Algorithm Problems Practices

X. DP
O(1) space - in place, reuse the input
-- multiple dp relations

LeetCode 375 - Wiggle Subsequence - link
return the length of the longest subsequence that is a wiggle sequence
up[i] - considering i​th element as the last element of the wiggle subsequence and ending with a rising wiggle.
down[i]
if nums[i] > nums[i-1] up[i] = down[i-1]+1 down[i]=down[i-1]
else if nums[i] < nums[i-1]  down[i]=up[i-1]+1 up[i] = up[i-1]
else down[i]= down[i-1] up[i] = up[i-1]
-- space optimization, only use up and down
2. Greedy

Leetcode 44 - Wildcard Matching
with support for '?' and '*'.
dp[s.length() + 1][p.length() + 1]
dp[i[j]: the first i characters in string s matches the first characters of string p.
dp[0][0] = true;
dp[i][0] = false;
dp[0][j] = dp [0][j - 1] if p.charAt(j - 1) == '*'
-- If p.charAt(j - 1) != '*', then dp[i][j] = dp[i - 1][j - 1] if s.charAt(i) == p.charAt(j) || p.charAt(j) == '?' (else false)
-- If p.charAt(j - 1) == '*', then
-- dp[i][j]
     = dp[i][j - 1] || // Match 0 character
     = dp[i - 1][j] // Match any sequence of characters
O(n) space - https://discuss.leetcode.com/topic/10794/my-java-dp-solution

LeetCode 10 - Regular Expression Matching: . and *
'*' Matches zero or more of the preceding element.
- DFS+Cache: dp[i][j] or DP
https://discuss.leetcode.com/topic/40371/easy-dp-java-solution-with-detailed-explanation
https://discuss.leetcode.com/topic/31974/java-4ms-dp-solution-with-o-n-2-time-and-o-n-space-beats-95
1, If p.charAt(j) == s.charAt(i) :  dp[i][j] = dp[i-1][j-1];
2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
3, If p.charAt(j) == '*': 
   here are two sub conditions:
               1   if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //in this case, a* only counts as empty
               2   if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
                              dp[i][j] = dp[i-1][j]    //in this case, a* counts as multiple a 
                           or dp[i][j] = dp[i][j-1]   // in this case, a* counts as single a
                           or dp[i][j] = dp[i][j-2]   // in this case, a* counts as empty

Leetcode 139 - Word Break
Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
1. DP: O(n^2)
Get words minLen, maxLen
boolean [] dp = new boolean[n+1];
dp[0] = true ;
dp[i] = dp[j] && dict.contains(sentences.subString(j,i)) 0<j<i

// Use trie to check whether subString is a word
2. DFS + Cache
3. BFS:

LeetCode 287 - Perfect Squares
dp(i) = min{1+dp(i-j*j)}
https://discuss.leetcode.com/topic/26400/an-easy-understanding-dp-solution-in-java
bfs -  q.offer(0); visited.add(0);
https://leetcode.com/discuss/58056/summary-of-different-solutions-bfs-static-and-mathematics

https://discuss.leetcode.com/topic/2545/a-solution-using-bfs/6
The vertices of the graph are simply the positions of the first characters of the words and each edge actually represents a word

Word Break II
https://leetcode.com/discuss/91894/java-6ms-simple-solution-beating-88%25
http://www.jiuzhang.com/solutions/word-break-ii/
- avoid to use substring
boolean[][] isWord
boolean[] possible

Lintcode: K Sum I
http://www.cnblogs.com/yuzhangcmu/p/4279676.html
int[][][] D = new int[len + 1][k + 1][target + 1];
D[i][j][t] = D[i - 1][j][t];
if (t - A[i - 1] >= 0) {
    D[i][j][t] += D[i - 1][j - 1][t - A[i - 1]];
}

space op: int[][] D = new int[k + 1][target + 1];

Lintcode: k Sum II
- DFS

LeetCode 97 - Interleaving String
dp[i][j]  =
str3.charAt(i + j) == str1.charAt(i) && dp[ i - 1][j] ||
str3.charAt(i + j) == str2.charAt(j) && dp[i][j - 1]

LeetCode 132 - Palindrome Partitioning II
http://happycoding2010.blogspot.com/2015/10/leetcode-132-palindrome-partitioning-ii.html
Return the minimum cuts needed for a palindrome partitioning of s
isPal[i][j] store if s[i..j] is palindrome or not. res[j] is the min cut of s[0..j].

