Merge pull request #518 from TonyKim9401/main

[Tony] WEEK 09 Solutions

find-minimum-in-rotated-sorted-array/TonyKim9401.java

@@ -0,0 +1,25 @@
+// TC: O(log n)
+// Using binary search, it takes `log n` time complexity, n indicates the length of the given array nums
+// SC: O(1)
+// constant space occupation
+class Solution {
+    public int findMin(int[] nums) {
+
+        int start = 0;
+        int end = nums.length - 1;
+        int min = Integer.MAX_VALUE;
+
+        while (start <= end) {
+            int mid = start + (end - start) / 2;
+
+            if (nums[start] <= nums[mid]) {
+                min = Math.min(min, nums[start]);
+                start = mid + 1;
+            } else if (nums[mid] <= nums[end]) {
+                min = Math.min(min, nums[mid]);
+                end = mid - 1;
+            }
+        }
+        return min;
+    }
+}

linked-list-cycle/TonyKim9401.java

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+// TC: O(n)
+// SC: O(1)
+public class Solution {
+    public boolean hasCycle(ListNode head) {
+        ListNode slow = head;
+        ListNode fast = head;
+
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+            if (slow == fast) return true;
+        }
+        return false;
+    }
+}

maximum-subarray/TonyKim9401.java

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+// TC: O(n)
+// visit all elements once for each
+// SC: O(1)
+// constant space occupation
+class Solution {
+    public int maxSubArray(int[] nums) {
+        int total = 0;
+        int output = nums[0];
+
+        for (int num : nums) {
+            if (total < 0) total = 0;
+            total += num;
+            output = total > output ? total : output;
+        }
+        return output;
+    }
+}

minimum-window-substring/TonyKim9401.java

@@ -0,0 +1,52 @@
+// TC: O(n)
+// using two pointer lef and right, it visits all elements only once each.
+// SC: O(n + m)
+// 2 hashmap used for checking the given Strings s and t, n is the size of s, m is the size of m
+class Solution {
+    public String minWindow(String s, String t) {
+        Map<Character, Integer> map = new HashMap<>();
+
+        for (char c : t.toCharArray()) {
+            map.put(c, map.getOrDefault(c, 0) + 1);
+        }
+
+        int required = map.size();
+        int formed = 0;
+
+        int left = 0;
+        int right = 0;
+        int[] ans = {-1, 0, 0};
+
+        Map<Character, Integer> windowCounts = new HashMap<>();
+
+        while (right < s.length()) {
+            char c = s.charAt(right);
+            windowCounts.put(c, windowCounts.getOrDefault(c, 0) + 1);
+
+            if (map.containsKey(c) &&
+                    windowCounts.get(c).intValue() == map.get(c).intValue()) {
+                formed += 1;
+            }
+
+            while (left <= right && formed == required) {
+                c = s.charAt(left);
+
+                if (ans[0] == -1 || right - left + 1 < ans[0]) {
+                    ans[0] = right - left + 1;
+                    ans[1] = left;
+                    ans[2] = right;
+                }
+
+                windowCounts.put(c, windowCounts.get(c) - 1);
+                if (map.containsKey(c) &&
+                        windowCounts.get(c).intValue() < map.get(c).intValue()) {
+                    formed -= 1;
+                }
+
+                left += 1;
+            }
+            right += 1;
+        }
+        return ans[0] == -1 ? "" : s.substring(ans[1], ans[2] + 1);
+    }
+}

pacific-atlantic-water-flow/TonyKim9401.java

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+// TC: O(n * m)
+// visit all elements
+// SC: O(n * m)
+// create result from all elements
+class Solution {
+    public List<List<Integer>> pacificAtlantic(int[][] heights) {
+        List<List<Integer>> output = new ArrayList<>();
+
+        if (heights.length == 0 || heights[0].length == 0) return output;
+
+        int rows = heights.length;
+        int cols = heights[0].length;
+
+        boolean[][] pac = new boolean[rows][cols];
+        boolean[][] atl = new boolean[rows][cols];
+
+        for (int j = 0; j < rows; j++) {
+            dfs(j, 0, pac, heights[j][0], heights);
+            dfs(j, cols-1, atl, heights[j][cols-1], heights);
+        }
+
+        for (int i = 0; i < cols; i++) {
+            dfs(0, i, pac, heights[0][i], heights);
+            dfs(rows-1, i, atl, heights[rows-1][i], heights);
+        }
+
+        for (int i = 0; i < rows; i++) {
+            for (int j = 0; j < cols; j++) {
+                if (pac[i][j] && atl[i][j]) output.add(List.of(i,j));
+            }
+        }
+        return output;
+    }
+
+    private void dfs(int i, int j, boolean[][] visit, int preValue, int[][] heights) {
+        if (i < 0 || j < 0 || i == heights.length || j == heights[0].length || visit[i][j] || preValue > heights[i][j]) return;
+
+        visit[i][j] = true;
+        dfs(i + 1, j, visit, heights[i][j], heights);
+        dfs(i - 1, j, visit, heights[i][j], heights);
+        dfs(i, j + 1, visit, heights[i][j], heights);
+        dfs(i, j - 1, visit, heights[i][j], heights);
+    }
+}