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Gibt alle Subarrays mit der Summe 0 aus

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Gegeben ein Array arr[] der Größe N Die Aufgabe besteht darin, alle Unterarrays im Array mit der Summe auszugeben .

Beispiele:  



Eingang: arr = [6 3 -1 -3 4 -2 2 4 6 -12 -7]
Ausgabe:

Subarray von Index 2 bis 4 gefunden
Subarray von Index 2 bis 6 gefunden
Subarray von Index 5 bis 6 gefunden
Subarray von Index 6 bis 9 gefunden
Subarray von Index 0 bis 10 gefunden

Eingang: arr = [1 2 -3 3 -1 -1]
Ausgabe:



Subarray von Index 0 bis 2 gefunden
Subarray von Index 2 bis 3 gefunden
Subarray von Index 3 bis 5 gefunden

[Naiver Ansatz] Durch Generieren aller möglichen Subarrays - O(n2) Zeit und O(1) Hilfsraum

Der ganz grundlegende Ansatz besteht darin, zu überlegen alle möglichen Subarrays und prüft, ob ihre Summe Null ist. Dieser Ansatz ist zwar einfach, aber auch für große Arrays ineffizient.

C++
// C++ program to print all subarrays // in the array which has sum 0 #include    using namespace std; vector<pair<int int> > findSubArrays(int arr[] int n) {  // Array to store all the start and end  // indices of subarrays with 0 sum  vector<pair<int int> > output;  for (int i = 0; i < n; i++) {  int prefix = 0;  for (int j = i; j < n; j++) {  prefix += arr[j];  if (prefix == 0)  output.push_back({ i j });  }  }  return output; } // Function to print all subarrays with 0 sum void print(vector<pair<int int> > output) {  for (auto it = output.begin(); it != output.end(); it++)  cout << 'Subarray found from Index ' << it->first  << ' to ' << it->second << endl; } // Driver code int main() {  // Given array  int arr[] = { 6 3 -1 -3 4 -2 2 4 6 -12 -7 };  int n = sizeof(arr) / sizeof(arr[0]);  // Function Call  vector<pair<int int> > output = findSubArrays(arr n);  // if we didn’t find any subarray with 0 sum  // then subarray doesn’t exists  if (output.size() == 0) {  cout << 'No subarray exists';  }  else {  print(output);  }  return 0; } 
Java
// Java program to print all subarrays // in the array which has sum 0 import java.io.*; import java.util.*; // User defined pair class class Pair {  int first second;  Pair(int a int b)  {  first = a;  second = b;  } } public class GFG {  static ArrayList<Pair> findSubArrays(int[] arr int n)  {  // Array to store all the start and end  // indices of subarrays with 0 sum  ArrayList<Pair> out = new ArrayList<>();  for (int i = 0; i < n; i++) {  int prefix = 0;  for (int j = i; j < n; j++) {  prefix += arr[j];  if (prefix == 0)  out.add(new Pair(i j));  }  }  return out;  }  // Function to print all subarrays with 0 sum  static void print(ArrayList<Pair> out)  {  for (int i = 0; i < out.size(); i++) {  Pair p = out.get(i);  System.out.println('Subarray found from Index '  + p.first + ' to '  + p.second);  }  }  // Driver code  public static void main(String args[])  {  // Given array  int[] arr  = { 6 3 -1 -3 4 -2 2 4 6 -12 -7 };  int n = arr.length;  // Function Call  ArrayList<Pair> out = findSubArrays(arr n);  // if we didn’t find any subarray with 0 sum  // then subarray doesn’t exists  if (out.size() == 0)  System.out.println('No subarray exists');  else  print(out);  } } 
