Thursday, February 26, 2015

LeetCode [93] Restore IP Addresses

Ref
[1] https://oj.leetcode.com/problems/restore-ip-addresses/
OJ
at No comments:

LeetCode [89] Gray Code



Ref
[1] https://oj.leetcode.com/problems/gray-code/
OJ
[2] https://www.cs.princeton.edu/~rs/AlgsDS07/24CombinatorialSearch.pdf
method1
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LeetCode [79] Word Search

 79. Word Search

Medium

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cells, where "adjacent" cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.

 

Example 1:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true

Example 2:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

 

Constraints:

  • board and word consists only of lowercase and uppercase English letters.
  • 1 <= board.length <= 200
  • 1 <= board[i].length <= 200
  • 1 <= word.length <= 10^3
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class Solution {
public:
    bool bt(vector<vector<char> > &board, string word, bool **passed, int i, int j, int m, int n, int len, int nw){
        if(len==nw) return true;
        //up
        if(i-1>=0 && board[i-1][j]==word[len] && !passed[i-1][j]){
                passed[i-1][j] = true;
                if(bt(board, word, passed, i-1, j, m, n, len+1, nw)) return true;
                passed[i-1][j] = false;
        }
        //low
        if(i+1<m && board[i+1][j]==word[len] && !passed[i+1][j]){
                passed[i+1][j] = true;
                if(bt(board, word, passed, i+1, j, m, n, len+1, nw)) return true;
                passed[i+1][j] = false;
        }   
        //left
        if(j-1>=0 && board[i][j-1]==word[len] && !passed[i][j-1]){
                passed[i][j-1] = true;
                if(bt(board, word, passed, i, j-1, m, n, len+1, nw)) return true;
                passed[i][j-1] = false;
        }       
        //left
        if(j+1<n && board[i][j+1]==word[len] && !passed[i][j+1]){
                passed[i][j+1] = true;
                if(bt(board, word, passed, i, j+1, m, n, len+1, nw)) return true;
                passed[i][j+1] = false;
        }   
        return false;
    }
    
    bool exist(vector<vector<char> > &board, string word) {
        int m = board.size();
        if(m==0) return false;
        int n = board[0].size();
        int nw = word.size();
        if(n==0||nw==0) return false;
        bool **passed = new bool *[m];
        for(int i=0; i<m; ++i){
          passed[i] = new bool[n];
        }

        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(board[i][j]==word[0]&&!passed[i][j]){
                    passed[i][j] = true;
                    if(bt(board, word, passed, i, j, m, n, 1, nw)) return true;
                    passed[i][j] = false;
                }
            }
        }
        return false;
    }
};

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(word.empty()) return true;
        int m = board.size();
        if(m==0) return false;
        int n = board[0].size();
        if(n==0) return false;
        
        string cur;
        int sz = word.size();
        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                char c = word[0];
                if(board[i][j]==c){
                    cur += c;
                    board[i][j] = '.';
                    if(bt(board, cur, word, 1, i, j, sz, m, n)) return true;
                    board[i][j] = c;
                }
            }
        }
        return false;
    }
    
    bool bt(vector<vector<char>>& board, string cur, string word, int len, int i, int j, int sz, int m, int n){
        if(len==sz) return true;
        char c = word[len];
        string cur1 = cur;
        cur1 += c;
        if(i-1>=0 && board[i-1][j]==c){
            board[i-1][j] = '.';
            if(bt(board, cur1, word, len+1, i-1, j, sz, m, n)) return true;
            board[i-1][j] = c;
        }
        if(i+1<m && board[i+1][j]==c){
            board[i+1][j] = '.';
            if(bt(board, cur1, word, len+1, i+1, j, sz, m, n)) return true;
            board[i+1][j] = c;
        }
        if(j-1>=0 && board[i][j-1]==c){
            board[i][j-1] = '.';
            if(bt(board, cur1, word, len+1, i, j-1, sz, m, n)) return true;
            board[i][j-1] = c;
        }
        if(j+1<n && board[i][j+1]==c){
            board[i][j+1] = '.';
            if(bt(board, cur1, word, len+1, i, j+1, sz, m, n)) return true;
            board[i][j+1] = c;
        }        
        return false;
    }
};

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size();
        if(m==0) return false;
        int n = board[0].size();
        
        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(dfs(board, word, 0, word.size(), i, j, m, n)) return true;
            }
        }
        return false;
    }
    
    bool dfs(vector<vector<char>>& board, string word, int pos, int sz, int i, int j, int m, int n){

            if(pos<sz && board[i][j]==word[pos]){
                if(pos+1==sz) return true;
                board[i][j]='#';
                if(i-1>=0) if(dfs(board, word, pos+1, sz, i-1, j, m, n)) return true;
                if(i+1<m)  if(dfs(board, word, pos+1, sz, i+1, j, m, n)) return true;
                if(j-1>=0) if(dfs(board, word, pos+1, sz, i, j-1, m, n)) return true;
                if(j+1<n)  if(dfs(board, word, pos+1, sz, i, j+1, m, n)) return true;
                board[i][j]=word[pos];
            }

        return false;
    }
};


