Directional Increase (Codeforces 1694C)
Tags: Greedy, greedy

We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.

Input: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$  — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$  — the size of array $$$a$$$.The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array.It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise.You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).

Note: In the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$

Sample Inputs:
7
2
1 0
4
2 -1 -1 0
4
1 -4 3 0
4
1 -1 1 -1
5
1 2 3 4 -10
7
2 -1 1 -2 0 0 0
1
0

Sample Outputs:
No
Yes
No
No
Yes
Yes
Yes

Solution (Python):
from itertools import permutations

def allPermutations(str):
    permList = permutations(str)
    return permList

def factors(x):
    result = []
    i = 1
    while i * i <= x:
        if x % i == 0:
            result.append(i)
            if x // i != i:
                result.append(x // i)
        i += 1
    return result

def fast_expo(val, power):
    result = pow(val, power // 2)
    result = result * result
    if power % 2 != 0:
        result = result * val
    return result

def is_prime(n):
    if n < 2:
        return False
    i = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += 1
    return True
t = int(input())
i = 0
while i < t:
    n = int(input())
    l = list(map(int, input().split()))
    flag = 1
    cnt = 0
    index = 0
    for j in range(n):
        cnt += l[j]
        if cnt < 0:
            flag = 0
            break
        elif cnt == 0:
            index = j
            break
    if cnt != 0:
        print('NO')
    else:
        for j in range(index + 1, n):
            if l[j] != 0:
                flag = 0
                break
        if flag == 0:
            print('NO')
        else:
            print('YES')
    i += 1

Solution (Python):
import math
import sys
sys.setrecursionlimit(2 * 10 ** 6)

def test():
    pass

def f(arr):
    if arr[0] < 0 or arr[-1] > 0:
        return False
    t = -arr[0]
    for a in arr[1:]:
        if a < t < 0:
            return False
        if t == 0 and a != 0:
            return False
        t -= a
    return t == 0

def main():
    N = int(input())
    for _ in range(N):
        n = int(input())
        arr = list(map(int, input().split()))
        if f(arr):
            print('YES')
        else:
            print('NO')

def init():
    import sys
    import os
    if 'CP_LOCAL_ENV' in os.environ:
        test()
        sys.stdin = open('test.txt')
if __name__ == '__main__':
    init()
    main()

Solution (Python):
I = input
for _ in [0] * int(I()):
    I()
    (p, z, zero) = (0, 1, 0)
    for v in I().split():
        p += int(v)
        if zero and p > 0:
            z = 0
            break
        if p == 0:
            zero = True
        if p < 0:
            z = 0
            break
    print(['NO', 'YES'][zero and z])