LeetCode 241 - Different Ways to Add Parentheses
DP: https://discuss.leetcode.com/topic/26076/java-recursive-9ms-and-dp-4ms-solution
- split string first
- dp[i][j] stores all possible results from the i-th integer to the j-th integer (inclusive)
- ArrayList<Integer>[][] dp
DFS: divide and conquer + cache
https://discuss.leetcode.com/topic/19901/a-recursive-java-solution-284-ms/12

LeetCode 446 - Arithmetic Slices II - Subsequence
T(i, d) = summation of (1 + T(j, d)) as long as 0 <= j < i && d == A[i] - A[j]
Map<Integer, Integer>[] map = new Map[A.length];

LintCode 395 - Coins in a Line II
dp[i] means the largest value you(the first player) can get when you start from values[i]
dp[i] = values[i] + Math.min(dp[i+2], dp[i+3]);
dp[i] = Math.max(dp[i], values[i] + values[i+1] + Math.min(dp[i+3], dp[i+4]));


LeetCode 73 - Unique Paths II
if some obstacles are added to the grids. How many unique paths would there be
dp[i][j] = dp[i−1][j]+dp[i][j−1] if grid[i][j]=0
O(1) space - in place, reuse the input
https://discuss.leetcode.com/topic/4987/java-solution-using-dynamic-programming-o-1-space
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
if(obstacleGrid[i][j] == 1)
    obstacleGrid[i][j] = 0;

LeetCode 72 - Edit Distance
f(i, j) = f(i - 1, j - 1)
f(i, j) = 1 + min { f(i, j - 1), f(i - 1, j), f(i - 1, j - 1) }

LeetCode - 403 Frog Jump
DP https://discuss.leetcode.com/topic/59423/simple-easy-to-understand-java-solution
map = new HashMap≪Integer, HashSet≪Integer>>(stones.length);
List≪HashSet≪Integer>> dp
DFS + cache
https://discuss.leetcode.com/topic/59439/straight-forward-9ms-7-line-c-solution-with-explanation

LeetCode 97 - Interleaving String
1. DP:
dp[0][0] = true;
dp[0][j] = str3.charAt(j) == str2.charAt(j)
dp[i][0] = str3.charAt(i) == str1.charAt(i)
2. BFS
https://discuss.leetcode.com/topic/6562/8ms-c-solution-using-bfs-with-explanation

X. BFS
When to use?
- to find minimum steps
- two-end BFS, multi-end bfs
- BFS + PriorityQueue

LeetCode 417 - Pacific Atlantic Water Flow
bfs or dfs
https://discuss.leetcode.com/topic/62379/java-bfs-dfs-from-ocean
if (pacific[i][j] && atlantic[i][j]) res.add(new int[] {i, j});

LeetCode - Word Ladder I
find the length of shortest transformation sequence from start to end
Two-end BFS
https://discuss.leetcode.com/topic/29303/two-end-bfs-in-java-31ms
beginSet.add(beginWord);
endSet.add(endWord);
HashSet≪String> visited
if (beginSet.size() > endSet.size()) beginSet, endSet = endSet, beginSet;
if (endSet.contains(target)) return len + 1;

LeetCode Word Ladder II
https://discuss.leetcode.com/topic/17975/super-fast-java-solution-two-end-bfs
http://www.jiuzhang.com/solutions/word-ladder-ii/
http://www.programcreek.com/2014/06/leetcode-word-ladder-ii-java/
LinkedList≪WordNode> queue
https://discuss.leetcode.com/topic/2857/share-two-similar-java-solution-that-accpted-by-oj
Map≪String,List≪String>> map: key is a word, the list is the word's pre words.
or:
unvisited = new HashSet≪String>(dict);
visited = new HashSet≪String>();

unvisited.removeAll(visited);
visited.clear();

LeetCode 407 - Trapping Rain Water II
https://discuss.leetcode.com/topic/60418/java-solution-using-priorityqueue
BFS + PriorityQueue

LeetCode [317] Shortest Distance from All Buildings
- do not go into a land, if it is not accessible by at least one of previous buildings.

