**Problem**

An array contains both positive and negative elements, find the largest subarray whose
sum equals 0.

int[] input = {4, 6, 3, -9, -5, 1, 3, 0, 2}

int output = {4, 6, 3, -9, -5, 1} of length 6

This is simple. Will write later (incomplete)

Given an

Each element of tmp will store the sum of the input up to that element.

Now if you check tmp, you'll notice that there might be values that are equal to each other.For example, take the element 4 at index 0 and 3, element 3 at index 6 and 7. So, it means sum between these 2 indices has remained the same, i.e. all the elements between them add upto 0. So, based on that we get {6, 3, -9} and {0}.

Also, we know tmp[-1] = 0. When we have not started the array we have no element added to it. So, if we find a zero inside the tmp array, that means all the numbers starting 0th index to 5th index(where 0 exists in temp) are all 0s, so our subarray becomes {4,6,3, -9,-5,1}.

Out of {6, 3, -9}, {0} and {4,6,3, -9,-5,1}, last one is our answer as it is the largest sub array.

We notice that some values are same in tmp array. Let's say that this values are at indexes

**Example**int[] input = {4, 6, 3, -9, -5, 1, 3, 0, 2}

int output = {4, 6, 3, -9, -5, 1} of length 6

### Solution

**Method 1 - Brute force**This is simple. Will write later (incomplete)

**Method 2 - Storing the sum upto ith element in temp array**Given an

`int[] input`

array, you can create an `int[] tmp`

array where ` `

tmp[i] = tmp[i - 1] + input[i];

Each element of tmp will store the sum of the input up to that element.

__Example__int[] input = {4 | 6| 3| -9| -5| 1| 3| 0| 2} int[] tmp = {4 | 10|13| 4| -1| 0| 3| 3| 5}

Now if you check tmp, you'll notice that there might be values that are equal to each other.For example, take the element 4 at index 0 and 3, element 3 at index 6 and 7. So, it means sum between these 2 indices has remained the same, i.e. all the elements between them add upto 0. So, based on that we get {6, 3, -9} and {0}.

Also, we know tmp[-1] = 0. When we have not started the array we have no element added to it. So, if we find a zero inside the tmp array, that means all the numbers starting 0th index to 5th index(where 0 exists in temp) are all 0s, so our subarray becomes {4,6,3, -9,-5,1}.

Out of {6, 3, -9}, {0} and {4,6,3, -9,-5,1}, last one is our answer as it is the largest sub array.

__To sum it up__We notice that some values are same in tmp array. Let's say that this values are at indexes

`j an k with j < k`

, then the sum of the input till `j`

is equal to the sum till `k`

and this means that the sum of the portion of the array between `j`

and `k`

is 0! Specifically the 0 sum subarray will be from index j + 1 to k.
- NOTE: if
`j + 1 == k`

, then`k is 0`

and that's it! ;) - NOTE: The algorithm should consider a virtual
`tmp[-1] = 0`

; - NOTE: An empty array has sum 0 and it's minimal and this special case should be brought up as well in an interview. Then the interviewer will say that doesn't count but that's another problem! ;)

Here is the code

public static string[] SubArraySumList(int[] array, int sum) { int tempsum; List<string> list = new List<string>(); for (int i = 0; i < array.Length; i++) { tempsum = 0; for (int j = i; j < array.Length; j++) { tempsum += array[j]; if (tempsum == sum) { list.Add(String.Format("[{0}-{1}]", i, j)); } } } return list.ToArray(); }

Here is the solution using Hashmap, iterate over it again to get the max subarray:

public static void subArraySumsZero() { int [] seed = new int[] {1,2,3,4,-9,6,7,-8,1,9}; int currSum = 0; HashMap<Integer, Integer> sumMap = new HashMap<Integer, Integer>(); for(int i = 0 ; i < seed.length ; i ++){ currSum += seed[i]; if(currSum == 0){ System.out.println("subset : { 0 - " + i + " }"); }else if(sumMap.get(currSum) != null){ System.out.println("subset : { " + (sumMap.get(currSum) + 1) + " - " + i + " }"); sumMap.put(currSum, i); }else sumMap.put(currSum, i); } System.out.println("HASH MAP HAS: " + sumMap); }

**References**

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