[박종훈] 4주차 답안 제출

counting-bits/dev-jonghoonpark.md

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+- https://leetcode.com/problems/counting-bits/
+- time complexity : O(n \* log n), logn 이 붙는 이유는 bit는 log를 따라 수가 결정되기 때문
+- space complexity : O(n)
+- https://algorithm.jonghoonpark.com/2024/04/23/leetcode-338
+
+```java
+public int[] countBits(int n) {
+    int result[] = new int[n + 1];
+    for (int i = 0; i <= n; i++) {
+        int num = i;
+        int count = 0;
+        while (num > 0) {
+            count += num & 1;
+            num = num >> 1;
+        }
+        result[i] = count;
+    }
+    return result;
+}
+```

group-anagrams/dev-jonghoonpark.md

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+- https://leetcode.com/problems/group-anagrams/
+- time complexity : O(n \* m log m), 여기서 m은 str 배열(strs)의 각 str의 평균이다.
+- space complexity : O(n \* m)
+- https://algorithm.jonghoonpark.com/2024/05/25/leetcode-49
+
+```java
+public List<List<String>> groupAnagrams(String[] strs) {
+    List<List<String>> result = new ArrayList<>();
+    HashMap<String, Integer> map = new HashMap<>();
+
+    for(String str: strs) {
+        char[] temp = str.toCharArray();
+        Arrays.sort(temp);
+        String sorted = String.valueOf(temp);
+        if(map.containsKey(sorted)) {
+            result.get(map.get(sorted)).add(str);
+        } else {
+            int newIndex = result.size();
+            List<String> newArrayList = new ArrayList<>();
+            result.add(newArrayList);
+            newArrayList.add(str);
+            map.put(sorted, newIndex);
+        }
+    }
+
+    return result;
+}
+```
+
+## TC, SC
+
+시간 복잡도는 O(n \* m log m)이고, 공간 복잡도는 O(n \* m)이다.
+여기서 m은 str 배열(strs)의 각 str의 평균이다.

missing-number/dev-jonghoonpark.md

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+- https://leetcode.com/problems/missing-number/
+- time complexity : O(n)
+- space complexity : O (n)
+- https://algorithm.jonghoonpark.com/2024/05/25/leetcode-268
+
+```java
+public int missingNumber(int[] nums) {
+    int[] counts = new int[nums.length + 1];
+
+    for(int num : nums) {
+        counts[num] = 1;
+    }
+
+    for(int i = 0; i < counts.length; i ++){
+        if(counts[i] == 0) {
+            return i;
+        }
+    }
+
+    return -1;
+}
+```
+
+등차수열로 푸는 방법도 있는 재밌는 문제.
+
+```java
+public int missingNumber(int[] arr) {
+    int sum = 0;
+    int max = (arr.length * (arr.length + 1)) / 2;
+    for (int i = 0; i < arr.length; i++) {
+        sum += arr[i];
+    }
+    return max - sum;
+}
+```

number-of-1-bits/dev-jonghoonpark.md

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+- https://leetcode.com/problems/number-of-1-bits/
+- time complexity : O(logn)
+- space complexity : O(1)
+- https://algorithm.jonghoonpark.com/2024/02/20/leetcode-191
+
+```java
+public int hammingWeight(int n) {
+    int count = 0;
+    while (n != 0) {
+        count += n % 2;
+        n = n >> 1;
+    }
+    return count;
+}
+```

reverse-bits/dev-jonghoonpark.md

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+- https://leetcode.com/problems/reverse-bits/
+- time complexity : O(1)
+- space complexity : O(1)
+- https://algorithm.jonghoonpark.com/2024/04/23/leetcode-190
+
+```java
+public class Solution {
+    public int reverseBits(int n) {
+        int ans = 0;
+        for (int i = 0; i < 32; i++) {
+            ans <<= 1;
+            ans |= (n & 1);
+            n >>= 1;
+        }
+        return ans;
+    }
+}
+```