Given an array of positive and negative numbers, find if there is a subarray (of size at-least one) with 0 sum.
Examples :
Input: {4, 2, -3, 1, 6}
Output: true
There is a subarray with zero sum from index 1 to 3.
Input: {4, 2, 0, 1, 6}
Output: true
There is a subarray with zero sum from index 2 to 2.
Input: {-3, 2, 3, 1, 6}
Output: false
There is no subarray with zero sum.
A simple solution is to consider all subarrays one by one and check the sum of every subarray. We can run two loops: the outer loop picks a starting point i and the inner loop tries all subarrays starting from i (See this for implementation). Time complexity of this method is O(n2).
We can also use hashing. The idea is to iterate through the array and for every element arr[i], calculate sum of elements form 0 to i (this can simply be done as sum += arr[i]). If the current sum has been seen before, then there is a zero sum array. Hashing is used to store the sum values, so that we can quickly store sum and find out whether the current sum is seen before or not.
Example :
arr[] = {1, 4, -2, -2, 5, -4, 3}
If we consider all prefix sums, we can
notice that there is a subarray with 0
sum when :
1) Either a prefix sum repeats or
2) Or prefix sum becomes 0.
Prefix sums for above array are:
1, 5, 3, 1, 6, 2, 5
Since prefix sum 1 repeats, we have a subarray
with 0 sum.
Following is implementation of the above approach.
C++
// A C++ program to find if there is a zero sum // subarray #include <bits/stdc++.h> using namespace std; bool subArrayExists(int arr[], int n) { unordered_set<int> sumSet; // Traverse through array and store prefix sums int sum = 0; for (int i = 0 ; i < n ; i++) { sum += arr[i]; // If prefix sum is 0 or it is already present if (sum == 0 || sumSet.find(sum) != sumSet.end()) return true; sumSet.insert(sum); } return false; } // Driver code int main() { int arr[] = {-3, 2, 3, 1, 6}; int n = sizeof(arr)/sizeof(arr[0]); if (subArrayExists(arr, n)) cout << "Found a subarray with 0 sum"; else cout << "No Such Sub Array Exists!"; return 0; } |
Java
// A Java program to find if there is a zero sum subarray import java.util.HashMap; class ZeroSumSubarray { // Returns true if arr[] has a subarray with sero sum static Boolean subArrayExists(int arr[]) { // Creates an empty hashMap hM HashMap<Integer, Integer> hM = new HashMap<Integer, Integer>(); // Initialize sum of elements int sum = 0; // Traverse through the given array for (int i = 0; i < arr.length; i++) { // Add current element to sum sum += arr[i]; // Return true in following cases // a) Current element is 0 // b) sum of elements from 0 to i is 0 // c) sum is already present in hash map if (arr[i] == 0 || sum == 0 || hM.get(sum) != null) return true; // Add sum to hash map hM.put(sum, i); } // We reach here only when there is // no subarray with 0 sum return false; } // driver code public static void main(String arg[]) { int arr[] = {-3, 2, 3, 1, 6}; if (subArrayExists(arr)) System.out.println("Found a subarray with 0 sum"); else System.out.println("No Such Sub Array Exists!"); } } |
Python3
# A python program to find if # there is a zero sum subarray def subArrayExists(arr,n): s = set() # raverse through array # and store prefix sums sum = 0 for i in range(n): sum += arr[i] # If prefix sum is 0 or # it is already present if sum == 0 or sum in s: return True s.add(sum) return False # Driver code arr = [-3, 2, 3, 1, 6] n = len(arr) if subArrayExists(arr, n) == True: print("Found a sunbarray with 0 sum") else: print("No Such sub array exits!") # This code is contributed by Shrikant13 |
C#
// A C# program to find if there // is a zero sum subarray using System; using System.Collections.Generic; class GFG { // Returns true if arr[] has // a subarray with sero sum static Boolean subArrayExists(int []arr) { // Creates an empty hashMap hM Dictionary<int, int> hM = new Dictionary<int, int>(); // Initialize sum of elements int sum = 0; // Traverse through the given array for (int i = 0; i < arr.Length; i++) { // Add current element to sum sum += arr[i]; // Return true in following cases // a) Current element is 0 // b) sum of elements from 0 to i is 0 // c) sum is already present in hash map if (arr[i] == 0 || sum == 0 ) return true; // Add sum to hash map hM[i] = sum; } // We reach here only when there is // no subarray with 0 sum return false; } // Driver Code public static void Main() { int []arr = {-3, 2, 3, 1, 6}; if (subArrayExists(arr)) Console.WriteLine("Found a subarray " + "with 0 sum"); else Console.WriteLine("No Such Sub " + "Array Exists!"); } } // This code is contributed by Sam007 |
Output :
No Such Sub Array Exists!
Time Complexity of this solution can be considered as O(n) under the assumption that we have good hashing function that allows insertion and retrieval operations in O(1) time.
Exercise:
Extend the above program to print starting and ending indexes of all subarrays with 0 sum.
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