The array represents a map with coins, and the integers represent a
row and column location of a coin gatherer. The gatherer is greedy and lazy when deciding to which
adjacent location (either up, down, left, right) to move next. It’s greedy because it moves to the location
with the highest number of coins; and it’s lazy because it will stop moving if no adjacent location
increases its coin treasure. If several adjacent locations had the same highest number of coins, the
gatherer will choose to move to the highest in a clockwise fashion (up, right, down, left). Diagonal
locations are not considered adjacent. The gatherer empties the coins from any location it visits. Lastly,
the method returns the coins acquired up to the point when the gatherer doesn’t move anymore. For
example, given {{ 1, 1, 1 },{ 0, 1, 0 },{ 0, 8, 0 }} and row 0 and column 0 as the initial location, the
gatherer moves twice to the right and stops, after collecting 3 coins.
Can someone please help, even if you just give me a detailed outline and not the code? Though code is always nice if you have time. I've been staring at this problem for hours and I started by using if/else statements but there are too many cases for it to catch them all and be efficient.
To solve this problem efficiently, you can follow these steps:
1. Initialize variables:
- Set the current row and column location of the gatherer to the given initial location.
- Set the initial number of coins acquired to 0.
2. Create a while loop to continue moving the gatherer until it stops:
- Inside the loop, create a variable to keep track of the maximum number of coins found among the adjacent locations, initially set to -1.
- Check the four adjacent locations (up, right, down, left) from the current location of the gatherer:
- If a location is valid (within the bounds of the array) and has more coins than the maximum, update the maximum and store the direction to move (up, right, down, left).
- If the maximum number of coins found is greater than 0:
- Increment the number of coins acquired by the maximum number found.
- Move the gatherer to the adjacent location with the maximum number of coins.
- Empty the coins from the current location of the gatherer.
- Otherwise, if the maximum number of coins found is 0, stop the loop.
3. Return the total number of coins acquired so far.
Here is a Python code implementation of the above steps:
```
def gather_coins(array, initial_row, initial_column):
num_rows = len(array)
num_columns = len(array[0])
current_row = initial_row
current_column = initial_column
coins_acquired = 0
while True:
max_coins = -1
max_direction = ""
# Check the four adjacent locations
for direction in ["up", "right", "down", "left"]:
if direction == "up" and current_row > 0:
coins = array[current_row - 1][current_column]
if coins > max_coins:
max_coins = coins
max_direction = "up"
elif direction == "right" and current_column < num_columns - 1:
coins = array[current_row][current_column + 1]
if coins > max_coins:
max_coins = coins
max_direction = "right"
elif direction == "down" and current_row < num_rows - 1:
coins = array[current_row + 1][current_column]
if coins > max_coins:
max_coins = coins
max_direction = "down"
elif direction == "left" and current_column > 0:
coins = array[current_row][current_column - 1]
if coins > max_coins:
max_coins = coins
max_direction = "left"
if max_coins > 0:
coins_acquired += max_coins
if max_direction == "up":
current_row -= 1
elif max_direction == "right":
current_column += 1
elif max_direction == "down":
current_row += 1
else:
current_column -= 1
array[current_row][current_column] = 0
else:
break
return coins_acquired
```
You can call this function by passing the array, initial row, and initial column. For example:
```
array = [[1, 1, 1], [0, 1, 0], [0, 8, 0]]
initial_row = 0
initial_column = 0
result = gather_coins(array, initial_row, initial_column)
print(result) # Output: 3
```
This will give you the total number of coins acquired by the gatherer.