September LeetCoding Challenge, Day 8: Sum of Root To Leaf Binary Numbers

September 9, 2020

This is part of a series of posts about the September LeetCoding Challenge. Check the first post for more information.

The problem for September 8 is Sum of Root to Leaf Binary Numbers. You’re given a binary tree in which each node has value 0 or 1. In this case, each path from the root to a leaf represents a binary number starting with the most significant bit. For example, a path from root to leaf \(0 \rightarrow 1 \rightarrow 1 \rightarrow 0 \rightarrow 1\) represents the number \(01101\) in binary, or \(13\) in decimal. We’re interested in the sum of numbers produced from the paths from the root to every leaf in the tree.

We’re told that the number of nodes in the tree doesn’t exceed 1000 and that the sum will not exceed \(2^{31} - 1\). Therefore, we can do a DFS on the tree while keeping track of the number at each path, accumulating the current number whenever we reach a leaf. Since the path follows the significance of the bits in the resulting number, we can keep track of the current number while traversing the tree by multiplying the current number by 2 whenever we go either left or right, and sum the value at the node we’re visiting. The following is an implementation of this idea:

class Solution {
private:
  void dfs(TreeNode* curr, int curr_value, int& curr_sum) {
    if (curr == nullptr)
      return;
    bool is_leaf = curr->left == nullptr && curr->right == nullptr;
    curr_value = curr_value * 2 + curr->val;
    if (is_leaf)
      curr_sum += curr_value;
    if (curr->left != nullptr)
      dfs(curr->left, curr_value, curr_sum);
    if (curr->right != nullptr)
      dfs(curr->right, curr_value, curr_sum);
  }

public:
  int sumRootToLeaf(TreeNode* root) {
    int ans = 0;
    dfs(root, 0, ans);
    return ans;
  }
};

September LeetCoding Challenge, Day 7: Word Pattern

September 8, 2020

This is part of a series of posts about the September LeetCoding Challenge. Check the first post for more information.

The problem for September 7 is Word Pattern. You’re given two strings. One of them consists of lowercase letters and the other consists of lowercase letters separated by spaces. You want to find out if a bijection exists between the characters of the first string and the words in the second string. In other words, you want to return true if there is a one-to-one correspondence between the characters of the first string and the characters of the second string, and false otherwise.

If the number of words is different from the number of characters, then you can be sure that no bijection exists. If the number of words is the same, we can go word by word and check if a word happens to be mapped to different characters. If it does, then no bijection exists. This is still not sufficient to determine a bijection, since the same character can still be mapped to different words. In order for a bijection to exist, the number of different words must be equal to the number of different characters. Combining all those checks lets us determine if there is a one-to-one correspondence between characters and words. The following is an implementation of the previous idea:

class Solution {
private:
  vector<string> split(string str) {
    istringstream ss(str);
    vector<string> ans;
    string curr;
    while (ss >> curr)
      ans.push_back(curr);
    return ans;
  }

public:
  bool wordPattern(string pattern, string str) {
    vector<string> words = split(str);
    if (words.size() != pattern.size())
      return false;
    unordered_map<string, char> pat;
    set<char> used;
    int N = words.size();
    for (int i = 0; i < N; ++i) {
      if (pat.find(words[i]) != pat.end() && pat[words[i]] != pattern[i])
        return false;
      pat[words[i]] = pattern[i];
      used.insert(pattern[i]);
    }
    return pat.size() == used.size();
  }
};

September LeetCoding Challenge, Day 6: Image Overlap

September 8, 2020

This is part of a series of posts about the September LeetCoding Challenge. Check the first post for more information.

The problem for September 6 is Image Overlap. You’re given two images represented as binary, square matrices, of the same size. You want to translate one of the images by sliding it left, right, up or down any number of units such that, when placed on top of the other image, the number of 1s that overlap in both images is maximized. The length of the side of the images is at most 30.

