[Cyjin-jani] WEEK 03 solutions

number-of-1-bits/Cyjin-jani.js

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+// tc: O(logn)
+// sc: O(logn)
+const hammingWeight_simple = function (n) {
+  const binaryStr = n.toString(2);
+  let answer = 0;
+
+  for (char of binaryStr) {
+    if (+char === 1) {
+      answer++;
+    }
+  }
+
+  return answer;
+};
+
+// toString(2)를 쓰지 않고 처리
+// tc: O(logn)
+// sc: O(1)
+const hammingWeight = function (n) {
+  let answer = 0;
+
+  while (n >= 0) {
+    const remains = n % 2;
+    if (remains === 1) answer++;
+    n = Math.floor(n / 2);
+  }
+
+  return answer;
+};

valid-palindrome/Cyjin-jani.js

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+// naive한 풀이
+// tc: O(n)
+// sc: O(n)
+const isPalindromeNaive = function (s) {
+  const str = s.toLowerCase().replace(/[^a-z0-9]/g, ''); // 이 부분에서 공간복잡도가 O(n)
+  let leftIdx = 0;
+  let rightIdx = str.length - 1;
+
+  while (leftIdx <= rightIdx) {
+    if (str[leftIdx] !== str[rightIdx]) {
+      return false;
+    } else {
+      leftIdx++;
+      rightIdx--;
+    }
+  }
+  return true;
+};
+
+// 좀 더 최적화 된 풀이
+// tc: O(n)
+// cs: O(1)
+const isPalindrome = function (s) {
+  let leftIdx = 0;
+  let rightIdx = s.length - 1;
+
+  while (leftIdx < rightIdx) {
+    while (leftIdx < rightIdx && !isAlphaNumeric(s[leftIdx])) leftIdx++;
+    while (leftIdx < rightIdx && !isAlphaNumeric(s[rightIdx])) rightIdx--;
+
+    if (s[leftIdx].toLowerCase() !== s[rightIdx].toLowerCase()) return false;
+
+    leftIdx++;
+    rightIdx--;
+  }
+  return true;
+};
+
+function isAlphaNumeric(char) {
+  return /[a-zA-Z0-9]/.test(char);
+}