Using a map to find subarray with given sum (with negative numbers)

Question

Consider the following problem statement:

Given an unsorted array of integers, find a subarray which adds to a given number. If there are more than one subarrays with sum as the given number, print any of them.

Examples:

Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33
Ouptut: Sum found between indexes 2 and 4

Input: arr[] = {10, 2, -2, -20, 10}, sum = -10
Ouptut: Sum found between indexes 0 to 3

Input: arr[] = {-10, 0, 2, -2, -20, 10}, sum = 20
Ouptut: No subarray with given sum exists

On this site, the following linear time solution was suggested involving the use of a map to store the sums of the current subsets as the algorithm iterates through the array:

// Function to print subarray with sum as given sum
void subArraySum(int arr[], int n, int sum)
{
    // create an empty map
    unordered_map<int, int> map;

    // Maintains sum of elements so far
    int curr_sum = 0;

    for (int i = 0; i < n; i++)
    {
        // add current element to curr_sum
        curr_sum = curr_sum + arr[i];

        // if curr_sum is equal to target sum
        // we found a subarray starting from index 0
        // and ending at index i
        if (curr_sum == sum)
        {
            cout << "Sum found between indexes "
                 << 0 << " to " << i << endl;
            return;
        }

        // If curr_sum - sum already exists in map
        // we have found a subarray with target sum
        if (map.find(curr_sum - sum) != map.end())
        {
            cout << "Sum found between indexes "
                 << map[curr_sum - sum] + 1
                 << " to " << i << endl;
            return;
        }

        map[curr_sum] = i;
    }

    // If we reach here, then no subarray exists
    cout << "No subarray with given sum exists";
}
// Driver program to test above function
int main()
{
    int arr[] = {10, 2, -2, -20, 10};
    int n = sizeof(arr)/sizeof(arr[0]);
    int sum = -10;

    subArraySum(arr, n, sum);

    return 0;
}

With the test case that is provided in the post, the second if statement checking whether current_sum - sum is never entered. Instead, the sum is found and is printed in the first if statement. So there a few points of confusion here:

1: What is the purpose of looking up current_sum-sum in the map?

2: In which cases would the second if statement even be entered to put the map to use in solving the problem?

kandhan kandhan · Accepted Answer · 2016-09-04T22:40:30

    curr_sum = curr_sum + arr[i];

Here:

curr_sum holds the sum of all the entries(up to arr[i]) starting from arr[0]

if (curr_sum == sum)
{
    cout << "Sum found between indexes "
         << 0 << " to " << i << endl;
    return;
}

Here:

The if condition checks if the curr_sum is equal to the given sum. Remember, curr_sum holds the sum starting from arr[0].

So, if the indexes for your result subarray is 2 and 4, the above condition is not enough.

Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33
Ouptut: Sum found between indexes 2 and 4

In that example,

when i = 4, curr_sum would be 38. But if you take 1 and 4 out(arr[0] and arr[1] which is a subarray), you get 33.

    if (map.find(curr_sum - sum) != map.end())
    {
        cout << "Sum found between indexes "
             << map[curr_sum - sum] + 1
             << " to " << i << endl;
        return;
    }

    map[curr_sum] = i;

The above code does that.

 map[curr_sum] = i;

map stores the index value (i) with the subarray sum ( sum of the arr between indexes 0 and i ) as key. So for the example, it would be:

map[1]=0
map[5]=1
map[25]=2
map[28]=3
map[38]=4

This code:

    map.find(curr_sum - sum)

searches the map is if it has an entry with key (curr_sum - sum).

In our example, when i = 4, curr_sum would be 38 and sum as given would be 33. If we eliminate the subarray arr[0,2], we get 33. So, map.find(curr_sum - sum) => map.find(38-33) is a hit since we have entry map[5] = 1.

Hope this clears you

Using a map to find subarray with given sum (with negative numbers)

3 Answers

-5, 2, 10, 20, 3, 8