LeetCode 286 - Walls and Gates
Fill each empty room with the distance to its nearest gate.
- Multi end bfs - O(n^2)
- Naive BFS or dfs - O(n^4)

LeetCode 433 - Minimum Genetic Mutation
https://discuss.leetcode.com/topic/65780/java-solution-using-bfs

Leetcode 200: Find the number of islands
- bfs, dfs, union-find
- number of lakes
- different shapes of number Island
http://www.1point3acres.com/bbs/thread-175393-1-1.html

LeetCode 247 - Strobogrammatic Number II
Find all strobogrammatic numbers that are of length = n.
- Use char[] to avoid string concatenation
iterative, bfs - https://discuss.leetcode.com/topic/28990/simple-java-solution-without-recursion
dfs - https://discuss.leetcode.com/topic/21579/accepted-java-solution-using-recursion
https://discuss.leetcode.com/topic/20733/my-concise-java-solution-using-dfs
public void findStrobogrammaticHelper(List res, char[] a, int l, int r)

LeetCode 248 - Strobogrammatic Number III
Count the total strobogrammatic numbers that exist in the range of low <= num <= high
https://discuss.leetcode.com/topic/44718/java-1ms-solution-with-comments-in-the-code

X. DFS
- Backtrack: don't forget to clear state
- dfs+cache
LeetCode 212 - Word Search II
Given a 2D board and a list of words from the dictionary, find all words in the board.
- dfs + trie
https://discuss.leetcode.com/topic/33246/java-15ms-easiest-solution-100-00/
- pass TrieNode

LeetCode 395 - Longest Substring with At Least K Repeating Characters
https://discuss.leetcode.com/topic/57265/java-d-c-solution

Leetcode 291 Word Pattern II
Map<Character, String> forwardMap, invertedMap

LeetCode124 - Word Search
find if the word exists in the grid
https://discuss.leetcode.com/topic/7907/accepted-very-short-java-solution-no-additional-space/39

LeetCode 22 – Generate Parentheses
generate all combinations of well-formed parentheses.
helper(List res, StringBuilder sb, int open, int close, int n)
DFS: https://discuss.leetcode.com/topic/5866/my-accepted-java-solution
-- dfs(ArrayList<String> result, String s, int left, int right)
DFS+cache
https://discuss.leetcode.com/topic/33127/java-easy-to-understand-recursive-dp-method-with-explanations/
helper(int n, List<String>[] lists)

BSF, DP
f(n) = (f(0))f(n-1) + (f(1))f(n-2) + ... + (f(n-2))f(1) + (f(n-1))f(0)
https://discuss.leetcode.com/topic/3474/an-iterative-method

LeetCode 320 - Generalized Abbreviation
DFS/Backtracking
https://discuss.leetcode.com/topic/32765/java-14ms-beats-100
Divide &Conquer
Bit String/Mark
- 2^n combinations

LeetCode 329. Longest Increasing Path in a Matrix
- DFS+Cache: dfs(int[][] matrix, int i, int j, int[][] memo)

LeetCode 267 - Palindrome Permutation II
void getPerm(List<Character> list, String mid, boolean[] used, StringBuilder sb, List<String> res)
private void helper(char[] array, List<String> result, Map<Character, Integer> map, int left, int right)

LeetCode - Subsets
res = new ArrayList<List>();
3. bit string
Math.pow(2, n)
if (((1 << j) & i) != 0) subset.add(nums[j]);

LeetCode 46 - Permutations
LeetCode 47 - Permutations II
if(i>0 &&nums[i-1]==nums[i] && !used[i-1]) continue;

Google interview - max vacation days


X. Slide Window
LeetCode 30 - Substring with Concatenation of All Words
https://discuss.leetcode.com/topic/6617/an-o-n-solution-with-detailed-explanation/11
countMap, curMap, foundCount,
toFindMap, foundMap
O(KN)

LeetCode 67 - Minimum Window Substring
Given a set T of characters and a string S, find the minimum window in S which will contain all the characters in T in complexity O(n).
https://leetcode.com/discuss/90376/o-n-5ms-java-solution-beats-93-18%25
Map<Character, Integer> needToFill, hasFound

Two pointers
3Sum
- presort, O(N^2)
- how to avoid duplicate results
https://discuss.leetcode.com/topic/28857/easiest-java-solution

3Sum Smaller
https://discuss.leetcode.com/topic/23421/simple-and-easy-understanding-o-n-2-java-solution
nums[i] + nums[j] + nums[k] < target
count += right-left;

3Sum Closest
if (sum <= target) j++; else k--;

X. Trie
- use trie so no need to scan every candidate(O(n))
LeetCode 421 - Maximum XOR of Two Numbers in an Array
- use bit trie for int
- children = new Trie[2];
https://discuss.leetcode.com/topic/63207/java-o-n-solution-using-trie
https://discuss.leetcode.com/topic/63213/java-o-n-solution-using-bit-manipulation-and-hashmap
2. 32n
https://discuss.leetcode.com/topic/63213/java-o-n-solution-using-bit-manipulation-and-hashmap
if(set.contains(tmp ^ prefix)) max = tmp;