Python
# User defined pair class class Pair: first = 0 second = 0 def __init__(self a b): self.first = a self.second = b class GFG: @staticmethod def findSubArrays(arr n): # Array to store all the start and end # indices of subarrays with 0 sum out = [] i = 0 while (i < n): prefix = 0 j = i while (j < n): prefix += arr[j] if (prefix == 0): out.append(Pair(i j)) j += 1 i += 1 return out # Function to print all subarrays with 0 sum @staticmethod def print(out): i = 0 while (i < len(out)): p = out[i] print('Subarray found from Index ' + str(p.first) + ' to ' + str(p.second)) i += 1 # Driver code @staticmethod def main(args): # Given array arr = [6 3 -1 -3 4 -2 2 4 6 -12 -7] n = len(arr) # Function Call out = GFG.findSubArrays(arr n) # if we didn't find any subarray with 0 sum # then subarray doesn't exists if (len(out) == 0): print('No subarray exists') else: GFG.print(out) if __name__ == '__main__': GFG.main([]) 
C#
using System; using System.Collections.Generic; class GFG {    // Array to store all the start and end  // indices of subarrays with 0 sum  static List<Tuple<int int>> findSubArrays(int[] arr int n)  {  var output = new List<Tuple<int int>>();  for (int i = 0; i < n; i++)  {  int prefix = 0;  for (int j = i; j < n; j++)  {  prefix += arr[j];  if (prefix == 0)  output.Add(Tuple.Create(i j));  }  }  return output;  }  // Function to print all subarrays with 0 sum  static void print(List<Tuple<int int>> output)  {  foreach (var subArray in output)  Console.Write('Subarray found from Index ' + subArray.Item1 + ' to ' + subArray.Item2+'n');  }  // Driver code  public static void Main()  {  // Given array  int[] arr = { 6 3 -1 -3 4 -2 2 4 6 -12 -7 };  int n = arr.Length;  // Function Call  List<Tuple<int int>> output = findSubArrays(arr n);  // if we didn’t find any subarray with 0 sum  // then subarray doesn’t exists  if (output.Count == 0)  {  Console.WriteLine('No subarray exists');  }  else  {  print(output);  }  } } 
JavaScript
// Javascript program to print all subarrays // in the array which has sum 0 function findSubArrays(arr n) {  // Array to store all the start and end  // indices of subarrays with 0 sum  let out =[];  for (let i = 0; i < n; i++) {  let prefix = 0;  for (let j = i; j < n; j++) {  prefix += arr[j];  if (prefix == 0)  out.push([i j]);  }  }  return out; } // Function to print all subarrays with 0 sum function print(out) {  for (let it of out)  console.log('Subarray found from Index ' + it[0]  + ' to ' + it[1]); } // Driver code // Given array let arr = [ 6 3 -1 -3 4 -2 2 4 6 -12 -7 ]; let n = arr.length ; // Function Call let out = findSubArrays(arr n); // if we didn’t find any subarray with 0 sum // then subarray doesn’t exists if (out.length == 0) {  console.log('No subarray exists'); } else {  print(out); }   

