🧠 Check if Subarray with 0 Sum Exists

✅ Simple Statement

Given an array, check if there’s a contiguous subarray (1 or more elements) whose sum is zero.

🧩 Example

Input: [4, -6, 3, -1, 4]
Output: True
(4 + -6 + 3 + -1 = 0)

🧾 Pseudocode (Optimal – Using Prefix Sum Set)

initialize empty set
initialize prefix_sum = 0
for each element in array:
    add to prefix_sum
    if prefix_sum is 0 or already exists in set:
        return true
    add prefix_sum to set
return false

🧪 Stack Trace

Input: [4, -6, 3, -1, 4]

Index	Value	Prefix Sum	Set (Seen Sums)	Found?
0	4	4	{}	❌
1	-6	-2	{4}	❌
2	3	1	{4, -2}	❌
3	-1	0	{4, -2, 1}	✅ Zero!

🐍 Python Code

Brute-force (O(n²))

def has_zero_sum_subarray_brute(nums):
    for i in range(len(nums)):
        total = 0
        for j in range(i, len(nums)):
            total += nums[j]
            if total == 0:
                return True
    return False

Optimal (O(n))

def has_zero_sum_subarray_optimal(nums):
    seen = set()
    prefix_sum = 0
    for num in nums:
        prefix_sum += num
        if prefix_sum == 0 or prefix_sum in seen:
            return True
        seen.add(prefix_sum)
    return False

📜 JavaScript Code

Brute-force

function hasZeroSumSubarrayBrute(arr) {
  for (let i = 0; i < arr.length; i++) {
    let sum = 0;
    for (let j = i; j < arr.length; j++) {
      sum += arr[j];
      if (sum === 0) return true;
    }
  }
  return false;
}

Optimal

function hasZeroSumSubarrayOptimal(arr) {
  const seen = new Set();
  let prefixSum = 0;
  for (let i = 0; i < arr.length; i++) {
    prefixSum += arr[i];
    if (prefixSum === 0 || seen.has(prefixSum)) return true;
    seen.add(prefixSum);
  }
  return false;
}