[byteho0n] Week4

coin-change/ekgns33.java

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+/*
+input : array of integers each element represents coins, single integer amount
+output : fewest number of coin to make up the amount
+constraint
+1) elements are positive?
+yes
+2) coins are in integer range?
+yes
+3) range of amoutn
+integer range?
+[0, 10^4]
+4) at least one valid answer exists?
+nope. if there no exist than return -1
+
+edge. case
+1) if amount == 0 return 0;
+
+ solution 1) top-down approach
+
+amount - each coin
+    until amount == 0
+    return min
+tc : O(n * k) when n is amount, k is unique coin numbers
+sc : O(n) call stack
+
+solution 2) bottom-up
+tc : O(n*k)
+sc : O(n)
+let dp[i] the minimum number of coins to make up amount i
+
+ */
+class Solution {
+  public int coinChange(int[] coins, int amount) {
+    //edge
+    if(amount == 0) return 0;
+    int[] dp = new int[amount+1];
+    dp[0] = 0;
+    for(int i = 1; i<= amount; i++) {
+      dp[i] = amount+1;
+      for(int coin: coins) {
+        if(i - coin >= 0) {
+          dp[i] = Math.min(dp[i], dp[i-coin] + 1);
+        }
+      }
+    }
+    return dp[amount] == amount+1 ? -1 : dp[amount];
+  }
+}

merge-two-sorted-lists/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 : two sorted integer linked list
+ output : merged single linked list with sorted order
+ constraints:
+ 1) both linked list can be empty?
+ yes
+ 2) both given lists are sorted?
+ yes
+ edge :
+ 1) if both list are empty return null
+ 2) if one of input lists is empty return the other one
+
+ solution 1)
+ using two pointer
+
+ compare the current pointer's value.
+ get the smaller one, set to the next node
+
+ until both pointer reach the end
+
+ tc : O(n) sc : O(1)
+ */
+class Solution {
+  public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+    ListNode dummy = new ListNode();
+    ListNode p1 = list1;
+    ListNode p2 = list2;
+    ListNode currNode = dummy;
+    while(p1 != null && p2 != null) {
+      if(p1.val < p2.val){
+        currNode.next = p1;
+        p1 = p1.next;
+      } else {
+        currNode.next = p2;
+        p2 = p2.next;
+      }
+      currNode = currNode.next;
+    }
+
+    if(p1 == null) {
+      currNode.next = p2;
+    } else {
+      currNode.next = p1;
+    }
+
+    return dummy.next;
+  }
+}

missing-number/ekgns33.java

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+/**
+ input : integer of array
+ output : return the only missing number
+ constraints:
+ 1) is there any duplicate?
+ no.
+ 2) range of element
+ [0, n]
+
+ solution 1)
+ iterate through the array
+ save to hash set
+ iterate i from 0 to n
+ check if exists
+
+ tc : O(n) sc : O(n)
+
+ solution 2)
+ iterate throught the array
+ add all the elements
+ get sum of integer sequence 0 .. n with. n(n+1)/2
+ get subtraction
+ tc : O(n) sc : O(1)
+ */
+class Solution {
+  public int missingNumber(int[] nums) {
+    int sum = 0;
+    int n = nums.length;
+    for(int num : nums) {
+      sum += num;
+    }
+    int totalSum = (n+1) * n / 2;
+    return totalSum - sum;
+  }
+}

palindromic-substrings/ekgns33.java

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+/*
+input : string s
+output: the number of palindromic substrings of given string
+tc : O(n^2) sc : o(n^2) when n is the length of string s
+
+optimize?
+maybe nope.. atleast n^2 to select i and j
+
+ */
+class Solution {
+  public int countSubstrings(String s) {
+    int n = s.length();
+    int cnt = 0;
+    boolean[][] dp = new boolean[n][n];
+    for(int i = n-1; i >= 0; i--) {
+      for(int j = i; j < n; j++) {
+        dp[i][j] = s.charAt(i) == s.charAt(j) && (j-i +1 < 3 || dp[i+1][j-1]);
+        if(dp[i][j]) cnt++;
+      }
+    }
+    return cnt;
+  }
+}

word-search/ekgns33.java

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+/*
+input : m x n matrix and string word
+output : return true if given word can be constructed
+
+solution 1) brute force
+tc : O(n * 4^k) when n is the number of cells, k is the length of word
+sc : O(k) call stack
+ */
+class Solution {
+  private int[][] directions = new int[][] {{-1,0}, {1,0}, {0,1}, {0,-1}};
+  public boolean exist(char[][] board, String word) {
+    //edge case
+    int m = board.length;
+    int n = board[0].length;
+
+    if(m * n < word.length()) return false;
+
+
+    //look for the starting letter and do dfs
+    for(int i = 0; i < m; i++) {
+
+      for(int j = 0; j < n; j++) {
+
+        if(board[i][j] == word.charAt(0)) {
+          //do dfs and get answer
+          board[i][j] = '0';
+          boolean res = dfsHelper(board, word, i, j, 1);
+          board[i][j] = word.charAt(0);
+          if(res) return true;
+        }
+      }
+    }
+
+    return false;
+  }
+
+  public boolean dfsHelper(char[][] board, String word, int curR, int curC, int curP) {
+
+    //endclause
+    if(curP == word.length()) return true;
+
+    boolean ret = false;
+
+    for(int[] direction : directions) {
+      int nextR = curR + direction[0];
+      int nextC = curC + direction[1];
+
+      if(nextR < 0 || nextR >= board.length || nextC < 0 || nextC >= board[0].length) continue;
+
+      if(board[nextR][nextC] == word.charAt(curP)) {
+        board[nextR][nextC] = '0';
+        ret = ret || dfsHelper(board, word, nextR, nextC, curP + 1);
+        board[nextR][nextC] = word.charAt(curP);
+      }
+    }
+    return ret;
+  }
+}