Ausgabe
Subarray found from Index 0 to 10 Subarray found from Index 2 to 4 Subarray found from Index 2 to 6 Subarray found from Index 5 to 6 Subarray found from Index 6 to 9 

Zeitkomplexität: AN2), da wir 2 Schleifen verwenden.
Hilfsraum: O(1) da konstanter zusätzlicher Platz erforderlich ist.



[Erwarteter Ansatz] Verwenden Hashing - O(n) Zeit und O(n) Hilfsraum

Ein effizienterer Ansatz besteht darin, Hashing zu verwenden, um die kumulative Summe der Elemente und ihrer Indizes zu speichern. Dies ermöglicht die Überprüfung, ob in konstanter Zeit ein Subarray mit der Nullsumme existiert.

Nachfolgend finden Sie detaillierte Schritte der Intuition:

  1. Erstellen Sie eine Hash-Map, um die kumulative Summe und die entsprechenden Indizes zu speichern.
  2. Initialisieren Sie die kumulative Summe auf Null.
  3. Durchlaufen Sie das Array:
    • Addieren Sie das aktuelle Element zur kumulativen Summe.
    • Wenn die kumulative Summe Null ist, wird ein Subarray vom Anfang bis zum aktuellen Index gefunden.
    • Wenn die kumulative Summe bereits in der Hash-Map vorhanden ist, bedeutet dies, dass es ein Subarray mit der Nullsumme gibt.
    • Speichern Sie die kumulative Summe und den Index in der Hash-Map.
C++
// C++ program to print all subarrays // in the array which has sum 0 #include    using namespace std; // Function to print all subarrays in the array which // has sum 0 vector<pair<int int> > findSubArrays(int arr[] int n) {  // create an empty map  unordered_map<int vector<int> > map;  // create an empty vector of pairs to store  // subarray starting and ending index  vector<pair<int int> > out;  // Maintains sum of elements so far  int sum = 0;  for (int i = 0; i < n; i++) {  // add current element to sum  sum += arr[i];  // if sum is 0 we found a subarray starting  // from index 0 and ending at index i  if (sum == 0)  out.push_back(make_pair(0 i));  // If sum already exists in the map there exists  // at-least one subarray ending at index i with  // 0 sum  if (map.find(sum) != map.end()) {  // map[sum] stores starting index of all  // subarrays  vector<int> vc = map[sum];  for (auto it = vc.begin(); it != vc.end(); it++)  out.push_back(make_pair(*it + 1 i));  }  // Important - no else  map[sum].push_back(i);  }  // return output vector  return out; } // Utility function to print all subarrays with sum 0 void print(vector<pair<int int> > out) {  for (auto it = out.begin(); it != out.end(); it++)  cout << 'Subarray found from Index ' << it->first  << ' to ' << it->second << endl; } // Driver code int main() {  int arr[] = { 6 3 -1 -3 4 -2 2 4 6 -12 -7 };  int n = sizeof(arr) / sizeof(arr[0]);  vector<pair<int int> > out = findSubArrays(arr n);  // if we didn’t find any subarray with 0 sum  // then subarray doesn’t exists  if (out.size() == 0)  cout << 'No subarray exists';  else  print(out);  return 0; } 
Java
// Java program to print all subarrays // in the array which has sum 0 import java.io.*; import java.util.*; // User defined pair class class Pair {  int first second;  Pair(int a int b)  {  first = a;  second = b;  } } public class GFG {  // Function to print all subarrays in the array which  // has sum 0  static ArrayList<Pair> findSubArrays(int[] arr int n)  {  // create an empty map  HashMap<Integer ArrayList<Integer> > map  = new HashMap<>();  // create an empty vector of pairs to store  // subarray starting and ending index  ArrayList<Pair> out = new ArrayList<>();  // Maintains sum of elements so far  int sum = 0;  for (int i = 0; i < n; i++) {  // add current element to sum  sum += arr[i];  // if sum is 0 we found a subarray starting  // from index 0 and ending at index i  if (sum == 0)  out.add(new Pair(0 i));  ArrayList<Integer> al = new ArrayList<>();  // If sum already exists in the map there exists  // at-least one subarray ending at index i with  // 0 sum  if (map.containsKey(sum)) {  // map[sum] stores starting index of all  // subarrays  al = map.get(sum);  for (int it = 0; it < al.size(); it++) {  out.add(new Pair(al.get(it) + 1 i));  }  }  al.add(i);  map.put(sum al);  }  return out;  }  // Utility function to print all subarrays with sum 0  static void print(ArrayList<Pair> out)  {  for (int i = 0; i < out.size(); i++) {  Pair p = out.get(i);  System.out.println('Subarray found from Index '  + p.first + ' to '  + p.second);  }  }  // Driver code  public static void main(String args[])  {  int[] arr  = { 6 3 -1 -3 4 -2 2 4 6 -12 -7 };  int n = arr.length;  ArrayList<Pair> out = findSubArrays(arr n);  // if we did not find any subarray with 0 sum  // then subarray does not exists  if (out.size() == 0)  System.out.println('No subarray exists');  else  print(out);  } } 
Python
