Learn and practice the Sliding Window, and Union Find patterns to solve coding interview problems.

Speedrun the Oracle Coding Interview: Top Problems in Go

Each pattern in this module exemplifies the power of dissecting a problem in detail and figuring out the least expensive operations required to solve it. In the Sliding Window pattern, we’ll gain practice in various techniques to ensure that a solution that seems to require nested loops actually executes in linear time. Using the Union Find pattern, we’ll learn to solve a diverse set of problems using the disjoint-set data structure.

Elegant by Design

# Statement

Let’s consider a scenario with an $(m \times n)$ 2D grid containing binary numbers, where `'0'` represents water and `'1'` represents land. If any `'1'` cells are connected to each other horizontally or vertically (not diagonally), they form an island. Your task is to return the total number of islands in the grid.

**Constraints:**

* $1 \leq$ `grid.length` $\leq 50$

* $1 \leq$ `grid[i].length` $\leq 50$

* `grid[i][j]` is either `'0'` or `'1'`.
# Examples

# Understand the problem

Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:

What is the output if the following grid is given as input?

$$ grid =
\begin{matrix}
[[1, 0, 0], \\
[0, 0, 0], \\
[0, 0, 1]]
\end{matrix}
$$

What is the output if the following grid is given as input?
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>g</mi><mi>r</mi><mi>i</mi><mi>d</mi><mo>=</mo><mtable rowspacing="0.16em" columnalign="center" columnspacing="1em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>0</mn><mo separator="true">,</mo><mn>0</mn><mo stretchy="false">]</mo><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mn>0</mn><mo separator="true">,</mo><mn>0</mn><mo separator="true">,</mo><mn>0</mn><mo stretchy="false">]</mo><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mn>0</mn><mo separator="true">,</mo><mn>0</mn><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">]</mo></mrow></mstyle></mtd></mtr></mtable></mrow><annotation encoding="application/x-tex"> grid =
\begin{matrix}
[[1, 0, 0], \\
[0, 0, 0], \\
[0, 0, 1]]
\end{matrix}
</annotation></semantics></math>grid=[[1,0,0],[0,0,0],[0,0,1]]​

What is the output if the following grid is given as input?

$$ grid =
\begin{matrix}
[[1, 1, 1], \\
[1, 1, 1], \\
[1, 1, 1]]
\end{matrix}
$$

What is the output if the following grid is given as input?
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>g</mi><mi>r</mi><mi>i</mi><mi>d</mi><mo>=</mo><mtable rowspacing="0.16em" columnalign="center" columnspacing="1em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">]</mo></mrow></mstyle></mtd></mtr></mtable></mrow><annotation encoding="application/x-tex"> grid =
\begin{matrix}
[[1, 1, 1], \\
[1, 1, 1], \\
[1, 1, 1]]
\end{matrix}
</annotation></semantics></math>grid=[[1,1,1],[1,1,1],[1,1,1]]​

What is the output if the following grid is given as input?

$$ grid =
\begin{matrix}
[[1, 0, 0], \\
[1, 1, 1], \\
[1, 0, 1]]
\end{matrix}
$$

What is the output if the following grid is given as input?
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>g</mi><mi>r</mi><mi>i</mi><mi>d</mi><mo>=</mo><mtable rowspacing="0.16em" columnalign="center" columnspacing="1em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>0</mn><mo separator="true">,</mo><mn>0</mn><mo stretchy="false">]</mo><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo stretchy="false">[</mo><mn>1</mn><mo separator="true">,</mo><mn>0</mn><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">]</mo></mrow></mstyle></mtd></mtr></mtable></mrow><annotation encoding="application/x-tex"> grid =
\begin{matrix}
[[1, 0, 0], \\
[1, 1, 1], \\
[1, 0, 1]]
\end{matrix}
</annotation></semantics></math>grid=[[1,0,0],[1,1,1],[1,0,1]]​

Number of Islands

Number of Islands

# Figure it out!

We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.



# Try it yourself

Implement your solution in `main.go` in the following coding playground. You will need the provided supporting code to implement your solution.

package main



func driverCode(ds TwoDArray) int {
    return numIslands(ds.parsedData)
}

package main

type UnionFind struct {
	parent []int
	rank []int
	count int
}

// Initializing the parent list and count variable by traversing the grid
func NewUnionFind(grid [][]byte) *UnionFind {
	uf := &UnionFind{}
	m := len(grid)
	n := len(grid[0])

	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == '1' {
				uf.parent = append(uf.parent, i*n+j)
				uf.count++
			} else {
				uf.parent = append(uf.parent, -1)
			}
			uf.rank = append(uf.rank, 1)
		}
	}

	return uf
}

// Function to find the root parent of a node
func (uf *UnionFind) Find(v int) int {
	if uf.parent[v] != v {
		uf.parent[v] = uf.Find(uf.parent[v])
	}
	return uf.parent[v]
}

