[bky373] 10th week solutions

graph-valid-tree/bky373.java

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+/*
+    time: O(n + m), where n is the number of nodes and m is the number of edges in the graph.
+    space: O(n + m)
+ */
+class Solution {
+
+    public boolean validTree(int n, int[][] edges) {
+
+        if (edges.length != n - 1) {
+            return false;
+        }
+
+        List<List<Integer>> adjList = new ArrayList<>();
+        for (int i = 0; i < n; i++) {
+            adjList.add(new ArrayList<>());
+        }
+        for (int[] edge : edges) {
+            adjList.get(edge[0])
+                   .add(edge[1]);
+            adjList.get(edge[1])
+                   .add(edge[0]);
+        }
+
+        Stack<Integer> stack = new Stack<>();
+        Set<Integer> visited = new HashSet<>();
+        stack.push(0);
+        visited.add(0);
+
+        while (!stack.isEmpty()) {
+            int curr = stack.pop();
+            for (int adj : adjList.get(curr)) {
+                if (visited.contains(adj)) {
+                    continue;
+                }
+                visited.add(adj);
+                stack.push(adj);
+            }
+        }
+
+        return visited.size() == n;
+    }
+}

house-robber-ii/bky373.java

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+/*
+    time: O(N)
+    time: O(1)
+ */
+class Solution {
+
+    public int rob(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+        if (nums.length == 1) {
+            return nums[0];
+        }
+
+        return Math.max(
+                dp(nums, 0, nums.length - 2),
+                dp(nums, 1, nums.length - 1)
+        );
+    }
+
+    private static int dp(int[] nums, int start, int end) {
+        int maxOfOneStepAhead = nums[end];
+        int maxOfTwoStepsAhead = 0;
+
+        for (int i = end - 1; i >= start; --i) {
+            int curr = Math.max(maxOfOneStepAhead, maxOfTwoStepsAhead + nums[i]);
+
+            maxOfTwoStepsAhead = maxOfOneStepAhead;
+            maxOfOneStepAhead = curr;
+        }
+        return maxOfOneStepAhead;
+    }
+}

house-robber/bky373.java

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+/*
+    time: O(N)
+    space: O(1)
+ */
+class Solution {
+
+    public int rob(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+
+        int maxOfTwoStepsAhead = 0;
+        int maxOfOneStepAhead = nums[nums.length - 1];
+
+        for (int i = nums.length - 2; i >= 0; --i) {
+            int curr = Math.max(maxOfOneStepAhead, maxOfTwoStepsAhead + nums[i]);
+
+            maxOfTwoStepsAhead = maxOfOneStepAhead;
+            maxOfOneStepAhead = curr;
+        }
+        return maxOfOneStepAhead;
+    }
+}

longest-palindromic-substring/bky373.java

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+class Solution {
+
+    /*
+     * Approach 1.
+     * time: O(n^3)
+     * space: O(1)
+     */
+    public String longestPalindrome1(String s) {
+        String longest = "";
+        for (int i = 0; i < s.length(); i++) {
+            for (int j = i; j < s.length(); j++) {
+                String ss = s.substring(i, j + 1);
+                if (isPalindrome(ss)) {
+                    if (ss.length() > longest.length()) {
+                        longest = ss;
+                    }
+                }
+            }
+        }
+        return longest;
+    }
+
+    // Approach 1.
+    private boolean isPalindrome(String s) {
+        int left = 0;
+        int right = s.length() - 1;
+        while (left < right) {
+            if (s.charAt(left) != s.charAt(right)) {
+                return false;
+            }
+            left++;
+            right--;
+        }
+        return true;
+    }
+
+    /*
+     * Approach 2-1.
+     * time: 시간 복잡도는 평균적으로 O(n)이고, palindrome 의 길이가 n 에 가까워지면 시간 복잡도 역시 O(n^2) 에 가까워 진다.
+     * space: O(1)
+     */
+    class Solution {
+
+        public String longestPalindrome(String s) {
+            int maxStart = 0;
+            int maxEnd = 0;
+
+            for (int i = 0; i < s.length(); i++) {
+                int start = i;
+                int end = i;
+                while (0 <= start && end < s.length() && s.charAt(start) == s.charAt(end)) {
+                    if (maxEnd - maxStart < end - start) {
+                        maxStart = start;
+                        maxEnd = end;
+                    }
+                    start--;
+                    end++;
+                }
+
+                start = i;
+                end = i + 1;
+                while (0 <= start && end < s.length() && s.charAt(start) == s.charAt(end)) {
+                    if (maxEnd - maxStart < end - start) {
+                        maxStart = start;
+                        maxEnd = end;
+                    }
+                    start--;
+                    end++;
+                }
+            }
+            return s.substring(maxStart, maxEnd + 1);
+        }
+    }
+
+    /*
+     * Approach 2-2.
+     * time: 시간 복잡도는 평균적으로 O(n)이고, palindrome 의 길이가 n 에 가까워지면 시간 복잡도 역시 O(n^2) 에 가까워 진다.
+     * space: O(1)
+     */
+    class Solution {
+        int maxStart = 0;
+        int maxEnd = 0;
+
+        public String longestPalindrome(String s) {
+            for (int i = 0; i < s.length(); i++) {
+                calculateMaxLength(i, i, s);
+                calculateMaxLength(i, i+1, s);
+            }
+            return s.substring(maxStart, maxEnd + 1);
+        }
+
+        public void calculateMaxLength(int start, int end, String s){
+            while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
+                if (this.maxEnd - this.maxStart < end - start) {
+                    this.maxStart = start;
+                    this.maxEnd = end;
+                }
+                start--;
+                end++;
+            }
+        }
+
+    }
+}

number-of-connected-components-in-an-undirected-graph/bky373.java

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+/*
+    time: O(n+m), where n is the number of nodes and m is the number of edges in the graph.
+    space: O(n)
+ */
+class Solution {
+
+    public int countComponents(int n, int[][] edges) {
+        boolean[] visited = new boolean[n];
+
+        Map<Integer, List<Integer>> map = new HashMap<>();
+        for (int[] edge : edges) {
+            map.computeIfAbsent(edge[0], k -> new ArrayList<>())
+               .add(edge[1]);
+            map.computeIfAbsent(edge[1], k -> new ArrayList<>())
+               .add(edge[0]);
+        }
+
+        int ans = 0;
+        for (int i = 0; i < n; i++) {
+            if (visited[i]) {
+                continue;
+            }
+            dfs(i, map, visited);
+            ans++;
+        }
+        return ans;
+    }
+
+    public void dfs(int k, Map<Integer, List<Integer>> map, boolean[] visited) {
+        if (visited[k]) {
+            return;
+        }
+        visited[k] = true;
+        if (!map.containsKey(k)) {
+            return;
+        }
+        List<Integer> values = map.get(k);
+        for (int v : values) {
+            dfs(v, map, visited);
+        }
+    }
+}