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class Solution {
    int m, n;
    int[][] dirs = {{-1,0},{1,0},{0,-1},{0,1}};
    public boolean exist(char[][] board, String word) {
        m = board.length;
        n = board[0].length;
        
        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(board[i][j]==word.charAt(0)){
                    board[i][j] = '.';
                    if(find(board, word, 1, i, j)) return true;
                    board[i][j] = word.charAt(0);
                }
            }
        }
        return false;
    }
    
    boolean find(char[][] board, String word, int p, int i, int j){
        if(p==word.length()) return true;
        for(int[] d : dirs){
            int ii = i+d[0], jj = j+d[1];
            if(ii>=0 && ii<m && jj>=0 && jj<n && board[ii][jj]==word.charAt(p)){
                board[ii][jj] = '.';
                if(find(board, word, p+1, ii, jj)) return true;
                board[ii][jj] = word.charAt(p);
            }
        }
        return false;
    }
}
at No comments:

LeetCode [77] Combinations

 77. Combinations

Medium

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

You may return the answer in any order.

 

Example 1:

Input: n = 4, k = 2
Output:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

Example 2:

Input: n = 1, k = 1
Output: [[1]]

 

Constraints:

  • 1 <= n <= 20
  • 1 <= k <= n
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//C++
class Solution {
public:
    vector<vector<int>> combine(int n, int k) {
        vector<vector<int>> res;
        if(!n||!k) return res;
        vector<int> cur;
        bt(n, k, res, cur, 0, 0);
    }
    void bt(int n, int k, vector<vector<int>> &res, vector<int> cur, int pos, int size){
        if(size==k){
            res.push_back(cur);
        }else if(pos<=n-k+size){
            bt(n, k, res, cur, pos+1, size);
            cur.push_back(pos+1);
            bt(n, k, res, cur, pos+1, size+1);
        }
    }
};

//Java
class Solution {
    List<List<Integer>> ret = new ArrayList<>();
    public List<List<Integer>> combine(int n, int k) {
        helper(new ArrayList<>(), n, k, 1);
        return ret;
    }

    void helper(List<Integer> cur, int n, int k, int i) {
        if(cur.size()==k){
            ret.add(new ArrayList<Integer>(cur));
        }else{
            for(int j=i; j<=n; ++j){
                if(n-j+1+cur.size()<k) break;
                cur.add(j);
                helper(cur, n, k, j+1);
                cur.remove(cur.size()-1);
            }
        }
    }
}
at No comments:

LeetCode [52] N-Queens II


Ref
OJ
at No comments:

LeetCode [51] N-Queens


Ref
[1] https://oj.leetcode.com/problems/n-queens/
OJ
at No comments:

Wednesday, February 25, 2015

LeetCode [47] Permutations II

47. Permutations II
Medium

Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.

 

Example 1:

Input: nums = [1,1,2]
Output:
[[1,1,2],
 [1,2,1],
 [2,1,1]]

Example 2:

Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

 

Constraints:

  • 1 <= nums.length <= 8
  • -10 <= nums[i] <= 10



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class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> ret;
        sort(nums.begin(), nums.end());
        bt(nums, ret, 0, nums.size());
        return ret;
    }

    //nums cannot be reference so that nums[pos+1..n-1] is in increasing order
    void bt(vector<int> nums, vector<vector<int>> &ret, int pos, int n){
        if(pos==n){
            ret.push_back(nums);
        }else{
            for(int i=pos; i<n; ++i){
                if(i>pos && nums[i]==nums[pos]) continue;
                swap(nums[pos], nums[i]);
                bt(nums, ret, pos+1, n);
                //cannot swap back; otherwise nums[pos+1..n-1] will not be in increasing order
            }
        }
    }
};

=========== 

Note
It is important to keep the increasing order of the non-determined portion of the vector, ie., nums[pos+1, n-1], such that we can conveniently skip the duplicate cases by line 17.

An example for the recursion of nums. pos=0. Note that nums[1, 4] are in increasing order.
0 1 2 3 4 -- index
1 2 3 4 5
2 1 3 4 5
3 1 2 4 5
4 1 2 3 5
5 1 2 3 4

If nums is swapped back at line20. nums[1, 4] are no longer in increasing order.
0 1 2 3 4 -- index
1 2 3 4 5
2 1 3 4 5
3 2 1 4 5
4 2 3 1 5
5 2 3 4 1
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