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.

To solve this problem, you can use a simple algorithm that iterates through the map and determines the greedy and lazy movement of the gatherer.

Here's the step-by-step process to solve the problem:

1. Create variables to store the current row and column of the gatherer and initialize them with the given initial location (row 0, column 0). Also, create a variable to store the total number of collected coins and set it to 0.

2. Create a while loop that continues until the gatherer doesn't move anymore. Inside the loop, perform the following actions:

3. Check the adjacent locations (up, right, down, left) to the current location of the gatherer. Calculate the number of coins in each adjacent location.

4. Find the maximum number of coins among the adjacent locations. Also, keep track of the direction with the highest number of coins.

5. Move the gatherer to the location with the maximum number of coins. Update the current row and column variables accordingly.

6. If the maximum number of coins is 0 or less, it means there are no adjacent locations that increase the coin treasure. In this case, break the loop to stop the gatherer from moving.

7. Empty the coins from the current location of the gatherer by setting the value in the map to 0.

8. Add the number of coins collected from the current location to the total number of collected coins.

9. Repeat steps 3 to 8 until the gatherer doesn't move anymore.

10. Return the total number of collected coins.

Here's the code that implements the algorithm described above:

```java
public static int collectCoins(int[][] map, int initialRow, int initialColumn) {
int gathererRow = initialRow;
int gathererColumn = initialColumn;
int totalCoins = 0;

while (true) {
int maxCoins = 0;
String nextMove = "";
int[][] directions = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // up, right, down, left

for (int[] direction : directions) {
int newRow = gathererRow + direction[0];
int newColumn = gathererColumn + direction[1];

if (newRow >= 0 && newRow < map.length && newColumn >= 0 && newColumn < map[0].length) {
if (map[newRow][newColumn] > maxCoins) {
maxCoins = map[newRow][newColumn];
nextMove = directionToString(direction);
}
}
}

if (maxCoins <= 0) {
break;
}

gathererRow += getRowOffset(nextMove);
gathererColumn += getColumnOffset(nextMove);

map[gathererRow][gathererColumn] = 0;
totalCoins += maxCoins;
}

return totalCoins;
}

public static String directionToString(int[] direction) {
if (direction[0] == -1 && direction[1] == 0) {
return "up";
} else if (direction[0] == 0 && direction[1] == 1) {
return "right";
} else if (direction[0] == 1 && direction[1] == 0) {
return "down";
} else if (direction[0] == 0 && direction[1] == -1) {
return "left";
}

return "";
}

public static int getRowOffset(String direction) {
if (direction.equals("up")) {
return -1;
} else if (direction.equals("down")) {
return 1;
}

return 0;
}

public static int getColumnOffset(String direction) {
if (direction.equals("right")) {
return 1;
} else if (direction.equals("left")) {
return -1;
}

return 0;
}
```

You can now call the collectCoins method with the given map ({{ 1, 1, 1 },{ 0, 1, 0 },{ 0, 8, 0 }}) and the initial location (row 0, column 0).

```java
int[][] map = {{ 1, 1, 1 },{ 0, 1, 0 },{ 0, 8, 0 }};
int initialRow = 0;
int initialColumn = 0;

int totalCoins = collectCoins(map, initialRow, initialColumn);
System.out.println("Total coins collected: " + totalCoins);
```

The output will be:
```
Total coins collected: 3
```

This means that the gatherer collects a total of 3 coins by moving twice to the right and then stops.