[nb] Commit

.index

@@ -0,0 +1,12 @@
+zig-zag-conversion
+util
+string-to-integer-aoi
+roman-to-integer
+reverse-integer
+rec-permute
+pow-x-n
+permutations
+integer-to-roman
+group-anagrams
+all-elements-in-two-binary-search-tree
+dynamic-programming

dynamic-programming/22-generate-parenthesis/golang/main.go

@@ -0,0 +1,56 @@
+package main
+/*
+Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
+Example 1:
+
+Input: n = 3
+Output: ["((()))","(()())","(())()","()(())","()()()"]
+Example 2:
+  ()()(); (())();  ((())); 
+  ()(());
+  (()())
+
+Input: n = 1
+Output: ["()"]
+ 
+  ()(); (())
+
+Constraints:
+
+1 <= n <= 8
+*/
+
+func rec(int index, dp map[string]string) {
+}
+
+
+func generateParenthesis(n int)[] string {
+
+  if n == 1 {
+    return []string{"()"}
+  }
+
+  if n == 2 {
+    return []string{"()()", "(())"}
+  }
+
+  var output = make([]string, 0) 
+  var dp = map[string]string{}
+  var start_string string = ""
+  for i := 0; i < n - 1; i++ {
+    start_string = start_string + "()"
+    out_arr = a
+  }
+  output = append(output, start_string)
+
+
+  for i := 1; i < n - 1; i++ {
+    result := rec(i+1, dp) + rec(i-1) 
+  }
+
+
+}
+
+func main() {
+
+}

dynamic-programming/22-generate-parenthesis/js/index.js

@@ -0,0 +1,17 @@
+function generateParenthesis(n) {
+  if (n == 1) {
+    return ['()'];
+  }
+
+  if (n == 2) {
+    return ['()()', '(())'];
+  }
+
+  const input = [];
+  for (i = 0; i < n; i++) {
+    input.push('()');
+  }
+
+  for(i = 0; i < n; i++; i++) {
+  }
+}

dynamic-programming/5-longest-palindromic-substring/golang/main.go

@@ -0,0 +1,102 @@
+package main
+import (
+  "fmt"
+  "strings"
+)
+/*
+Given a string s, return the longest palindromic substring in s.
+
+Example 1:
+
+Input: s = "babad"
+Output: "bab"
+Note: "aba" is also a valid answer.
+Example 2:
+
+Input: s = "cbbd"
+Output: "bb"
+Example 3:
+
+Input: s = "a"
+Output: "a"
+Example 4:
+
+Input: s = "ac"
+Output: "a"
+ 
+
+Constraints:
+
+1 <= s.length <= 1000
+s consist of only digits and English letters.
+*/
+
+/*
+  Hints: If the string is 2 characters and they are both the same character then it is a palindrome (ie: aa or 11)
+
+  So you can use create a table to calculate a table where if table[i+1][j-1] is true and string[i+1] == string[j-1] 
+  then the substring string is a palindrome
+
+  You can see that i+1, j-1 is like two pointers with one going forward and the other going backwards
+
+
+    a b a 
+  a 0 0 1
+  b 0 1 0
+  a 1 0 1
+*/
+func longestPalindrome(s string) string {
+  /*
+    If string s has a len = 1 then the longest palindrome is s
+
+    If string s has len = 2 and the two characters are the same then the longest palindrome is s
+  */
+
+  var n = len(s)
+
+  if n == 1 {
+    return s 
+  }
+
+  if n == 2 && s[0] == s[1] {
+    return s
+  }
+
+  // Generate memory table table[i][i] = true 
+  var output string
+  dp := make([][]bool, n)
+  for i:= 0; i < n - 1; i++ {
+    dp[i] = make([]bool, n)
+    dp[i][i] = true
+  }
+
+  // Need to do this to access by index
+  // see https://stackoverflow.com/questions/15018545/how-to-index-characters-in-a-golang-string
+  strarr := strings.Split(s, "")
+  // Also check for palidrome where length == 2 using s[i] == s[i+1] 
+  for i := 0; i < n - 1; i++ {
+    if (strarr[i] == strarr[i+1]) {
+      dp[i][i+1] = true
+      output = s[i:i+1]
+    }
+  }
+
+   // n = 5 
+   // i = 2; x = 0; y = 1
+
+  // Find substring where n > 2
+  for i := 2; i < n - 1; i++ {
+    for x := 0; x < n - i; x++ {
+      // starting at x find the end idx (len substring - 1 )
+      y :=  x + i - 1 
+      if (dp[x+1][y-1] && strarr[x] == strarr[y]) {
+        output = s[x:(y+1)] 
+      }
+    }
+  }
+  return output
+}
+
+func main () {
+  fmt.Println(longestPalindrome("babad"))
+}