BFS

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BFS

Breadth First Search
One starts at the root (selecting some arbitrary node as the root in the case of a graph) and explores the neighbor nodes first, before moving to the next level neighbors.
Find minimum path/distance/...
- Minimum Tree Depth
- Word Edit Distance
- etc...
No Recursion, QUEUE

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Queue in Java

Queue<> queue = new LinkedList<>();
- boolean offer(Element e); (add)
  - offer can return false while add can only cause exception when the element cannot be added due to capacity restriction.
- Element poll(); (remove)
  - poll returns null when empty while remove will throw exception.
- Element peek(); (element)
  - peek returns null when empty while element will throw exception.

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BFS for Binary Trees

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Binary Tree Level Order Traversal

/ \

B C

/ \ / \

D E F G

/ \

H I

while (level != end) {

visit (level);

level++;

}

How to visit the next level? How to know what next level is?

Put the child nodes into queue while visiting a certain line.

Given a binary tree, return its level order traversal.

A, B, C, D, E, F, G, H, I

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Binary Tree Level Order Traversal

/ \

B C

/ \ / \

D E F G

/ \

H I

Queue<TreeNode> queue = new LinkedList<>();

queue.offer(root);

while (!queue.isEmpty()) {

Queue<TreeNode> queue2 = new LinkedList<>();

while (!queue.isEmpty()) {

TreeNode top = queue.poll();

// visit top.

if (top.left != null) queue2.offer(top.left);

if (top.right != null) queue2.offer(top.right);

}

queue = queue2;

}

Given a binary tree, return its level order traversal.

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Binary Tree Level Order Traversal

/ \

B C

/ \ / \

D E F G

/ \

H I

Queue<TreeNode> queue = new LinkedList<>();

queue.offer(root);

queue.offer(null);

while (!queue.isEmpty()) {

TreeNode top = queue.poll();

if (top == null) {

// do something to the current level.

queue.offer(null);

} else {

// visit top.

if (top.left != null) queue.offer(top.left);

if (top.right != null) queue.offer(top.right);

}

Given a binary tree, return its level order traversal.

Dummy Node

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Binary Tree Level Order Traversal

/ \

B C

/ \ / \

D E F G

/ \

H I

Queue<TreeNode> queue = new LinkedList<>();

queue.offer(root);

while (!queue.isEmpty()) {

TreeNode top = queue.poll();

// visit top.

if (top.left != null) {

queue.offer(top.left);

}

if (top.right != null) {

queue.offer(top.right);

}

Given a binary tree, return its level order traversal.

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Binary Tree Level Order Traversal

Given a binary tree, return its level order traversal.

[
  [3],
  [9,20],
  [15,7]
]

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Binary Tree Level Order Traversal

Given a binary tree, return its level order traversal.

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> results = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<>();
    if (root != null) {
        queue.offer(root);
    }
    while (!queue.isEmpty()) {
        List<Integer> oneResult = new ArrayList<>();
        Queue<TreeNode> queue2 = new LinkedList<>();
        while (!queue.isEmpty()) {
            TreeNode top = queue.poll();
            if (top.left != null) {
                queue2.offer(top.left); 
            }
            if (top.right != null) {
                queue2.offer(top.right);
            }
            oneResult.add(top.val);
        }
        results.add(oneResult);
        queue = queue2;
    }
    return results;
}

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Binary Tree Zigzag Level Order Traversal

Given a binary tree, return its zigzag level order traversal.

[
  [3],
  [20,9],
  [15,7]
]

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Binary Tree Zigzag Level Order Traversal

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    List<List<Integer>> results = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<>();
    if (root != null) {
        queue.offer(root);
    }
    boolean isLeft = true;
    while (!queue.isEmpty()) {
        List<Integer> oneResult = new ArrayList<>();
        Queue<TreeNode> queue2 = new LinkedList<>();
        while (!queue.isEmpty()) {
            TreeNode top = queue.poll();
            if (top.left != null) {
                queue2.offer(top.left); 
            }
            if (top.right != null) {
                queue2.offer(top.right);
            }
            oneResult.add(top.val);
        }
        if (!isLeft) {
            Collections.reverse(oneResult);
        }
        results.add(oneResult);
        isLeft = !isLeft;
        queue = queue2;
    }
    return results;
}

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Binary Tree Zigzag Level Order Traversal

Reversing an ArrayList for every loop is time consuming.
Visit order is opposite from level to level
- FILO
- Stack

