[HerrineKim] Week 4

coin-change/HerrineKim.js

@@ -0,0 +1,21 @@
+// 시간 복잡도: O(n * m)
+// 공간 복잡도: O(n)
+
+/**
+ * @param {number[]} coins
+ * @param {number} amount
+ * @return {number}
+ */
+var coinChange = function(coins, amount) {
+  const dp = new Array(amount + 1).fill(Infinity);
+  dp[0] = 0;
+
+  for (let coin of coins) {
+      for (let i = coin; i <= amount; i++) {
+          dp[i] = Math.min(dp[i], dp[i - coin] + 1);
+      }
+  }
+
+  return dp[amount] === Infinity ? -1 : dp[amount];
+};
+

missing-number/HerrineKim.js

@@ -0,0 +1,17 @@
+// 시간 복잡도: O(n)
+// 공간 복잡도: O(1)
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var missingNumber = function(nums) {
+  const n = nums.length;
+  // 0부터 n까지의 합 공식: n * (n + 1) / 2
+  const expectedSum = (n * (n + 1)) / 2;
+  // 실제 배열의 합
+  const actualSum = nums.reduce((sum, num) => sum + num, 0);
+
+  return expectedSum - actualSum;
+};
+

palindromic-substrings/HerrineKim.js

@@ -0,0 +1,25 @@
+// 시간 복잡도: O(n^2)
+// 공간 복잡도: O(1)
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var countSubstrings = function(s) {
+  let count = 0;
+
+  const countPalindrome = (left, right) => {
+      while (left >= 0 && right < s.length && s[left] === s[right]) {
+          count++;
+          left--;
+          right++;
+      }
+  };
+
+  for (let i = 0; i < s.length; i++) {
+      countPalindrome(i, i);
+      countPalindrome(i, i + 1);
+  }
+
+  return count;
+};