The length of the side of the images is small enough for us to try all possible translations in \(\mathcal{O}(n^4)\). The following is an implementation of that strategy:

class Solution {
public:
  int largestOverlap(vector<vector<int>>& A, vector<vector<int>>& B) {
    int L = A.size(), ans = 0;
    for (int di = -L + 1; di < L; ++di) {
      for (int dj = -L + 1; dj < L; ++dj) {
        int curr = 0;
        for (int i = 0; i < L; ++i) {
          for (int j = 0; j < L; ++j) {
            int bi = i + di, bj = j + dj, vb = 0;
            if (bi >= 0 && bi < L && bj >= 0 && bj < L)
              vb = B[bi][bj];
            if (A[i][j] == 1 && vb == 1)
              curr++;
          }
        }
        ans = max(ans, curr);
      }
    }
    return ans;
  }
};

September LeetCoding Challenge, Day 5: All Elements in Two Binary Search Trees

September 6, 2020

This is part of a series of posts about the September LeetCoding Challenge. Check the first post for more information.

The problem for September 5 is All Elements in Two Binary Search Trees. We’re interested in, given two binary search trees of integers, returning a sorted list of all the integers from both trees.

Since we’re dealing with binary search trees here, each node is guaranteed to store a key greater than all the keys in the node’s left subtree and less than those in its right subtree. As such, an in-order traversal of the tree is guaranteed to produce a sorted list of its elements. We can do an in-order traversal of a binary tree with \(n\) nodes in \(\mathcal{O}(n)\). Having two sorted lists of size \(a\) and \(b\), we can merge them in a new sorted list in \(\mathcal{O}(a + b)\). With these two building blocks, we can produce an algorithm to return the sorted list of all integers of both trees in \(\mathcal{O}(n + m)\), \(n\) being the number of nodes in the first tree, and \(m\) being the number of nodes in the second tree. The following is an implementation of that idea:

class Solution {
private:
  void in_order_traverse(TreeNode* curr, vector<int>& list) {
    if (curr != nullptr) {
      in_order_traverse(curr->left, list);
      list.push_back(curr->val);
      in_order_traverse(curr->right, list);
    }
  }

public:
  vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
    vector<int> list1, list2;
    in_order_traverse(root1, list1);
    in_order_traverse(root2, list2);
    vector<int> ans;
    int N1 = list1.size(), N2 = list2.size(), i1 = 0, i2 = 0;
    while (i1 < N1 || i2 < N2) {
      if (i1 >= N1)
        ans.push_back(list2[i2++]);
      else if (i2 >= N2)
        ans.push_back(list1[i1++]);
      else if (list1[i1] < list2[i2])
        ans.push_back(list1[i1++]);
      else
        ans.push_back(list2[i2++]);
    }
    return ans;
  }
};

September LeetCoding Challenge, Day 4: Partition Labels

September 5, 2020

This is part of a series of posts about the September LeetCoding Challenge. Check the first post for more information.

The problem for September 4 is Partition Labels. According to the problem statement, we have a string \(S\) of lowercase English characters which we want to partition in as many parts as possible. The catch is that each letter must appear in at most one part. Each part is a substring, not a subsequence of string \(S\). For example, given string \(S\) equal to "ababcbacadefegdehijhklij", the valid partition that produces most parts is "ababcbaca", "defegde" and "hijhklij".

Once we select a given character to belong to a part, then all characters with the same letter as the chosen character must belong to that part. In sum, each part must go as far as the last occurrence of each letter in the part. We can solve this in \(\mathcal{O}(|S|)\) by first identifying the indices of the last occurrences of each letter, and then greedily collect characters for each part until the previous restriction is satisfied. The following is an implementation of that strategy:

class Solution {
public:
  vector<int> partitionLabels(string S) {
    vector<int> ans;
    unordered_map<char, int> last;
    int N = S.size();
    for (int i = 0; i < N; ++i)
      last[S[i]] = i;
    int curr = 0;
    int prev = -1;
    for (int i = 0; i < N; ++i) {
      curr = max(curr, last[S[i]]);
      if (curr == i) {
        ans.push_back(curr - prev);
        prev = curr;
      }
    }
    return ans;
  }
};