to check if we have // successfully crossed the // street without any energy // loss <= o at any checkpoint let flag = 0 ; // Traverse each check // point linearly for ( let i = 0 ; i < n ; i ++ ) { currEnergy += arr [ i ]; // If current energy becomes // negative or 0 increment // initial minimum energy by // the negative value plus 1. // to keep current energy // positive (at least 1). Also // update current energy and flag. if ( currEnergy <= 0 ) { initMinEnergy += Math . abs ( currEnergy ) + 1 ; currEnergy = 1 ; flag = 1 ; } } // If energy never became // negative or 0 then // return 1. Else return // computed initMinEnergy return ( flag == 0 ) ? 1 : initMinEnergy ; } // Driver Code let arr = new Array ( 4 - 10 4 4 4 ); let n = arr . length ; document . write ( minInitialEnergy ( arr n )); // This code is contributed // by Saurabh Jaiswal < /script>
Ausgabe7Zeitkomplexität: O(n)
Hilfsraum: O(1)
