[byteho0n] - Week 7

longest-substring-without-repeating-characters/ekgns33.java

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+/**
+ input : String s
+ output: length of the longest substring without repeating chararcters
+
+ example)
+
+ abccba >> abc, cba >> 3
+ abcdeffg >> abcdef >> 6
+ constraints
+ 1) length of string s?
+ [0, 5 * 10^4]
+ 2) alphanumeric characters?
+ english, digit, symbol, space. >> 256 ascii characters
+  else >> hashmap
+
+ solution1)
+
+ iterate through string from index i = 0 to n-1
+ iterate through string s from index j = i + 1, j to n
+ if duplicate character comes up, then compare length
+ tc : O(n^2)
+ sc : O(1)
+
+ solution 2) sliding window
+
+ ds : array or hashmap << when character range is ascii, we can use array instead
+
+ use two pointer l and r
+ read until there is no repeating character in range [l, r]
+ if duplicate exists
+ move l until every character is unique
+ ex)
+
+ abaabcdef
+ l
+ r
+ ^^    ^
+ tc : O(n)
+ sc : O(1)
+
+
+ */
+
+class Solution {
+  public int lengthOfLongestSubstring(String s) {
+    int[] freq = new int[256];
+    int l = 0, r = 0;
+    int n = s.length();
+    int maxLength = 0;
+    if(n <= 1) return n;
+
+    while(r < n && l <= r) {
+      char curC = s.charAt(r);
+      freq[curC]++;
+      while(l < r && freq[curC] > 1) {
+        char prevC = s.charAt(l);
+        freq[prevC]--;
+        l++;
+      }
+      maxLength = Math.max(maxLength, r - l + 1);
+      r++;
+    }
+    maxLength = Math.max(maxLength, r- l);
+    return maxLength;
+  }
+}

number-of-islands/ekgns33.java

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+/**
+ input : 2d matrix filled with '1', '0'
+ output : number of islands
+
+ 1 indicates land, 0 means water.
+ if lands are surrounded by water we call it island
+
+ example
+
+ 1 1 1.  >> 3 islands
+ 0 0 0
+ 1 0 1
+
+ 1 1 1 1 0
+ 1 1 0 1 0 >> 1 island
+ 0 0 0 0 0
+
+ constraints:
+ 1) m, n range
+ [1, 300]
+ 2) can we change the input grid cell?
+ there is no restriction.
+
+ solution 1)
+
+ ds : array
+ algo : bfs + in-place
+
+ read every cell in grid.
+ if cell is '1'
+ do bfs, change every visiting cell into '2' << if changing input is allowed.
+ increase count
+
+ tc : O(mn), visiting each cell exactly once.
+ sc : O(mn) for queue
+
+ optimize?? maintain visited matrix, set or something else..
+ if we are allowed to modify given input matrix
+ we don't have to maintain visited matrix, but simply change values
+ */
+class Solution {
+  public int numIslands(char[][] grid) {
+    int m = grid.length, n = grid[0].length;
+    int numberOfIslands = 0;
+    for(int i = 0; i < m; i++) {
+      for(int j = 0; j < n; j++) {
+        if(grid[i][j] == '1') {
+          bfsHelper(grid, i, j, m, n);
+          numberOfIslands++;
+        }
+      }
+    }
+    return numberOfIslands;
+  }
+  private static int[][] directions = {
+      {1, 0}, {-1, 0}, {0, 1}, {0, -1}
+  };
+
+  private void bfsHelper(char[][] grid, int x, int y, int m, int n) {
+    Queue<int[]> queue = new LinkedList<>();
+    queue.add(new int[]{x,y});
+    grid[x][y] = '2';
+    while(!queue.isEmpty()) {
+      int[] cur = queue.poll();
+      for(int[] direction : directions) {
+        int nx = direction[0] + cur[0];
+        int ny = direction[1] + cur[1];
+        if(nx < 0 || nx >= m || ny < 0 || ny >= n || grid[nx][ny] != '1') continue;
+        grid[nx][ny] = '2';
+        queue.add(new int[]{nx, ny});
+      }
+    }
+  }
+}

reverse-linked-list/ekgns33.java

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+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ *
+ *
+ * input : head of linked list
+ * output : reversed linked list
+ *
+ * solution1)
+ * set fast pointer and slow pointer
+ * iterate through the linked list
+ *  set fast pointer's next to slow pointer
+ *
+ * tc : O(n)
+ * sc : O(1)
+ */
+class Solution {
+  public ListNode reverseList(ListNode head) {
+    if(head == null) return head;
+    ListNode curr = head;
+    ListNode prev = null;
+    while(curr.next != null) {
+      ListNode next = curr.next;
+      curr.next = prev;
+      prev = curr;
+      curr = next;
+    }
+    curr.next = prev;
+    return curr;
+  }
+}

set-matrix-zeroes/ekgns33.java

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+/**
+
+ sol1 ) brute force
+ save zero points, replace row
+
+ read O(mn)
+ save zero points
+ replace row,cols
+ tc : O(mn)
+ sc : O(mn)
+
+ sol2) use bitwise
+ length of matrix < integer range
+ use bit wise
+ tc : O(mn)
+ sc : O(m + n)
+
+ sol3) use first row, col as memo
+ read matrix, check to row or col if zero
+
+ replace if row, col is marked as zero
+ tc : O(mn)
+ sc : O(1) in-place
+ */
+class Solution {
+  public void setZeroes(int[][] matrix) {
+    int m = matrix.length;
+    int n = matrix[0].length;
+    boolean firstRowZero = false;
+    boolean firstColZero = false;
+    for(int i = 0; i < m; i++) {
+      for(int j = 0; j < n; j++) {
+        if(matrix[i][j] == 0) {
+          if(i == 0) firstRowZero = true;
+          if(j == 0) firstColZero = true;
+          matrix[i][0] = 0;
+          matrix[0][j] = 0;
+        }
+      }
+    }
+    for(int i = 1; i <m; i++) {
+      for(int j = 1; j < n; j++) {
+        if(matrix[i][0] == 0 || matrix[0][j] == 0) {
+          matrix[i][j] = 0;
+        }
+      }
+    }
+    if(firstRowZero) {
+      Arrays.fill(matrix[0], 0);
+    }
+    if(firstColZero) {
+      for(int j = 0; j <m; j++) {
+        matrix[j][0] = 0;
+      }
+    }
+
+
+  }
+}

unique-paths/ekgns33.java

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+/*
+input : 2D matrix 
+output : the number of possible unique paths
+constraints : 
+1) range of m and n
+[1, 100]
+
+solution 1) dfs?
+move right or move down.
+tc : O(mn)
+sc : O(mn)
+
+solution 2) dp + tabulation
+let dp[i][j] the number of unique paths from starting point.
+    there are only 2 way to get coordinate i, j 
+        1) from i-1, j
+        2) from i, j-1
+dp[i][j] = dp[i-1][j] + dp[i][j-1]
+
+tc : O(mn)
+sc : O(mn)
+ */
+
+class Solution {
+  public int uniquePaths(int m, int n) {
+    int[][] dp = new int[m][n];
+    for(int i = 0; i < m; i++) {
+      for(int j = 0; j < n; j++) {
+        if(i == 0 || j == 0) {
+          dp[i][j] = 1;
+          continue;
+        }
+        dp[i][j] = dp[i][j-1] + dp[i-1][j];
+      }
+    }
+    return dp[m-1][n-1];
+  }
+}