Zero-Sum Sub-Arrays

October 13, 2017

There is an obvious O(n2) algorithm, but this problem is similar to the well-known maximum sum subsequence problem, and admits a similar linear solution:

(define (count xs)
  (let ((counts (make-eq-hashtable))
        (sum 0) (result 0))
    (do ((xs xs (cdr xs))) ((null? xs) result)
      (set! sum (+ sum (car xs)))
      (when (hashtable-contains? counts sum)
        (set! result (+ result (hashtable-ref counts sum 0))))
      (hashtable-update! counts sum add1 1))))

Our solution keeps a running sum as it scans the array, counting the number of times the sum has been seen previously in the counts hashtable. Each time a sum is the same as one previously found, we have found a zero-sum sub-array. Here’s an example:

> (count '(-1 1 -1 1))
4

You can run the program at https://ideone.com/hZ0Dag.

Pages: 1 2

Posted by programmingpraxis
Filed in Exercises
6 Comments »

6 Responses to “Zero-Sum Sub-Arrays”

  1. nobody said
    October 13, 2017 at 7:02 PM

    Please consider the input ‘(-1 1).

  2. programmingpraxis said
    October 13, 2017 at 7:05 PM

    Bah! Humbug.

  3. V said
    October 14, 2017 at 3:54 AM

    Solution in Ruby

    
def zero_sum_sub_arrays(xs)
  return 0 if xs.size < 2

  (2..xs.size)
    .flat_map { |n| xs.each_cons(n).to_a }
    .select { |xs| xs.sum == 0 }
    .size
end

# Test

5.times do |n|
  [1, -1].repeated_permutation(n).map do |input|
    puts "#{input} -> #{zero_sum_sub_arrays(input)}"
  end
end

    produces:

    
[] -> 0
[1] -> 0
[-1] -> 0
[1, 1] -> 0
[1, -1] -> 1
[-1, 1] -> 1
[-1, -1] -> 0
[1, 1, 1] -> 0
[1, 1, -1] -> 1
[1, -1, 1] -> 2
[1, -1, -1] -> 1
[-1, 1, 1] -> 1
[-1, 1, -1] -> 2
[-1, -1, 1] -> 1
[-1, -1, -1] -> 0
[1, 1, 1, 1] -> 0
[1, 1, 1, -1] -> 1
[1, 1, -1, 1] -> 2
[1, 1, -1, -1] -> 2
[1, -1, 1, 1] -> 2
[1, -1, 1, -1] -> 4
[1, -1, -1, 1] -> 3
[1, -1, -1, -1] -> 1
[-1, 1, 1, 1] -> 1
[-1, 1, 1, -1] -> 3
[-1, 1, -1, 1] -> 4
[-1, 1, -1, -1] -> 2
[-1, -1, 1, 1] -> 2
[-1, -1, 1, -1] -> 2
[-1, -1, -1, 1] -> 1
[-1, -1, -1, -1] -> 0
  4. Paul said
    October 15, 2017 at 3:36 PM

    In Python.

    def number_of_pairs(n): return n * (n - 1) // 2

def zerosum(arr):
    C = Counter(accumulate([0]+arr))
    return sum(number_of_pairs(C[c]) for c in C)
  5. Daniel said
    October 17, 2017 at 10:18 PM

    Here’s a solution in C++11.

    #include <stdio.h>
#include <unordered_map>
#include <vector>

int zero_sum_count(std::vector<int>& vec) {
    int count = 0;
    int running_sum = 0;
    std::unordered_map<int, int> lookup;
    lookup[0] = 1;
    for (int value : vec) {
        running_sum += value;
        auto search = lookup.find(running_sum);
        if (search == lookup.end()) {
            lookup[running_sum] = 1;
        } else {
            count += search->second;
            lookup[running_sum] += 1;
        }
    }
    return count;
}

int main(int argc, char* argv[]) {
    std::vector<int> vec = {-1, 1, -1, 1};
    int count = zero_sum_count(vec);
    printf("count: %d\n", count);
}

    Output:

    $ g++ -std=c++11 -o zero_sum_count zero_sum_count.cpp
$ ./zero_sum_count

    count: 4
  6. Sarthak Gupta said
    October 21, 2017 at 7:51 AM

    Here’s a solution in Python3-

    c = 0
    l = [1, -1, 1, -1, -1, 1]
    for i in range(len(l)):
    for j in range(i+1,len(l)):
    l1 = l[i:j+1]
    if sum(l1) == 0:
    print(l1)
    c += 1
    print(‘Total number of sub-arrays-‘, c)

    Here’s the output-

    [1, -1]
    [1, -1, 1, -1]
    [1, -1, 1, -1, -1, 1]
    [-1, 1]
    [1, -1]
    [1, -1, -1, 1]
    [-1, 1]
    Total number of sub-arrays- 7

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