Sum Zero Problem

Sum Zero Problem Statement

Given an array of integers, return any non-empty subarray whose elements sum up to zero.

Example One

{
"arr": [5, 1, 2, -3, 7, -4]
}

Output:

[1, 3]

Sum of [1, 2, -3] subarray is zero. It starts at index 1 and ends at index 3 of the given array, so [1, 3] is a correct answer. [3, 5] is another correct answer.

Example Two

{
"arr": [1, 2, 3, 5, -9]
}

Output:

[-1]

There is no non-empty subarray with sum zero.

Notes

• The output is an array of type [start_index, end_index] of a non-empty zero sum subarray; zero-based indices; both start_index and end_index are included in the subarray.
• If there are multiple such subarrays, you can return any.
• If no zero sum subarray is found, return [-1].

Constraints:

• 1 <= length of the input array <= 5 * 10⁵
• -10⁹ <= number in the array <= 10⁹

We provided two solutions.

We will refer to the length of the input array as n.

Sum Zero Solution 1: Brute Force

A naive approach would be to iterate over all possible subarrays of input array arr, such that while on subarray[i, j], i.e. subarray starting from i-th index and ending at j-th index, find sum of elements in it and if it’s zero, return [i, j]. If no such subarray is found, return [-1].

Time Complexity

O(n²).

As we are iterating over all possible subarrays of input array arr, time complexity will be O(n²).

Auxiliary Space Used

O(1).

We are not storing anything extra.

Space Complexity

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(1).

So, total space complexity: O(n) + O(1) + O(1) = O(n).

Code For Sum Zero Solution 1: Brute Force

    /*
    * Asymptotic complexity in terms of size of `arr` `n`:
    * Time: O(n^2).
    * Auxiliary space: O(1).
    * Total space: O(n).
    */

    static ArrayList<Integer> sum_zero(ArrayList<Integer> arr) {
        // To store current sum
        long sum = 0;
        // We are trying to find sum of each subarray[i, n - 1] where 0 <= i <= (n - 1).
        for (int i = 0; i < arr.size(); i++) {
            sum = 0;
            // To calculate sum of subarray starting from i and ending at n-1
            for (int j = i; j < arr.size(); j++) {
                sum += arr.get(j);
                // If sum == 0 means we found our subarray having sum as 0 with start index i
                // and end index j
                if (sum == 0) {
                    return new ArrayList<Integer>(Arrays.asList(i, j));
                }
            }
        }
        // If no subarray as sum equal to zero found then we will return [-1]
        return new ArrayList<Integer>(Arrays.asList(-1));
    }

Sum Zero Solution 2: Optimal

Notice that if there exists a zero sum subarray[i, j] in a given input array arr, then prefix sum (denote it as prefix where prefix[k] = arr[0] + arr[1] + arr[2] + ... + arr[k]) prefix[j] should be equal to prefix[i - 1], as prefix[j] = prefix[i - 1] + (arr[i] + arr[i + 1] + ... + arr[j]), where the term in bracket is sum of subarray[i, j], which is 0.

Considering this fact, build prefix sum array prefix. If for some i, j, 0 <= i <= j < n, prefix[i - 1] = prefix[j], then subarray[i, j] is the zero sum subarray.

Time Complexity

O(n).

To find out if any two sums of subarrays are equal or not we will store them in HashMap as prefix[k] (i.e. sum) as key and k as value. To maintain a hashmap it will take O(n) time complexity in the worst case to get and store n sums.

Auxiliary Space Used

O(n).

We are using hashmap to store sums. It will take O(n) of space.

Space Complexity

O(n).

Space used for input: O(n).

Auxiliary space used: O(n).

Space used for output: O(1).

So, total space complexity: O(n) + O(n) + O(1) = O(n).

Code For Sum Zero Solution 2: Optimal

    /*
    * Asymptotic complexity in terms of size of `arr` `n`:
    * Time: O(n).
    * Auxiliary space: O(n).
    * Total space: O(n).
    */

    static ArrayList<Integer> sum_zero(ArrayList<Integer> arr) {
        // To store prefix sum i.e. sum of subarray starting at index 0 and ending at index i
        // Key of hashmap will be sum and value will be index i for prefix sum
        HashMap<Long, Integer> map = new HashMap<>();
        // To check whether prefix sum it self is equal to zero
        map.put(0l, -1);
        // To store current sum
        long sum = 0;
        for (int i = 0; i < arr.size(); i++) {
            // To check if we encountered with value which itself is zero
            if(arr.get(i) == 0){
                return new ArrayList<Integer>(Arrays.asList(i, i));
            }
            // Adding current value in current sum
            sum += arr.get(i);
            // If we found value sum in our hashmap means we have encountered with this sum before
            // means arr[0, map.get(sum)] = sum and
            // arr[0, map.get(sum)] + arr[map.get(sum) + 1, i] = sum
            // which implies arr[map.get(sum) + 1, i] = 0 and hence interval we are looking for is
            // start = map.get(sum) + 1 and end = i
            if (map.containsKey(sum)) {
                return new ArrayList<Integer>(Arrays.asList(map.get(sum) + 1, i));
            } else {
                map.put(sum, i);
            }
        }
        // If no subarray having sum = 0 found then we will return [-1]
        return new ArrayList<Integer>(Arrays.asList(-1));
    }

We hope that these solutions to zero sum problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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