LeetCode 212 - Word Search II

logN related
Binary search/BiSection
- split to half
LeetCode 222 -  Count Complete Tree Nodes
O(log(n)^2) https://discuss.leetcode.com/topic/15533/concise-java-solutions-o-log-n-2

LeetCode 287 - Find the Duplicate Number
1. nlogn - binary search/bisection
https://discuss.leetcode.com/topic/25580/two-solutions-with-explanation-o-nlog-n-and-o-n-time-o-1-space-without-changing-the-input-array/2
2. O(n)
- graph: index -> as graph
- find cycle in linkedlist
http://keithschwarz.com/interesting/code/?dir=find-duplicate
https://discuss.leetcode.com/topic/18864/simple-java-solution-in-o-n-without-extra-space
3. 32n
https://discuss.leetcode.com/topic/38493/o-32-n-solution-using-bit-manipulation-in-10-lines

Using Binary search tree
LeetCode 327 - Count of Range Sum
Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.
1. Merge Sort
2. Binary search tree
private class TreeNode {
    long val = 0;
    int count = 1;
    int leftSize = 0;
    int rightSize = 0;
    TreeNode left = null;
    TreeNode right = null;
    countSmaller(TreeNode root, long val)
    countLarger(TreeNode root, long val)
    private int rangeSize(TreeNode root, long lower, long upper) {
        int total = root.count + root.leftSize + root.rightSize;
        int smaller = countSmaller(root, lower);    // Exclude everything smaller than lower
        int larger = countLarger(root, upper);      // Exclude everything larger than upper
        return total - smaller - larger;
    }
}
for(int i = 1; i < sums.length; i++) {
    output += rangeSize(root, sums[i] - upper, sums[i] - lower);
    insert(root, sums[i]);
}

LeetCode 315 - Count of Smaller Numbers After Self
1. Binary search tree
https://leetcode.com/discuss/73762/9ms-short-java-bst-solution-get-answer-when-building-bst
class TreeNode{
    int smallCount;
    int val;
    TreeNode left;
    TreeNode right;
}

2. Merge sort
http://algobox.org/count-of-smaller-numbers-after-self/
class Tuple {
    public int val;
    public int idx;
}
3. Use TreeSet
https://discuss.leetcode.com/topic/31422/easiest-java-solution/12

Binary search
LeetCode 153 - Find the minimum element in a sorted and rotated array
mid = low + (high - low) / 2;
Compare num[mid] with num[high]

LeetCode 33 - Searching an Element in a Rotated Sorted Array

LeetCode 162 - Find Peak Element: num[i] ≠ num[i+1]
- Compare mid with mid+1
nums[left - 1] < nums[left] && nums[right] > nums[right + 1]
while (right - left > 1) {
    int mid = left + (right - left) / 2;
    if (nums[mid] < nums[mid + 1]) {
        left = mid + 1;
    } else {
        right = mid;
    }
}
return (left == N - 1 || nums[left] > nums[left + 1]) ? left : right;

Bisection
Given a result, it is easy to test whether it is valid or not.
LeetCode 410 - Split Array Largest Sum
- binary(nums, m, sum, max);
LeetCode 367 - Valid Perfect Square

X. Interval
- First sort these intervals somehow
- Think as different events - start, end
- sort interval + greedy
- merge interval

LeetCode 218 - The Skyline Problem
- (start/end, height) tuple, set the start segment as negative height or Edge:(x, height, isStart)
- sort interval
- TreeMap<Integer, Integer> heightMap: key is the height, value is the count
- prevHeight, currHeight = heightMap.firstKey();
heightMap = new TreeMap<>(Collections.reverseOrder());
heightMap.put(0,1);
pq = new PriorityQueue<>((a, b) -> (b - a));
pq.offer(0);
https://discuss.leetcode.com/topic/38065/java-solution-using-priority-queue-and-sweepline/
http://www.programcreek.com/2014/06/leetcode-the-skyline-problem-java/
https://discuss.leetcode.com/topic/22482/short-java-solution

LeetCode 56 - Merge Intervals
- Sort the intervals based on increasing order of starting time
- if prev.end >= curr.start  prev = merged else result.add(prev); prev = curr;