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public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    List<List<Integer>> results = new ArrayList<>();
    if (root == null)
        return results;
    Stack<TreeNode> stack = new Stack<>();
    stack.push(root);
    boolean left = true;
    while (!stack.isEmpty()) {
        List<Integer> oneResult = new ArrayList<>();
        Stack<TreeNode> stack2 = new Stack<>();
        while (!stack.isEmpty()) {
            TreeNode top = stack.pop();
            if (!left) {
                if (top.right != null)
                    stack2.push(top.right);
                if (top.left != null)
                    stack2.push(top.left);
            } else {
                if (top.left != null)
                    stack2.push(top.left);
                if (top.right != null)
                    stack2.push(top.right);
            }
            oneResult.add(top.val);
        }
        results.add(oneResult);
        stack = stack2;
        left = !left;
    }
    return results;
}

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How to level order traverse

If level is important, most cases
- Two queues.
- Dummy node, different from all other nodes to be a flag.
- Maintain size of current level.
If level is unimportant
- One queue.

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Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

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Minimum Depth of Binary Tree

DFS

Minimum depth of Root
- if root is leaf, 1.
- if root.left is null, return right.
- if root.right is null, return left.
- return the minimum of (left, right).

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Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Minimum Depth of Binary Tree

BFS

Level Traverse the tree, return the depth of the first leaf node.

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Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Minimum Depth of Binary Tree

public int minDepth(TreeNode root) {
    if (root == null) {
        return 0;
    }
    Queue<TreeNode> queue = new LinkedList<TreeNode>();
    queue.offer(root);
    queue.offer(null);
    int minDepth = 1;
    while (!queue.isEmpty()) {
        TreeNode top = queue.poll();
        if (top == null) {
            minDepth++;
            queue.offer(null);
            continue;
        }
        if (top.left == null && top.right == null) {
            return minDepth;
        }
        if (top.left != null) {
            queue.offer(top.left);
        }
        if (top.right != null) {
            queue.offer(top.right);
        }
    }
    return minDepth;
}

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Minimum Depth of Binary Tree

public int minDepth(TreeNode root) {
    if (root == null) {
        return 0;
    }
    Queue<TreeNode> queue = new LinkedList<TreeNode>();
    queue.offer(root);
    int minDepth = 1;
    while (!queue.isEmpty()) {
        Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
        while (!queue.isEmpty()) {
            TreeNode top = queue.poll();
            if (top.left == null && top.right == null) {
                return minDepth;
            }
            if (top.left != null) {
                queue2.offer(top.left);
            }
            if (top.right != null) {
                queue2.offer(top.right);
            }
        }
        minDepth++;
        queue = queue2;
    }
    return minDepth;
}

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Level is important, most cases
- Two queues.
- Dummy node, different from all other nodes to be a flag.
- Maintain size of current level.
Level is unimportant
- One queue.
Don't add null (except dummy) into queues.
Left/Right null pointer check.

BFS for Binary Trees

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BFS for Word Distance

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Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the word list

For example,
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

Word Ladder

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DFS

begin: hot, end: hip, list: [hog, hig, dig, dip, hip, hit]
hot -> hog -> hig -> dig -> dip -> hip
Much deeper than needed, time consuming.
Error prone.

BFS

Keep a set of all words that can be transformed to.
Put all words with 1 edit distance into the set every time.
When the end word is met, then the minimum distance is found.

Word Ladder

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BFS
- Keep a set of all words that can be transformed to.
- Put all words with 1 edit distance into the set every time.
- When the end word is met, then the minimum distance is found.

Word Ladder

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

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BFS
- Keep a set of all words that can be transformed to.
- Put all words with 1 edit distance into the set every time.
- When the end word is added, then the minimum distance is found.

Word Ladder

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

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BFS
- Keep a set of all words that can be transformed to.
- Put all words with 1 edit distance into the set every time.
- When the end word is added, then the minimum distance is found.

Word Ladder

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

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BFS
- Keep a set of all words that can be transformed to.
- Put all words with 1 edit distance into the set every time.
- When the end word is added, then the minimum distance is found.

Word Ladder

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

dog

log

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BFS
- Keep a set of all words that can be transformed to.
- Put all words with 1 edit distance into the set every time.
- When the end word is added, then the minimum distance is found.

Word Ladder

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

dog

log

cog

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Word Ladder

transformed_words <- beginWord

while (transformed_words is not empty) {

for (word_b in transformed_words) {

if (endWord is within 1 distance) {

return distance

}

if (some word in dict is within 1 distance

&& it is not in transformed_words) {

transformed_words.add(word)

}

update distance

}

return 0

How to check?

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Word Ladder

transformed_words <- beginWord

while (transformed_words is not empty) {

for (word_b in transformed_words) {

if (endWord is within 1 distance) {

return distance

}

if (some word in dict is within 1 distance

&& it is not in transformed_words) {

transformed_words.add(word)

}

update distance

}

return 0

How to check?

