Merge pull request #978 from ekgns33/main

[byteho0n] Week 8

Author: TonyKim9401

clone-graph/ekgns33.java

@@ -0,0 +1,42 @@
+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public List<Node> neighbors;
+    public Node() {
+        val = 0;
+        neighbors = new ArrayList<Node>();
+    }
+    public Node(int _val) {
+        val = _val;
+        neighbors = new ArrayList<Node>();
+    }
+    public Node(int _val, ArrayList<Node> _neighbors) {
+        val = _val;
+        neighbors = _neighbors;
+    }
+}
+*/
+
+class Solution {
+  public Node cloneGraph(Node node) {
+    Map<Integer, Node> nodeMap = new HashMap<>();
+    if(node == null) return null;
+    boolean[] visited = new boolean[101];
+    dfsHelper(nodeMap, node, visited);
+    return nodeMap.get(1);
+  }
+
+  private void dfsHelper(Map<Integer, Node> map, Node node, boolean[] visited) {
+    if(!map.containsKey(node.val)) {
+      map.put(node.val, new Node(node.val));
+    }
+    visited[node.val] = true;
+    Node cur = map.get(node.val);
+    for(Node adjacent : node.neighbors) {
+      map.putIfAbsent(adjacent.val, new Node(adjacent.val));
+      cur.neighbors.add(map.get(adjacent.val));
+      if(!visited[adjacent.val]) dfsHelper(map, adjacent, visited);
+    }
+  }
+}

longest-common-subsequence/ekgns33.java

@@ -0,0 +1,76 @@
+/**
+ input : string text1, text2
+ output : length of lcs else 0
+ constraints:
+ 1) both input strings are not empty
+ 2) input strings are consist of lowercase eng. letters
+
+ solution 1) brute force
+
+ nested for loop
+ tc : O(n^2 * m) where n is min (len(text1), len(text2)), m is max
+ sc : O(1)
+
+ solution 2) dp
+
+ at each step we have 3 conditions
+
+ 1. current characters are same. compare next characters
+ abcde. ith index
+ ^
+ abc.  jth index
+ ^
+ then go to next and maximum length also increases
+ move (i + 1, j+1)
+ 2, 3. current characters are different. move i or j to compare
+
+ abcde
+ ^
+ abc
+ ^
+ move (i +1, j) or (i, j+1)
+
+
+ let dp[i][j] max length of lcs
+
+ a b c d e
+ a 1 1 1 1 1
+ c 1 1 2 2 2
+ e 1 1 2 2 3
+
+ a b d e f g h
+ a 1 1 1 1 1 1 1
+ f 1 1 1 1 2 2 2
+ h 1 1 1 1 2 2 3
+ d 1 1 2 2 2 2 3
+
+ if equals
+ dp[i][j] = dp[i-1][j-1] + 1
+ else
+ dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
+ tc : O(mn)
+ sc : O(mn) > O(min(m, n)) if optimize space to 2-Row array
+ */
+class Solution {
+  public int longestCommonSubsequence(String text1, String text2) {
+    int m = text1.length(), n = text2.length();
+    if(m < n) {
+      return longestCommonSubsequence(text2, text1);
+    }
+    int[][] dp = new int[2][n+1];
+    int prevRow = 0;
+    int curRow = 1;
+    for(int i = 1; i <= m; i++) {
+      for(int j = 1; j <= n; j++) {
+        if(text1.charAt(i-1) == text2.charAt(j-1)) {
+          dp[curRow][j] = dp[prevRow][j-1] + 1;
+        } else {
+          dp[curRow][j] = Math.max(dp[prevRow][j], dp[curRow][j-1]);
+        }
+      }
+      curRow = prevRow;
+      prevRow = (prevRow + 1) % 2;
+    }
+    return dp[prevRow][n];
+  }
+}

longest-repeating-character-replacement/ekgns33.java

@@ -0,0 +1,119 @@
+/**
+
+ input : String s
+ output : after k times operation, return length of longest substring
+ that consist of same letters
+
+ example
+ AAAA << 4
+ AAAB << AAA << 3
+
+ constraints:
+ 1) input string can be empty?
+ nope. at least one letter
+ 2) string s contains whitespace? non-alphabetical chars?
+ nope. only uppercase English letters
+ 3) range of k?
+ [0, n], when n is the length of string s
+
+ ABAB
+ AAAB
+ AAAA
+
+ AABABBA
+ AAAABBA
+ AABBBBA
+
+ AABABBA
+ l
+ r
+ count : 1
+ 4
+
+ solution 1: brute force
+ ds : array
+ algo : for loop
+
+ nested for loop
+ iterate through the string s from index i = 0 to n
+ set current character as 'target'
+ set count = 0
+ iterate through the string s from index j = i + 1 to n
+ if next character is identical to 'target'
+ continue;
+ else
+ increase count;
+
+ if count > k
+ break and save length
+
+ if count <=k compare length with max
+ ABAB
+ i
+ j
+ target = A
+ count = 1
+ ABAB << count = 2, length = 4
+ tc : O(n^2)
+ sc : O(1)
+
+ solution 2: two pointer
+
+ AABCABBA
+ l
+ r
+ A : 1
+ B : 1
+ C : 1
+ count = 1
+ res = 4
+ set two pointer l and r
+ set count to 0;
+ set maxFrequency character s[0];
+
+ move r pointer to right
+ if current char is not maxFreqChar
+ count + 1;
+ else
+ continue;
+
+ if count > k
+ while count <= k
+ move leftpointer right;
+ update frequency
+ set max Frequency character at each iteration O(26)
+ update count
+
+ tc: O(n)
+ sc : O(26) ~= O(1)
+ */
+
+class Solution {
+  public int characterReplacement(String s, int k) {
+    int l = 0, r = 1, maxFreq = 1, res = 1;
+    int n = s.length();
+    int[] freq = new int[26];
+    // base case;
+    char maxFreqChar = s.charAt(0);
+    freq[maxFreqChar - 'A']++;
+    // loop
+    while (r < n) {
+      char next = s.charAt(r);
+      maxFreq = Math.max(maxFreq, ++freq[next - 'A']);
+
+      while(r - l + 1 - maxFreq > k) {
+        freq[s.charAt(l) - 'A']--;
+        l++;
+        for(int i = 0; i <26; i++) {
+          if(freq[i] > maxFreq) {
+            maxFreq = freq[i];
+          }
+        }
+      }
+      res = Math.max(res, r - l+1);
+      r++;
+    }
+
+    return res;
+  }
+}

number-of-1-bits/ekgns33.java

@@ -0,0 +1,19 @@
+class Solution {
+  public int hammingWeight(int n) {
+    int curN = n;
+    int cnt = 0;
+    while(curN > 0) {
+      if(curN %2 == 1) {
+        cnt++;
+      }
+      curN /= 2;
+    }
+    return cnt;
+  }
+}
+/**
+
+ 1 2 3 4 5 6 7 8 9 10 11
+ 0 1 2 1 2 2 3 1 2  2  3
+
+ */

sum-of-two-integers/ekgns33.java

@@ -0,0 +1,28 @@
+/*
+
+bit manipulation
+
+find carry of each binary digits sum
+example
+
+  1 0 1 1
+ +0 0 1 0
+-----------
+      0 1
+      ^carry 1  (1<<2)
+    1
+   1 
+
+
+ */
+class Solution {
+  public int getSum(int a, int b) {
+    int carryBitSet;
+    while(b != 0) {
+      carryBitSet = a & b;
+      a = a ^ b;
+      b = carryBitSet << 1;
+    }
+    return a;
+  }
+}