LeetCode 253 - Meeting Rooms II
https://discuss.leetcode.com/topic/35253/explanation-of-super-easy-java-solution-beats-98-8-from-pinkfloyda
- int[] starts, ends

Use TreeMap to store intervals
LeetCode 352 - Data Stream as Disjoint Intervals
- TreeMap<Integer, Interval> treeMap: key is the start point
- lowerKey(), higherKey()

LettCode 452 - Minimum Number of Arrows to Burst Balloons
- sort interval and greedy
https://discuss.leetcode.com/topic/66709/c-easy-understood-solution-sort/

LeetCode 435 - Non-overlapping Intervals
- sort interval by end
https://discuss.leetcode.com/topic/65828/java-solution-with-clear-explain

LeetCode 436 - Find Right Interval
TreeMap<Integer, Integer> intervalMap: key is the start, value is the index
- first put all intervals to the map: sorted by start
- intervalMap.ceilingEntry(intervals[i].end);

Google – Maximum Time Range Overlaps
https://reeestart.wordpress.com/2016/07/02/google-maximum-time-range-overlaps/

Google - remove alarm
https://reeestart.wordpress.com/2016/06/30/google-remove-alarm/
hash map - map priority to set of alarm id
max priority heap - PriorityQueue<Integer>

Google – Find a Path Among Circles
O(n^2) - https://reeestart.wordpress.com/2016/07/02/google-find-a-path-among-circles/


Bucket
LeetCode - 220 Contains Duplicate III
Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.
slide window + bucket
https://leetcode.com/discuss/38206/ac-o-n-solution-in-java-using-buckets-with-explanation
https://discuss.leetcode.com/topic/27608/java-python-one-pass-solution-o-n-time-o-n-space-using-buckets/
https://leetcode.com/discuss/68090/easy-ac-solution-using-treeset-long-in-java

X. Ordered Stack
LeetCode 42 - Trapping Rain Water
1. Two Pointers
comapre height[left] and height[right]
2. Left and right array
int[] leftMax, rightMax
3. Ordered Stack - decreasing
s.push(i++); - store index
result += stack.isEmpty()? 0:(Math.min(A[stack.peek()],A[i])-A[bot])*(i-stack.peek()-1);

Next Greater Element
http://www.ideserve.co.in/learn/next-great-element-in-an-array


TreeMap
LeetCode 456 - 132 Pattern
Use TreeMap/TreeSet instead of PriorityQueue if it works and you need remove item from them

Union Find
http://algs4.cs.princeton.edu/15uf/UF.java.html

Graph
topological sort
-- Put indegree=0 nodes to queue, iterate queue,  and decrease indegree of the node's neighbors by 1, if indegree is now 0, put into queue
- time complexity O(V+E), space O(V)
Leetcode 310 - Minimum Height Trees
List<Set<Integer>> adj
LeetCode 210 - Course Schedule II
LeetCode 269 -  Alien Dictionary
https://discuss.leetcode.com/topic/32900/java-bfs-solution
- build graph
dfs
https://discuss.leetcode.com/topic/33565/3ms-clean-java-solution-dfs
visited[i] = -1. Not even exist.
visited[i] = 0. Exist. Non-visited.
visited[i] = 1. Visiting.
visited[i] = 2. Visited.

LeetCode 444 - Sequence Reconstruction
https://discuss.leetcode.com/topic/65948/java-solution-using-bfs-topological-sort/5
- build the edge for adjacent ints in the array
Map<Integer, List<Integer>> graph

Eulerian path/circuit
LeetCode 332 - Reconstruct Itinerary
https://discuss.leetcode.com/topic/36370/short-ruby-python-java-c
public List<String> findItinerary(String[][] tickets) {
    for (String[] ticket : tickets)
        targets.computeIfAbsent(ticket[0], k -> new PriorityQueue()).add(ticket[1]);
    visit("JFK");
    return route;
}
Map<String, PriorityQueue<String>> targets = new HashMap<>();
List<String> route = new LinkedList();
void visit(String airport) {
    while(targets.containsKey(airport) && !targets.get(airport).isEmpty())
        visit(targets.get(airport).poll());
    route.add(0, airport);
}