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Word Ladder

transformed_words <- beginWord

while (transformed_words is not empty) {

for (word_b in transformed_words) {

if (endWord is within 1 distance) {

return distance

}

if (some word in dict is within 1 distance

&& it is not in transformed_words) {

transformed_words.add(word)

}

update distance

}

return 0

How to check?

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Word Ladder

public int ladderLength(String beginWord, String endWord,
                        Set<String> wordList) {
    wordList.add(endWord);
    Queue<String> queue = new LinkedList<>();
    queue.offer(beginWord);
    Set<String> visited = new HashSet<>();
    visited.add(beginWord);
    int distance = 1;
    while (!queue.isEmpty()) {
        Queue<String> queue2 = new LinkedList<>();
        distance++;
        while (!queue.isEmpty()) {
            String top = queue.poll();
            List<String> wordsWithinDistance =
                getWordsWithinDistance(wordList, top);
            for (String word : wordsWithinDistance) {
                if (word.equals(endWord)) {
                    return distance;
                }
                if (!visited.contains(word)) {
                    queue2.add(word);
                    visited.add(word);
                }
            }
        }
        queue = queue2;
    }
    return 0;
}

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Word Ladder

public ArrayList<String> getWordsWithinDistance(
    Set<String> wordList, String word) {
    ArrayList<String> results = new ArrayList<>();
    char[] wordCharArr = word.toCharArray();
    for (int i = 0; i < word.length(); i++) {
        char oriChar = wordCharArr[i];
        for (char c = 'a'; c <= 'z'; c++) {
            if (c == oriChar) {
                continue;
            }
            wordCharArr[i] = c;
            String newStr = new String(wordCharArr);
            if (wordList.contains(newStr)) {
                results.add(newStr);
            }
        }
        wordCharArr[i] = oriChar;
    }
    return results;
}

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Word Ladder

public int ladderLength(String beginWord, String endWord,
                        Set<String> wordList) {
    wordList.add(endWord);
    wordList.remove(beginWord); // This is optional
    Queue<String> queue = new LinkedList<>();
    queue.offer(beginWord);
    int distance = 1;
    while (!queue.isEmpty() && !wordList.isEmpty()) {
        Queue<String> queue2 = new LinkedList<>();
        distance++;
        while (!queue.isEmpty()) {
            String top = queue.poll();
            ArrayList<String> wordsWithinDistance =
                getWordsWithinDistance(wordList, top);
            if (wordsWithinDistance.contains(endWord)) {
                return distance;
            }
            queue2.addAll(wordsWithinDistance);
        }
        queue = queue2;
    }
    return 0;
}

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Word Ladder

public ArrayList<String> getWordsWithinDistance(Set<String> wordList,
                                                String word) {
    ArrayList<String> results = new ArrayList<>();
    char[] wordCharArr = word.toCharArray();
    for (int i = 0; i < word.length(); i++) {
        char oriChar = wordCharArr[i];
        for (char c = 'a'; c <= 'z'; c++) {
            if (c == oriChar) {
                continue;
            }
            wordCharArr[i] = c;
            String newStr = new String(wordCharArr);
            if (wordList.contains(newStr)) {
                results.add(newStr);
                wordList.remove(newStr);
            }
        }
        wordCharArr[i] = oriChar;
    }
    return results;
}

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Return

  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

http://www.zhitongguigu.com/

Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

dog

log

http://www.zhitongguigu.com/

Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

dog

log

cog

http://www.zhitongguigu.com/

Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

dog

log

cog

Visited needs to be updated after one level.
wordList need to be updated after one level.

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

Example:

beginWord = "hit"

endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

hit

hot

dot

lot

dog

log

cog

Path needs to be recorded backwards.
HashMap<String, ArrayList<String>>

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

BFS to get shortest transformation
Record the path information during transformation so that the sequence can get regenerated later.
- with visited set, all nodes on the path can guarantee the shortest distance.
DFS to get all the transformation sequences
- backward from endWord instead of forward from startWord

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public List<List<String>> findLadders(String beginWord, String endWord,
                                      Set<String> wordList) {
    List<List<String>> results = new ArrayList<>();
    HashMap<String, ArrayList<String>> preList = new HashMap<>();
    
    wordList.add(endWord);
    Queue<String> queue = new LinkedList<>();
    queue.offer(beginWord);
    Set<String> visited = new HashSet<>();
    visited.add(beginWord);

