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Sherlock and Array

  • emivulpe + 2 comments

    Python solution. Comments would be appreciated

    # Enter your code here. Read input from STDIN. Print output to STDOUT
def exists_element(size, arr):
    sum_left = 0
    sum_right = sum(arr)
    previous = 0
    for idx in range(0, size):
        current = arr[idx]
        sum_left += previous
        sum_right -= current
        if sum_left == sum_right:
            return "YES"
        previous = current
    return "NO"

test_cases = int(raw_input())
for i in range(0, test_cases):
    size = int(raw_input())
    arr = map(int, raw_input().split())
    print exists_element(size, arr)
    • liyuekui + 1 comment

      No need to sum(arr) at all, that would cause another uneccesary O(n), here's my code:

      def solve(a, size):
    # Complete this function
    difference = 0
    left_index = 0
    compare_count = size - 1
    for i in xrange(size+1):
        if difference > 0:
            difference -= a[compare_count + left_index - i]
        else:
            difference += a[left_index]
            left_index += 1
    return 'YES' if difference == 0 else 'NO'
      • reis_vitore + 1 comment

        Failed Testcase 1!

        • liyuekui + 0 comments

          Thanks for feedback, code was updated and it passed all cases.

    • rjack2 + 0 comments

      Very nice. This helped me get through my timeouts.