[j2h30728] WEEK 03 Solutions

combination-sum/j2h30728.ts

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+/**
+ * 시간 복잡도: O(k * n^k) (n: candidates 길이, k: target / min(candidates) 최대 깊이)
+ * 공간 복잡도: O(k)
+ */
+function combinationSum(candidates: number[], target: number): number[][] {
+  const result: number[][] = [];
+  const current: number[] = [];
+
+  function backtracking(index: number, remain: number) {
+    if (remain === 0) {
+      result.push([...current]);
+      return;
+    }
+
+    if (remain < 0) {
+      return;
+    }
+
+    for (let i = index; i < candidates.length; i++) {
+      current.push(candidates[i]);
+      backtracking(i, remain - candidates[i]);
+      current.pop();
+    }
+  }
+
+  backtracking(0, target);
+  return result;
+}

decode-ways/j2h30728.ts

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+/**
+ * 시간 복잡도: O(n)
+ * 공간 복잡도: O(n)
+ */
+function numDecodings(s: string): number {
+  const dp = new Array(s.length + 1).fill(0);
+  dp[0] = 1;
+  dp[1] = parseInt(s[0]) === 0 ? 0 : 1;
+
+  for (let i = 2; i <= s.length; i++) {
+    dp[i] = 0;
+
+    if (parseInt(s[i - 1]) !== 0) {
+      dp[i] += dp[i - 1];
+    }
+    if (
+      parseInt(s.slice(i - 2, i)) <= 26 &&
+      parseInt(s.slice(i - 2, i)) >= 10
+    ) {
+      dp[i] += dp[i - 2];
+    }
+  }
+  return dp[s.length];
+}

maximum-subarray/j2h30728.ts

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+/**
+ * 시간 복잡도: O(n)
+ * 공간 복잡도: O(n)
+ */
+function maxSubArray1(nums: number[]): number {
+  const dp = new Array(nums.length).fill(0);
+  dp[0] = nums[0];
+
+  for (let i = 1; i < nums.length; i++) {
+    dp[i] += Math.max(dp[i - 1] + nums[i], nums[i]);
+  }
+  return Math.max(...dp);
+}
+
+/**
+ * 시간 복잡도: O(n)
+ * 공간 복잡도: O(1)
+ */
+function maxSubArray2(nums: number[]): number {
+  let max = nums[0];
+  let current = nums[0];
+
+  for (let i = 1; i < nums.length; i++) {
+    current = Math.max(current + nums[i], nums[i]);
+    max = Math.max(max, current);
+  }
+  return max;
+}

number-of-1-bits/j2h30728.ts

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+/**
+ * 시간 복잡도: O(n) (n: 입력값의 비트 수)
+ * 공간 복잡도: O(1)
+ */
+function hammingWeight(n: number): number {
+  let sum = 0;
+
+  while (n > 0) {
+    sum += Math.abs(n % 2);
+    n = Math.floor(n / 2);
+  }
+  return (sum += n);
+}

valid-palindrome/j2h30728.ts

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+/**
+ * 시간 복잡도: O(n)
+ * 공간 복잡도: O(n)
+ */
+function isPalindrome(s: string): boolean {
+  let string = "";
+  for (const ch of s) {
+    if ((ch >= "a" && ch <= "z") || (ch >= "A" && ch <= "Z")) {
+      string += ch.toLowerCase();
+    }
+  }
+
+  for (let i = 0; i < string.length / 2; i++) {
+    if (string[i] !== string[string.length - i - 1]) {
+      return false;
+    }
+  }
+  return true;
+}