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Word Ladder II

    while (!queue.isEmpty() && !wordList.isEmpty()) {
        Set<String> queue2 = new HashSet<>();
        boolean isFinished = false;
        while (!queue.isEmpty()) {
            String top = queue.poll();
            Set<String> wordsWithinDistance =
                getWordsWithinDistance(wordList, top);

            for (String word : wordsWithinDistance) {
                if (word.equals(endWord)) {
                    isFinished = true;
                }
                if (!visited.contains(word)) {
                    updatePreList(preList, word, top);
                    queue2.add(word);
                }
            }
        }
        visited.addAll(queue2);
        if (isFinished) {
            getPaths(results, preList, new ArrayList<String>(), endWord);
            break;
        }
        queue.addAll(queue2);
    }
    return results;
}

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Word Ladder II

public void updatePreList(HashMap<String, ArrayList<String>> preList, 
                          String cur, String pre) {
    if (!preList.containsKey(cur)) {
        preList.put(cur, new ArrayList<String>());
    }
    preList.get(cur).add(pre);
}

public Set<String> getWordsWithinDistance(Set<String> wordList, String word) {
    Set<String> results = new HashSet<>();
    char[] wordCharArr = word.toCharArray();
    for (int i = 0; i < word.length(); i++) {
        char oriChar = wordCharArr[i];
        for (char c = 'a'; c <= 'z'; c++) {
            if (c == oriChar) {
                continue;
            }
            wordCharArr[i] = c;
            String newStr = new String(wordCharArr);
            if (wordList.contains(newStr)) {
                results.add(newStr);
            }
        }
        wordCharArr[i] = oriChar;
    }
    return results;
}

Word Ladder II

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Word Ladder II

How to get paths backward from endWord?
- DFS (recursion)
  - Base Case: endWord.pre == null
  - Recursion rule: path from endWord = path from endWord.pre + endWord

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public void getPaths(List<List<String>> paths, 
                     HashMap<String, ArrayList<String>> preList,
                     List<String> curPath,
                     String end) {
    if (!preList.containsKey(end)) {
        curPath.add(end);
        Collections.reverse(curPath);
        paths.add(curPath);
        return;
    }
    for (String pre : preList.get(end)) {
        List<String> newPath = new ArrayList<String>(curPath);
        newPath.add(end);
        getPaths(paths, preList, newPath, pre);
    }
}

Word Ladder II

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public List<List<String>> findLadders(String beginWord, String endWord, Set<String> wordList) {
    List<List<String>> results = new ArrayList<>();
    HashMap<String, ArrayList<String>> preList = new HashMap<>();
    wordList.add(endWord);
    wordList.remove(beginWord);
    Queue<String> queue = new LinkedList<>();
    queue.offer(beginWord);
    
    while (!queue.isEmpty() && !wordList.isEmpty()) {
        Set<String> queue2 = new HashSet<>();
        boolean isFinished = false;
        while (!queue.isEmpty()) {
            String top = queue.poll();
            Set<String> wordsWithinDistance = getWordsWithinDistance(wordList, top);
            for (String word : wordsWithinDistance) {
                if (word.equals(endWord)) {
                    isFinished = true;
                }
                updatePreList(preList, word, top);
                queue2.add(word);
            }
        }
        wordList.removeAll(queue2);
        if (isFinished) {
            getPaths(results, preList, new ArrayList<String>(), endWord);
            break;
        }
        queue.addAll(queue2);
    }
    return results;
}

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public List<List<String>> findLadders(String beginWord, String endWord, Set<String> wordList) {
    List<List<String>> results = new ArrayList<>();
    HashMap<String, ArrayList<String>> preList = new HashMap<>();
    wordList.add(endWord);
    wordList.remove(beginWord);
    Queue<String> queue = new LinkedList<>();
    queue.offer(beginWord);
    
    while (!queue.isEmpty() && !wordList.isEmpty()) {
        Set<String> queue2 = new HashSet<>();
        boolean isFinished = false;
        while (!queue.isEmpty()) {
            String top = queue.poll();
            Set<String> wordsWithinDistance = getWordsWithinDistance(wordList, top);
            updatePreList(preList, wordsWithinDistance, top);
            queue2.addAll(wordsWithinDistance);
            isFinished = isFinished | wordsWithinDistance.contains(endWord);
        }
        wordList.removeAll(queue2);
        if (isFinished) {
            getPaths(results, preList, new ArrayList<String>(), endWord);
            break;
        }
        queue.addAll(queue2);
    }
    return results;
}
public void updatePreList(HashMap<String, ArrayList<String>> preList,
                          Set<String> curStrings, String pre) {
    for (String cur : curStrings) {
        if (!preList.containsKey(cur)) {
            preList.put(cur, new ArrayList<String>());
        }
        preList.get(cur).add(pre);
    }
}

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Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord

BFS to get shortest transformation
Record the paths from beginning while BFSing
Add only the valid paths into results.

