[ready-oun] Week 03 solutions

combination-sum/ready-oun.java

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+import java.util.*; 
+class Solution {
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        List<List<Integer>> result = new ArrayList<>(); 
+        List<Integer> temp = new ArrayList<>(); 
+
+        backtrack(candidates, target, 0, temp, result);
+        return result; 
+    }
+    private void backtrack(int[] candidates, int target, int start, List<Integer> temp, List<List<Integer>> result) {
+        if (target < 0) return; 
+        if (target == 0) {
+            result.add(new ArrayList<>(temp)); // deep copy
+            return; 
+        }
+
+        for (int i = start; i < candidates.length; i++) {
+            temp.add(candidates[i]); 
+            backtrack(candidates, target - candidates[i], i, temp, result); 
+            temp.remove(temp.size() -1); 
+        }
+
+    }
+}
+
+/**
+Return all unique combinations where the candidate num sum to target
+- each num in cand[] can be used unlimited times 
+- order of num in comb does NOT matter 
+
+1. use backtracking to explore all possible comb 
+2. backtrack, if current sum > target
+3. save comb, if current sum == target 
+4. avoid dupl -> only consider num from crnt idx onward (no going back)
+
+Time: O(2^target)
+Space: O(target)
+
+Learned: Backtracking vs DFS
+- DFS: search all paths deeply (no conditions, no rollback).
+- Backtracking = DFS + decision making + undo step.
+Explore, prune (if invalid), save (if valid), then undo last choice.
+
+    for (선택 가능한 숫자 하나씩) {
+        선택하고
+        target 줄이고 (목표 가까워짐)
+        재귀 호출로 다음 선택
+        실패하거나 성공하면 되돌리기 (백트래킹)
+    }
+
+DFS just visits everything,
+Backtracking visits only what’s promising — and turns back if not!
+ */

number-of-1-bits/ready-oun.java

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+class Solution {
+    public int hammingWeight(int n) {
+        int count = 0; 
+
+        while (n != 0) {
+            if ((n & 1) == 1) {
+                count++; 
+            }
+            n = n >>> 1;
+        }
+
+        return count; 
+    }
+}
+
+/**
+Time: O(1) – max 32 iterations (fixed bit size)
+Space: O(1)
+
+How it works: Shift each bit → Check → Count → Shift again
+    1. Shift each bit of n to the right
+    2. Check if the last bit is 1 using n & 1
+    3. If it is, increment the count
+    4. Shift n to the right using n = n >>> 1
+
+Learned:
+    (n & 1) isolates the least significant bit (LSB) to check if it’s 1
+    >>  : Arithmetic shift (fills in sign bit, so infinite loop for negatives)
+    >>> : Logical shift (always fills with 0, safe for negatives)
+    Java evaluates == before &, so use parentheses to control the order
+ */

valid-palindrome/ready-oun.java

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+class Solution {
+    public boolean isPalindrome(String s) {
+        StringBuilder cleaned = new StringBuilder(); 
+
+        for (char c : s.toCharArray()) {
+            if (Character.isLetterOrDigit(c)) {
+                cleaned.append(Character.toLowerCase(c)); 
+            }
+        }
+
+        int left = 0; 
+        int right = cleaned.length() - 1; 
+        while (left < right) {
+            // Fail fast: return false as soon as a mismatch is found
+            if (cleaned.charAt(left) != cleaned.charAt(right)) {
+                return false; 
+            }
+            left++;
+            right--; 
+        }
+
+        return true; 
+    }
+}
+
+/**
+converting all uppercase letters into lowercase letters
+removing all non-alphanumeric char
+
+1. cleaning 
+    0. str -> char with for-each loop  
+    1. check char if isLetterOrDigit
+    2. make char to LowerCase 
+    * Character Class static method 
+2. two ptrs comparison while left < right 
+    s[i] == s[n - 1 - i] 
+
+- Time: O(n) 
+
+ /** REMEMBER 
+1. length vs length() 
+
+arr.length => field 
+String(Builder).length() => method 
+
+2. “fail fast” approach: 
+As soon as we detect something wrong, we exit early.
+  */