# Python3 program to print all subarrays # in the array which has sum 0 # Function to get all subarrays # in the array which has sum 0 def findSubArrays(arr n): # create a python dict hashMap = {} # create a python list # equivalent to ArrayList out = [] # tracker for sum of elements sum1 = 0 for i in range(n): # increment sum by element of array sum1 += arr[i] # if sum is 0 we found a subarray starting # from index 0 and ending at index i if sum1 == 0: out.append((0 i)) al = [] # If sum already exists in the map # there exists at-least one subarray # ending at index i with 0 sum if sum1 in hashMap: # map[sum] stores starting index # of all subarrays al = hashMap.get(sum1) for it in range(len(al)): out.append((al[it] + 1 i)) al.append(i) hashMap[sum1] = al return out # Utility function to print # all subarrays with sum 0 def printOutput(output): for i in output: print('Subarray found from Index ' + str(i[0]) + ' to ' + str(i[1])) # Driver Code if __name__ == '__main__': arr = [6 3 -1 -3 4 -2 2 4 6 -12 -7] n = len(arr) out = findSubArrays(arr n) # if we did not find any subarray with 0 sum # then subarray does not exists if (len(out) == 0): print('No subarray exists') else: printOutput(out) 
C#
// C# program to print all subarrays // in the array which has sum 0 using System; using System.Collections.Generic; // User defined pair class class Pair {  public int first second;  public Pair(int a int b)  {  first = a;  second = b;  } } class GFG {  // Function to print all subarrays  // in the array which has sum 0  static List<Pair> findSubArrays(int[] arr int n)  {  // create an empty map  Dictionary<int List<int> > map  = new Dictionary<int List<int> >();  // create an empty vector of pairs to store  // subarray starting and ending index  List<Pair> outt = new List<Pair>();  // Maintains sum of elements so far  int sum = 0;  for (int i = 0; i < n; i++) {  // add current element to sum  sum += arr[i];  // if sum is 0 we found a subarray starting  // from index 0 and ending at index i  if (sum == 0)  outt.Add(new Pair(0 i));  List<int> al = new List<int>();  // If sum already exists in the map there exists  // at-least one subarray ending at index i with  // 0 sum  if (map.ContainsKey(sum)) {  // map[sum] stores starting index  // of all subarrays  al = map[sum];  for (int it = 0; it < al.Count; it++) {  outt.Add(new Pair(al[it] + 1 i));  }  }  al.Add(i);  if (map.ContainsKey(sum))  map[sum] = al;  else  map.Add(sum al);  }  return outt;  }  // Utility function to print all subarrays with sum 0  static void print(List<Pair> outt)  {  for (int i = 0; i < outt.Count; i++) {  Pair p = outt[i];  Console.WriteLine('Subarray found from Index '  + p.first + ' to '  + p.second);  }  }  // Driver code  public static void Main(String[] args)  {  int[] arr  = { 6 3 -1 -3 4 -2 2 4 6 -12 -7 };  int n = arr.Length;  List<Pair> outt = findSubArrays(arr n);  // if we did not find any subarray with 0 sum  // then subarray does not exists  if (outt.Count == 0)  Console.WriteLine('No subarray exists');  else  print(outt);  } } 
JavaScript
// JavaScript program to print all subarrays // in the array which has sum 0 // Function to print all subarrays in the array which // has sum 0 function findSubArrays(arr n) {  // create an empty map  let map = {};    // create an empty vector of pairs to store  // subarray starting and ending index  let out = [];    // Maintains sum of elements so far  let sum = 0;    for (var i = 0; i < n; i++)  {  // add current element to sum  sum += arr[i];    // if sum is 0 we found a subarray starting  // from index 0 and ending at index i  if (sum == 0)  out.push([0 i]);    // If sum already exists in the map there exists  // at-least one subarray ending at index i with  // 0 sum  if (map.hasOwnProperty(sum))  {  // map[sum] stores starting index of all subarrays  let vc = map[sum];  for (let it of vc)  out.push([it + 1 i]);  }  else  map[sum] = [];    // Important - no else  map[sum].push(i);  }    // return output vector  return out; }   // Utility function to print all subarrays with sum 0 function print(out) {  for (let it of out)  console.log('Subarray found from Index ' + it[0] + ' to ' + it[1]); }     // Driver code let arr = [6 3 -1 -3 4 -2 2 4 6 -12 -7]; let n = arr.length;   let out = findSubArrays(arr n);   // if we didn’t find any subarray with 0 sum // then subarray doesn’t exists if (out.length == 0)  console.log('No subarray exists'); else  print(out); 

Ausgabe
Subarray found from Index 2 to 4 Subarray found from Index 2 to 6 Subarray found from Index 5 to 6 Subarray found from Index 6 to 9 Subarray found from Index 0 to 10 

Zeitkomplexität: O(n) wobei n die Anzahl der Elemente im Array ist.
Hilfsraum: O(n) zum Speichern der Hash-Map.