Merge pull request #2129 from a93a/main

Create 0014-longest-common-prefix.kt

Author: a93a

Diff for: kotlin/0014-longest-common-prefix.kt

@@ -0,0 +1,84 @@
+/*
+* Solution as per the channel
+*/
+class Solution {
+    fun longestCommonPrefix(strs: Array<String>): String {
+        var len = 0
+        outerloop@ for(i in 0 until strs[0].length){
+            for(s in strs){
+                if(i == s.length || s[i] != strs[0][i]){
+                    break@outerloop
+                }
+            }
+            len++
+        }   
+        return strs[0].substring(0,len)
+    }
+}
+
+/*
+* Same solution as above but a little more in an idiomatic Kotlin way
+*/
+class Solution {
+    fun longestCommonPrefix(strs: Array<String>): String {
+        var res = ""
+        strs.minBy { it.length }?.forEachIndexed { i,c ->
+            if(strs.all { it[i] == c } ) res += c else return res      
+        }
+        return res
+    }
+}
+
+/*
+* Trie solution
+*/
+class TrieNode() {
+    val child = arrayOfNulls<TrieNode>(26)
+    var isEnd = false
+    fun childCount() = this.child?.filter{it != null}?.count()
+}
+
+class Solution {
+    fun longestCommonPrefix(strs: Array<String>): String {
+        
+        val root: TrieNode? = TrieNode()
+
+        for(word in strs) {
+            var current = root
+            for(c in word){
+                if(current?.child?.get(c - 'a') == null){
+                    current?.child?.set(c - 'a', TrieNode())
+                }         
+                current = current?.child?.get(c - 'a')
+            }
+            current?.isEnd = true
+        }
+        
+        var current = root
+        var len = 0
+        for (c in strs[0]){
+            println(c)
+            if (current?.childCount() == 1 && current?.isEnd != true) len++ else break
+            current = current?.child?.get(c - 'a')
+        }
+        println(len)
+        return strs[0].substring(0,len)
+    }
+}
+
+/*
+* Sorting solution
+*/
+class Solution {
+    fun longestCommonPrefix(strs: Array<String>): String { 
+        var len = 0
+        strs.sort()
+        val first = strs[0]
+        val last = strs[strs.size-1]
+        val minLen = strs.minBy {it.length}?.length!!
+        for(i in 0 until minLen){
+            if(first[i] == last[i]) len++ else break
+        }
+        return strs[0].substring(0,len)
+    }
+}