Create 0844-backspace-string-compare.kt

Author: a93a

kotlin/0844-backspace-string-compare.kt

@@ -0,0 +1,97 @@
+// Two pointer solution, time O(n) and space O(1)
+class Solution {
+    fun backspaceCompare(s: String, t: String): Boolean {
+        
+        fun nextValidChar(str: String, _index: Int): Int {
+            var backspace = 0
+            var index = _index
+            while (index >= 0) {
+                if (backspace == 0 && str[index] != '#')
+                    break
+                else if (str[index] == '#')
+                    backspace++
+                else
+                    backspace--
+                index--
+            }
+
+            return index
+        }
+
+        var index_s = s.lastIndex
+        var index_t = t.lastIndex
+        while (index_s >= 0 || index_t >= 0) {
+            index_s = nextValidChar(s, index_s)
+            index_t = nextValidChar(t, index_t)
+
+            val char_s = if (index_s >= 0) s[index_s] else '?'
+            val char_t = if (index_t >= 0) t[index_t] else '?'
+            if (char_s != char_t) 
+                return false
+            index_s--
+            index_t--
+        }
+
+        return true
+    }
+}
+
+// Stack solution, time O(n) and space O(n)
+class Solution {
+    fun backspaceCompare(s: String, t: String): Boolean {
+        val s1 = LinkedList<Char>()
+        val s2 = LinkedList<Char>()
+
+        for (c in s) {
+            if (c == '#') {
+                if (s1.isNotEmpty())
+                    s1.removeLast()
+            } else {
+                s1.addLast(c)
+            } 
+        }
+
+        for (c in t) {
+            if (c == '#') {
+                if (s2.isNotEmpty())
+                    s2.removeLast()
+            } else {
+                s2.addLast(c)
+            } 
+        }
+
+        return s1.joinToString("") == s2.joinToString("")
+    }
+}
+
+// Another style of two pointer solution, same idea, without calling methods, time O(n) and space O(1)
+class Solution {
+    fun backspaceCompare(s: String, t: String): Boolean {
+        var i = s.lastIndex
+        var j = t.lastIndex
+        var steps = 0
+
+        while (true) {
+            steps = 0
+            while (i >= 0 && (steps > 0 || s[i] == '#')) {
+                steps += if (s[i] == '#') 1 else -1
+                i--
+            }
+
+            steps = 0
+            while (j >= 0 && (steps > 0 || t[j] == '#')) {
+                steps += if (t[j] == '#') 1 else -1
+                j--
+            }
+
+            if (i >= 0 && j >= 0 && s[i] == t[j]) {
+                i--
+                j--
+            } else {
+                break
+            }
+        }
+
+        return i == -1 && j == -1
+    }
+}