Flower Growth Simulation — Problem Explanation

We are given an array of length $$$n$$$ that represents whether Petya watered his flower each day.
Each element $$$a_i$$$ is either:

• $$$1$$$ → the flower was watered
• $$$0$$$ → the flower was not watered

The flower begins at a height of $$$1$$$ cm on day 0 (before processing any input days).

Simulation Logic

We iterate through the array starting from index $$$1$$$ (the second day), and at each step, we examine the current element $$$a_i$$$ and the previous element $$$a_{i-1}$$$.
There are four possible scenarios:

• Case 1: $$$a_i = 1$$$ and $$$a_{i-1} = 1$$$
  → The flower was watered both yesterday and today
  → It grows by $$$5$$$ cm
  → height += 5

• Case 2: $$$a_i = 1$$$ and $$$a_{i-1} = 0$$$
  → The flower was watered today only
  → It grows by $$$1$$$ cm
  → height += 1

• Case 3: $$$a_i = 0$$$ and $$$a_{i-1} = 1$$$
  → The flower was not watered today, but was yesterday
  → It doesn't grow, but remains alive
  → height remains the same

• Case 4: $$$a_i = 0$$$ and $$$a_{i-1} = 0$$$
  → The flower was not watered for two consecutive days
  → The flower dies
  → return -1

Initialization

• The simulation starts with: height = 1

We loop through the array once and apply the above logic. If the flower survives all $$$n$$$ days, we return the final ⁣. If it dies due to two unwatered days in a row, we return -1.

Time Complexity

• This approach has a time complexity of $$$O(n)$$$, since we loop through the array once.

t = int(input())
for _ in range(t):
    n  = int(input())
    arr = list(map(int, input().split()))
    
    height = 1
    if arr[0] == 1:
        height += 1
    for i in range(1, n):
        if arr[i] == 1 and arr[i - 1] == 1:
            height += 5
        elif arr[i] == 1 and arr[i - 1] == 0:
            height += 1
        elif arr[i] == 0 and arr[i - 1] == 0:
            height =  -1  
            break
    print(height)