[juhui-jeong] WEEK 02 solutions

climbing-stairs/juhui-jeong.ts

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+/*
+
+시간 복잡도: O(n)
+공간 복잡도: O(1)
+
+해당 문제를 구할 때 바로 생각나지 않아 leet code의 코멘트들을 보며 힌트를 구했습니다.
+*/
+function climbStairs(n: number): number {
+  if (n <= 2) return n;
+
+  let a = 1;
+  let b = 2;
+  for (let i = 3; i <= n; i++) {
+    let c = a + b;
+    a = b;
+    b = c;
+  }
+  return b;
+}

product-of-array-except-self/juhui-jeong.ts

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+/*
+시간 복잡도: O(n)
+공간 복잡도: O(1)
+*/
+function productExceptSelf(nums: number[]): number[] {
+  const n = nums.length;
+  const answer = new Array<number>(n);
+
+  // 왼쪽 곱 채우기
+  let leftProduct = 1;
+  for (let i = 0; i < n; i++) {
+    answer[i] = leftProduct;
+    leftProduct *= nums[i];
+  }
+
+  // 오른쪽 곱 채우기
+  let rightProduct = 1;
+  for (let i = n - 1; i >= 0; i--) {
+    answer[i] *= rightProduct;
+    rightProduct *= nums[i];
+  }
+  return answer;
+}
+
+/*
+// Time Limit Exceeded
+function productExceptSelf(nums: number[]): number[] {
+  let answer = [];
+  for (let i = 0; i < nums.length; i++) {
+    const mulOfNums = nums.reduce((accumulator, currentValue, currentIndex) => {
+      if (currentIndex !== i) {
+        return accumulator * currentValue;
+      }
+      return accumulator;
+    }, 1);
+    answer.push(mulOfNums);
+  }
+  return answer;
+}
+*/

valid-anagram/juhui-jeong.ts

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+// 시간 복잡도(Time Complexity): O(n)
+// 공간 복잡도(Space Complexity): O(1)
+function isAnagram(s: string, t: string): boolean {
+  if (s.length !== t.length) return false;
+
+  const count = new Array(26).fill(0);
+
+  for (let i = 0; i < s.length; i++) {
+    count[s.charCodeAt(i) - 97]++;
+    count[t.charCodeAt(i) - 97]--;
+  }
+
+  return count.every((n) => n === 0);
+}
+
+/*
+// 첫 번째 풀이
+// 시간 복잡도(Time Complexity): O(n log n)
+// 공간 복잡도(Space Complexity): O(n)
+
+function isAnagram(s: string, t: string): boolean {
+  let baseString = s.split('').sort().toString();
+  let targetString = t.split('').sort().toString();
+
+  return baseString === targetString ? true : false;
+}
+*/