// Function to connect components
func (uf *UnionFind) Union(v1, v2 int) {
	p1 := uf.Find(v1)
	p2 := uf.Find(v2)
	if p1 != p2 {
		if uf.rank[p1] > uf.rank[p2] {
			uf.parent[p2] = p1
		} else if uf.rank[p1] < uf.rank[p2] {
			uf.parent[p1] = p2
		} else {
			uf.parent[p2] = p1
			uf.rank[p1]++
		}
		uf.count--
	}
}

// Function to return the number of connected components consisting of "1"s
func (uf *UnionFind) GetCount() int {
	return uf.count
}

class UnionFind:

 # Initializing the parent list and count variable by traversing the grid
 def __init__(self, grid):
 self.parent = []
 self.rank = []
 self.count = 0
 m = len(grid)
 n = len(grid[0])
 for i in range(m):
 for j in range(n):
 if grid[i][j] == "1":
 self.parent.append(i * n + j)
 self.count += 1
 else:
 self.parent.append(-1)
 self.rank.append(0)

 # Function to find the root parent of a node
 def find(self, i):
 if self.parent[i] != i:
 self.parent[i] = self.find(self.parent[i])
 return self.parent[i]

 # Function to connect components
 def union(self, x, y):
 root_x = self.find(x)
 root_y = self.find(y)
 if root_x != root_y:
 if self.rank[root_x] > self.rank[root_y]:
 self.parent[root_y] = root_x
 elif self.rank[root_x] < self.rank[root_y]:
 self.parent[root_x] = root_y
 else:
 self.parent[root_y] = root_x
 self.rank[root_x] += 1
 self.count -= 1

 # Function to return the number of conencted components consisting of "1"s
 def get_count(self):
 return self.count

Help me debug only the function Number of Islands . . I don't need the corrected code.I don't want explanations or code review. Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Please help me understand why this code failed for the last failed test case. Don't give me the corrected code. . If yes, then Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Your current task is to review only the numIslands function and the other function(s) invoked in it/them. Do not review extra/helper files or functions not invoked in the solution function numIslands. Do not review the code of data structures objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. Also, Answer the following questions: Is the Union Find pattern used? Take your time and review the code.

Please provide a brief time complexity analysis of the function numIslands. While calculating time complexity, include the time complexity of built-in Go functions invoked by the code. The Answer should be to the point and in a few lines.

Please provide a brief Space complexity analysis of the function numIslands. While calculating space complexity, include the space complexity of built-in Go functions invoked by the code. Do not include the space used for the input parameters and the space used for just returning an output(not used as a scratch space). The Answer should be to the point and in a few lines

TwoDArray

package main

import (
    "fmt"
    "os"
    "encoding/json"
    //"reflect"
    "strconv"
)



type TwoDArray struct {
    // class attributes
    parsedData [][]byte
}

func (ds *TwoDArray) generateError(data string) {
    fmt.Fprintln(os.Stderr, "Parsing Error: ", data, " in MyCustomDS is not parsable")
    os.Exit(-1)
}

func (ds *TwoDArray) String() string {
    // override print() for your data structure
    return ""
}

// return value will be passed to user program
func (ds *TwoDArray) Construct(value string, type_of_arr int) any {
    // parse test case in json string 'value' to your desired data structure
    var parsedData1 [][]string
	err := json.Unmarshal([]byte(value), &parsedData1)
	if err != nil {
		fmt.Println("Error:", err)
	}
	
	intArray := make([][]byte, len(parsedData1))
	
    for i, row := range parsedData1 {
    		intRow := make([]byte, len(row))
    		for j, str := range row {
    			num, err := strconv.Atoi(str)
    			if err != nil {
    				fmt.Println("Error:", err)
    			}
                // To convert int to ASCII
    			intRow[j] = byte(num)+byte(48)
    		}
    		intArray[i] = intRow
    }
    // fmt.Println("intArray:", intArray)

    ds.parsedData = intArray

    // json.Unmarshal([]byte(value), &ds.parsedData)
    // valueType := reflect.TypeOf(*ds)
	// fmt.Printf("Value:: %v, Type: %v\n", *ds, valueType)
    return *ds
}

// input param contains return value of user program
func (ds *TwoDArray) Traverse(input any, type_of_arr int) string {
    // convert your data structure 'input' to a json string representing a test case output
    cast, ok := input.(TwoDArray)
    if !ok {
    }
    st, _ := json.Marshal(cast.parsedData)
    return string(st)
}

def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """
 # Union find class
 class UnionFind:
 # Initializing the parent list and count variable by traversing the grid
 def __init__(self, grid):
 self.parent = []
 self.rank = []
 self.count = 0
 m = len(grid)
 n = len(grid[0])
 for i in range(m):
 for j in range(n):
 if grid[i][j] == "1":
 self.parent.append(i * n + j)
 self.count += 1
 else:
 self.parent.append(-1)
 self.rank.append(0)

 # Function to find the root parent of a node
 def find(self, i):
 if self.parent[i] != i:
 self.parent[i] = self.find(self.parent[i])
 return self.parent[i]