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Distance means level important
- Two queues
  - Order is not important, so queue is not essential actually.
- Dummy node
Pruning for duplicate word
- HashSet for visited words
- Remove from dict.
Spell check
- Word, Google Docs, etc.

BFS for Word Distance

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Shortest path
- Dijkstra's Algorithm
Maze solving
- The minimum steps needed

BFS for Distance

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Both complete search
DFS is recursion
- stack
- better at checking connection, etc.
BFS is iteration
- queue
- better at minimum distance, etc.
Many problems are both solvable by DFS/BFS
- Surrounded Region
- Number of Islands
- etc.

DFS v.s. BFS

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Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Surrounded Region

X X X X
X O O X
X X O X
X O X X

X X X X
X X X X
X X X X
X O X X

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Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Surrounded Region

public void solve(char[][] board) {
    if (board.length == 0 || board[0].length == 0)
        return;

    for (int j = 0; j < board[0].length; j++) {
        search(board, 0, j);
        search(board, board.length-1, j);
    }
    for (int i = 0; i < board.length; i++) {
        search(board, i, 0);
        search(board, i, board[0].length-1);
    }
    
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            board[i][j] = board[i][j] == 'F' ? 'O' : 'X';
        }
    }
}

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Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Surrounded Region

// BFS
public void search(char[][] board, int x, int y) {
    if (board[x][y] == 'X') {
        return;
    }
    Queue<Integer> queue = new LinkedList<Integer>();
    int xLen = board.length, yLen = board[0].length;
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
    
    queue.offer(x * yLen + y);
    board[x][y] = 'F';

    while (!queue.isEmpty()) {
        int temp = queue.poll();
        for (int i = 0; i < 4; i++) {
            int nx = temp / yLen + dx[i], ny = temp % yLen + dy[i];
            if (nx >= 0 && nx < xLen && ny >= 0 && ny < yLen
                && board[nx][ny] == 'O') {
                board[nx][ny] = 'F';
                queue.offer(nx * yLen + ny);
	    }
	}
    }
}

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Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Surrounded Region

// DFS
public void search(char[][] board, int i, int j) {
    if (board[i][j] != 'O')
	return;
    board[i][j] = 'F';
    if (i > 1)
	search(board, i-1, j);
    if (i < board.length - 2)
	search(board, i+1, j);
    if (j > 1)
	search(board, i, j-1);
    if (j < board[i].length - 2)
	search(board, i, j+1);
}

http://www.zhitongguigu.com/

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Surrounded Region

// DFS
public void search(char[][] board, int i, int j) {
    if (board[i][j] != 'O')
	return;
    board[i][j] = 'F';
    if (i >= 1)
	search(board, i-1, j);
    if (i <= board.length - 2)
	search(board, i+1, j);
    if (j >= 1)
	search(board, i, j-1);
    if (j <= board[i].length - 2)
	search(board, i, j+1);
}

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Stack Overflow

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Number of Islands

11110
11010
11000
00000

11000
11000
00100
00011

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Number of Islands

public int numIslands(char[][] grid) {
    if (grid == null || grid.length == 0 || grid[0].length == 0) {
        return 0;
    }
    int nums = 0;
    for (int i = 0; i < grid.length; i++) {
        for (int j = 0; j < grid[0].length; j++) {
            if (grid[i][j] == '1') {
                search(grid, i, j);
                nums++;
            }
        }
    }
    return nums;
}

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Number of Islands

// DFS
public void search(char[][] grid, int x, int y) {
    if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
        return;
    }
    if (grid[x][y] == '1') {
        grid[x][y] = '0';
        search(grid, x + 1, y);
        search(grid, x - 1, y);
        search(grid, x, y + 1);
        search(grid, x, y - 1);
    }
}

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Number of Islands

// BFS
public void search(char[][] grid, int x, int y) {
    Queue<Integer> queue = new LinkedList<>();
    int xLen = grid.length, yLen = grid[0].length;
    queue.offer(x * yLen + y);
    grid[x][y] = '0';
    
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
    while (!queue.isEmpty()) {
        Integer top = queue.poll();
        
        for (int i = 0; i < 4; i++) {
            int nx = top/yLen + dx[i], ny = top%yLen + dy[i];
            if (nx >= 0 && nx < xLen && ny >= 0 && ny < yLen
                && grid[nx][ny] == '1') {
                grid[nx][ny] = '0';
                queue.offer(nx * yLen + ny);
            }
        }
    }
}

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Homework

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