#practiceLinkDiv { display: none !important; }Dots masīvs, kas satur pozitīvus un negatīvus skaitļus. Masīvs attēlo kontrolpunktus no viena ielas gala līdz otram galam. Pozitīvās un negatīvās vērtības norāda enerģijas daudzumu šajā kontrolpunktā. Pozitīvie skaitļi palielina enerģiju, un negatīvie skaitļi samazinās. Atrodiet minimālo sākotnējo enerģiju, kas nepieciešama, lai šķērsotu ielu, lai enerģijas līmenis nekad nebūtu 0 vai mazāks par 0.
Piezīme: Minimālās nepieciešamās sākotnējās enerģijas vērtība būs 1, pat ja mēs veiksmīgi šķērsosim ielu, nezaudējot enerģiju līdz nullei un vienāda ar 0 nevienā kontrolpunktā. 1 ir nepieciešams sākotnējam pārbaudes punktam.
Piemēri:
Input : arr[] = {4 -10 4 4 4}Recommended Practice Minimālā enerģija Izmēģiniet to!
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
Brutāla spēka pieeja:
- Katram iespējamajam sākuma enerģijas līmenim (sākot no 1) simulējiet ielas šķērsošanu, izmantojot šo enerģijas līmeni, un pārbaudiet, vai enerģijas līmenis vienmēr ir pozitīvs.
- Atgrieziet minimālo sākotnējo enerģijas līmeni, kas nodrošina, ka enerģijas līmenis nekad nekļūst par nulli vai negatīvu.
Tālāk ir norādīts iepriekš minētās pieejas kods:
C++
#include
import 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 akshitaguprzj3
# 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 AGARWAL
using 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)); } } }
// Function to check if energy level never becomes negative or zero function check(arr n initEnergy) { let energy = initEnergy; for (let 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 function minInitialEnergy(arr n) { let minEnergy = 1; while (!check(arr n minEnergy)) { minEnergy++; } return minEnergy; } // Driver code let arr = [4 -10 4 4 4]; let n = arr.length; console.log(minInitialEnergy(arr n));
Izvade:
7
Laika sarežģītība: O(2^n)
Palīgtelpa: O(n)
Ņemam sākotnējo minimālo enerģiju 0 t.i.; initMinEnergy = 0 un enerģija jebkurā kontrolpunktā kā currEnergy = 0. Tagad šķērsojiet katru kontrolpunktu lineāri un pievienojiet enerģijas līmeni katrā i'tajā kontrolpunktā, t.i.; currEnergy = currEnergy + arr[i]. Ja currEnergy kļūst nepozitīvs, mums ir nepieciešama vismaz “abs(currEnergy) + 1” papildu sākotnējā enerģija, lai šķērsotu šo punktu. Tāpēc mēs atjauninām initMinEnergy = (initMinEnergy + abs(currEnergy) + 1). Mēs arī atjauninām currEnergy = 1, jo mums tagad ir nepieciešamā papildu minimālā sākotnējā enerģija nākamajam punktam.
Zemāk ir iepriekš minētās idejas īstenošana.
C++// C++ program to find minimum initial energy to // reach end #include
// 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.
# 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)
// 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.
<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>
// PHP 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. $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 <= o at any checkpoint $flag = 0; // Traverse each check // point linearly for ($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 Code $arr = array(4 -10 4 4 4); $n = sizeof($arr); echo minInitialEnergy($arr $n); // This code is contributed // by nitin mittal. ?> Izvade
7
Laika sarežģītība: O(n)
Palīgtelpa: O(1)