X. Binary tree
- usually expected runtime for tree is O(n)
- PreOrder/InOrder/PostOrder/LevelOrder traverse
- Use PostOrder Traverse
Leetcode 124 - Binary Tree Maximum Path Sum
Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree.
private int maxPathDown(TreeNode node) {
    if (node == null) return 0;
    int left = Math.max(0, maxPathDown(node.left));
    int right = Math.max(0, maxPathDown(node.right));
    maxValue = Math.max(maxValue, left + right + node.val);
    return Math.max(left, right) + node.val;
}
LeetCode 437 - Path Sum III
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards
https://discuss.leetcode.com/topic/64526/17-ms-o-n-java-prefix-sum-method
- preorder tree traverse
- use hashmap to track: the prefix sum -> how many ways get to this prefix sum
- map.put(sum, map.get(sum)-1); // Remove the current node so it wont affect other path

LeetCode 250: Count Univalue Subtrees
- post order

LeetCode 230 - Find k-th smallest element in BST
- inorder search, return when count decrease to 0 from k
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently?

LeetCode 298 - Binary Tree Longest Consecutive Sequence
- preorder: count = (root.val - val == 1)?count+1:1;

LeetCode 107 - Binary Tree Level Order Traversal II
- bottom-up level order

LeetCode 103 - Printing a Binary Tree in Zig Zag Level-Order
If we only need print, it's easier to first get all level nodes in correct order then print
X. bfs https://discuss.leetcode.com/topic/42184/java-easy-understand-recursive-methods-beats-96-attach-easy-bfs-methods
Queue<TreeNode>, boolean leftToRight = !leftToRight;
if(leftToRight) levelNodes.add(n.val); else levelNodes.addFirst(n.val);
x. dfs - https://discuss.leetcode.com/topic/3413/my-accepted-java-solution
ArrayList<LinkedList<Integer>> result
x. 2 stacks
https://discuss.leetcode.com/topic/3318/my-ac-java-code
https://discuss.leetcode.com/topic/32427/java-double-stack-solution
Stack<TreeNode> oddSt, evenSt

X. Binary search tree
LeetCode 333 - Largest BST Subtree
class Wrapper{
    int size;
    int lower, upper;
    boolean isBST;
}

LeetCode 95 - Unique Binary Search Trees II
DFS - divide and conquer

X. Bit trick
- 23n, loop and check very bit
LeetCode 287 - Find the Duplicate Number
LeetCode 421 - Maximum XOR of Two Numbers in an Array

X. Game theory
If a player is able to move to a losing position then he is in a winning position.
If a player is able to move only to the winning positions then he is in a losing position.
LeetCode flip game II
DFS + Cache (top-down)
- use it's dp(bottom-up) to time complexity
- Map<String, Boolean> winMap
- if(!canWin(opponent, map))

Google – Initialize Board
radom + backtrack - https://reeestart.wordpress.com/2016/07/01/google-initialize-board/

Math
LeetCode 60 - Permutation Sequence
Given n and k, return the kth permutation sequence.
k--;
mod = mod / (n - i);
int curIndex = k / mod;
k = k % mod;
builder.append(list.remove(groupIndex));

X. Misc
LeetCode 158 - Read N Characters Given Read4 II
https://discuss.leetcode.com/topic/6402/clean-accepted-java-solution

char[] buffer, int offset, remaining, boolean isEndOfFile

LeetCode 289 - Game of Life
use 2 bits: [2nd bit, 1st bit] = [next state, current state]

LeetCode 361 - Bomb Enemy
1. v1[i][j] + v2[i][j] + v3[i][j] + v4[i][j]
2.
rowhits, colhits[n]
for (int k=j; k<n && grid[i][k] != 'W'; k++)
    rowhits += grid[i][k] == 'E';
for (int k=i; k<m && grid[k][j] != 'W'; k++)
    colhits[j] += grid[k][j] == 'E';
result = max(result, rowhits + colhits[j]);

LeetCode 300 - Longest Increasing Subsequence
1.
int i = Collections.binarySearch(dp, num);
dp.set((i<0) ? -i-1 : i, num);
-- O(nlogn)
2. DP O(N^2)

4Sum
https://discuss.leetcode.com/topic/12368/clean-accepted-java-o-n-3-solution-based-on-3sum
http://www.lifeincode.net/programming/leetcode-two-sum-3-sum-3-sum-closest-and-4-sum-java/
O(N^3)
HashMap<Integer, ArrayList<Pair>> hashMap

LeetCode 454 - 4Sum II
https://discuss.leetcode.com/topic/67593/clean-java-solution-o-n-2

LeetCode 148 - Sort List
http://www.jiuzhang.com/solutions/sort-list/
merge and quick sort on linkedlist

LCA
-- Node to Node Binary Tree Path

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