#practiceLinkDiv { 表示: なし !重要; }正の数値と負の数値を含む配列を指定します。配列は、通りの一方の端からもう一方の端までのチェックポイントを表します。正と負の値は、そのチェックポイントでのエネルギー量を表します。正の数はエネルギーを増加させ、負の数は減少します。エネルギー レベルが 0 または 0 未満にならないように、道路を横断するために必要な最小の初期エネルギーを見つけます。
注記 : たとえどのチェックポイントでもエネルギーを 0 以下に失わずに通りを横切ることができたとしても、必要な最小初期エネルギーの値は 1 になります。最初のチェックポイントには 1 が必要です。
例:
Input : arr[] = {4 -10 4 4 4}Recommended Practice 最小エネルギー 試してみてください!
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
ブルートフォースアプローチ:
- 考えられるすべての初期エネルギー レベル (1 から開始) について、そのエネルギー レベルを使用して道路の横断をシミュレートし、エネルギー レベルが常に正のままであるかどうかを確認します。
- エネルギー レベルがゼロまたは負にならないようにする最小の初期エネルギー レベルを返します。
以下は上記のアプローチのコードです。
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));
出力:
7
時間計算量 : O(2^n)
補助スペース: の上)
初期の最小エネルギー 0 を採用します。 initMinEnergy = 0、および任意のチェックポイントのエネルギーを currEnergy = 0 とします。次に、各チェックポイントを線形に走査し、各 i 番目のチェックポイントでエネルギー レベルを追加します。 currEnergy = currEnergy + arr[i]。 currEnergy が正でなくなった場合、この点を超えるには少なくとも 'abs(currEnergy) + 1' の追加の初期エネルギーが必要になります。したがって、initMinEnergy = (initMinEnergy + abs(currEnergy) + 1) を更新します。次のポイントに必要な追加の最小初期エネルギーが得られたため、currEnergy = 1 も更新します。
以下は上記のアイデアを実装したものです。
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. ?> 出力
7
時間計算量 : O(n)
補助スペース : O(1)