#practiceLinkDiv { 표시: 없음 !중요; }양수와 음수를 포함하는 배열이 제공됩니다. 배열은 거리의 한쪽 끝에서 다른 쪽 끝까지의 검문소를 나타냅니다. 양수 및 음수 값은 해당 체크포인트의 에너지 양을 나타냅니다. 양수는 에너지를 증가시키고 음수는 감소합니다. 에너지 수준이 0 이하가 되지 않도록 길을 건너는 데 필요한 최소 초기 에너지를 구합니다.
메모 : 필요한 최소 초기 에너지 값은 어떤 검문소에서도 0 이하의 에너지를 잃지 않고 성공적으로 길을 건너더라도 1이 됩니다. 초기 체크포인트에는 1이 필요합니다.
예:
Input : arr[] = {4 -10 4 4 4}Recommended Practice 최소 에너지 시도해 보세요!
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
무차별 접근 방식 :
- 가능한 모든 초기 에너지 수준(1부터 시작)에 대해 해당 에너지 수준을 사용하여 도로 횡단을 시뮬레이션하고 에너지 수준이 항상 양수로 유지되는지 확인합니다.
- 에너지 수준이 0이나 음수가 되지 않도록 보장하는 최소 초기 에너지 수준을 반환합니다.
다음은 위의 접근 방식에 대한 코드입니다.
C++
#include using 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; }
Java 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
Python3 # 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
C# 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)); } } }
JavaScript // 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));
출력 :
7
시간 복잡도: O(2^n)
보조 공간 : 에)
우리는 초기 최소 에너지를 0으로 취합니다. initMinEnergy = 0이고 모든 체크포인트의 에너지는 currEnergy = 0입니다. 이제 각 체크포인트를 선형으로 순회하고 각 i번째 체크포인트에 에너지 레벨을 추가합니다. currEnergy = currEnergy + arr[i]. currEnergy가 양수가 아닌 경우 이 지점을 통과하려면 최소한 'abs(currEnergy) + 1'의 추가 초기 에너지가 필요합니다. 따라서 initMinEnergy = (initMinEnergy + abs(currEnergy) + 1)을 업데이트합니다. 또한 다음 지점에 필요한 추가 최소 초기 에너지가 있으므로 currEnergy = 1을 업데이트합니다.
아래는 위의 아이디어를 구현한 것입니다.
C++// C++ program to find minimum initial energy to // reach end #include using 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; }
Java // 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.
Python3 # 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# // 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.
JavaScript <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 // 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. ?> 산출
7
시간복잡도 : O(n)
보조공간 : O(1)