Update 2663-lexicographically-smallest-beautiful-string.js

Author: everthis

2663-lexicographically-smallest-beautiful-string.js

@@ -3,40 +3,39 @@
  * @param {number} k
  * @return {string}
  */
-const smallestBeautifulString = function(s, k) {
-const chars = s.split('')
+const smallestBeautifulString = function (s, k) {
+  const chars = s.split('')
 
-for (let i = chars.length - 1; i >= 0; i--) {
-  chars[i] = String.fromCharCode(chars[i].charCodeAt(0) + 1)
-  while (containsPalindrome(chars, i)) {
+  for (let i = chars.length - 1; i >= 0; i--) {
     chars[i] = String.fromCharCode(chars[i].charCodeAt(0) + 1)
+    while (containsPalindrome(chars, i)) {
+      chars[i] = String.fromCharCode(chars[i].charCodeAt(0) + 1)
+    }
+    if (chars[i] < String.fromCharCode('a'.charCodeAt(0) + k)) {
+      // If s[i] is among the first k letters, then change the letters after
+      // s[i] to the smallest ones that don't form any palindrome substring.
+      return changeSuffix(chars, i + 1)
+    }
   }
-  if (chars[i] < String.fromCharCode('a'.charCodeAt(0) + k)) {
-    // If s[i] is among the first k letters, then change the letters after
-    // s[i] to the smallest ones that don't form any palindrome substring.
-    return changeSuffix(chars, i + 1)
-  }
-}
 
-return ''
+  return ''
 
-// Returns true if chars[0..i] contains palindrome.
-function containsPalindrome(chars, i) {
-  return (
-    (i > 0 && chars[i] == chars[i - 1]) || (i > 1 && chars[i] == chars[i - 2])
-  )
-}
+  // Returns true if chars[0..i] contains palindrome.
+  function containsPalindrome(chars, i) {
+    return (
+      (i > 0 && chars[i] == chars[i - 1]) || (i > 1 && chars[i] == chars[i - 2])
+    )
+  }
 
-// Returns string where chars[i..] replaced with the smallest letters that
-// don't form any palindrome substring.
-function changeSuffix(chars, i) {
-  for (let j = i; j < chars.length; j++) {
-    chars[j] = 'a'
-    while (containsPalindrome(chars, j)) {
-      chars[j] = String.fromCharCode(chars[j].charCodeAt(0) + 1)
+  // Returns string where chars[i..] replaced with the smallest letters that
+  // don't form any palindrome substring.
+  function changeSuffix(chars, i) {
+    for (let j = i; j < chars.length; j++) {
+      chars[j] = 'a'
+      while (containsPalindrome(chars, j)) {
+        chars[j] = String.fromCharCode(chars[j].charCodeAt(0) + 1)
+      }
     }
+    return chars.join('')
   }
-  return chars.join('')
 }
-
-};