 # Function to connect components
 def union(self, x, y):
 root_x = self.find(x)
 root_y = self.find(y)
 if root_x != root_y:
 if self.rank[root_x] > self.rank[root_y]:
 self.parent[root_y] = root_x
 elif self.rank[root_x] < self.rank[root_y]:
 self.parent[root_x] = root_y
 else:
 self.parent[root_y] = root_x
 self.rank[root_x] += 1
 self.count -= 1

 # Function to return the number of conencted components consisting of "1"s
 def get_count(self):
 return self.count

 # Solution
 def num_islands(grid):
 if not grid:
 return 0

 cols = len(grid[0])
 rows = len(grid)
 union_find = UnionFind(grid)

 for r in range(rows):
 for c in range(cols):
 if grid[r][c] == '1':
 grid[r][c] = '0'

 if r - 1 >= 0 and grid[r - 1][c] == '1':
 union_find.union(r * cols + c, (r - 1) * cols + c)
 if r + 1 < rows and grid[r + 1][c] == '1':
 union_find.union(r * cols + c, (r + 1) * cols + c)
 if c - 1 >= 0 and grid[r][c - 1] == '1':
 union_find.union(r * cols + c, r * cols + c - 1)
 if c + 1 < cols and grid[r][c + 1] == '1':
 union_find.union(r * cols + c, r * cols + c + 1)

 count = union_find.get_count()
 return count
 
 # Expected output
 expected = num_islands(self.__inputs[0])

 # Actual output
 actual = self.__actual_outputs[0]

 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = [] 

 # store the 1st expected output that should be displayed to the user 
 expected_outputs.append(expected)

 if json.dumps(expected) == json.dumps(actual):
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)

 else:
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

/*
⬅️ We have provided a union_find.go file under the "Files" tab 
of this widget. You can use this file to build your solution.
*/
package main

func numIslands(grid [][]byte) int {
	
	// Replace this placeholder return statement with your code
	return -1
}

PrivateTestCases.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [['1','1','1'],['0','1','0'],['1','0','0'],['1','0','1']] 
    output: 
    - 1 : 3
  - name: "testcase 2"
    inputs:
    - 1 : [['1', '1', '1', '1', '0'],['1', '0', '0', '0', '1'],['1', '0', '0', '1', '1'],['0', '1', '0', '1', '0'],['1', '1', '0', '1', '1']]
    output: 
    - 1 : 3
  - name: "testcase 3"
    inputs:
    - 1 : [['1', '1', '1', '1', '0'],['1', '0', '0', '0', '1'],['1', '1', '1', '1', '1'],['0', '1', '0', '1', '0'],['1', '1', '0', '1', '1']]
    output: 
    - 1 : 1
  - name: "testcase 4"
    inputs:
    - 1 : [['1', '0', '1', '0', '1'],['0', '1', '0', '1', '0'],['1', '0', '1', '0', '1'],['0', '1', '0', '1', '0'],['1', '0', '1', '0', '1']]
    output: 
    - 1 : 13
  - name: "testcase 5"
    inputs:
    - 1 : [['1','0','1'],['0','0','0'],['1','0','1']]
    output: 
    - 1 : 4


def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """

 grid = self.__inputs[0]

 lower_bound = 1
 upper_bound = 300

 def is_valid_2d_array(arr):
 if not isinstance(arr, list) or not all(isinstance(row, list) for row in arr):
 return False # Not a valid 2D array if it's empty or contains non-list elements
 
 for row in arr:
 if any(elem is None or isinstance(elem, bool) for elem in row):
 return False # Not a valid 2D array if any element in a row is None
 if not all(isinstance(elem, str) for elem in row):
 return False # Not a valid 2D array if any element in a row is not an integer
 
 return True # Valid 2D array

 def check_cols_upper(lower_bound, upper_bound, grid):
 cols = []
 for row in grid:
 cols.append(len(row))

 for col in cols:
 if not(lower_bound <= col <= upper_bound):
 return False
 return True
 
 def check_cols(grid):
 cols = []
 for row in grid:
 cols.append(len(row))
 return all(element == cols[0] for element in cols)

 def check_values(grid):
 for i in range(len(grid)):
 for j in range(len(grid[0])):
 if grid[i][j] not in ['0', '1']:
 return False
 return True

 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not is_valid_2d_array(grid):
 self.store_result ("Not a valid test case", ValidationFunction.FAIL)

 elif not(lower_bound <= len(grid) <= upper_bound):
 self.store_result ("expected 'grid' to have " + str(lower_bound) + " <= size <= " + str(upper_bound) + " but got " + str(len(grid)), ValidationFunction.FAIL)
 
 elif not check_cols_upper(lower_bound, upper_bound, grid):
 self.store_result ("expected rows of 'grid' to have " + str(lower_bound) + " <= size <= " + str(upper_bound), ValidationFunction.FAIL)

 elif not check_cols(grid):
 self.store_result ("expected `grid` to have equal length of rows", ValidationFunction.FAIL)

 elif not check_values(grid):
 self.store_result("'grid' must consist of values from 0 to 1 only", ValidationFunction.FAIL)
 
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Number of Islands problem.

Sliding Window

Union Find

Conclusion

Number of Islands

Statement

Examples

Understand the problem

Figure it out!

Try it yourself