diff --git a/work/Blind/deduction.answers.js b/work/Blind/deduction.answers.js
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+// ============================================================
+// ANSWER KEY — open only after narrating all 6 out loud.
+// Solutions match your repo style exactly:
+// ~/Developer/github.com/prdlk/leetcode/work/Default
+// (object freq maps, (freq[n] || 0) + 1, same shapes)
+// ============================================================
+
+// problem1 = LC 1365 — How Many Numbers Are Smaller Than Current
+// (your file: Easy/Array/1365...js — this IS your solution)
+// RULE: "output[i] = count of elements strictly smaller than arr[i];
+//        ties don't count as smaller (that's the [7,7,7,7] example)."
+function problem1(nums) {
+  // Step 1: Begin by initializing a [Frequency Map]
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  // Step 2: Sort the numbers by ascending order
+  const sorted = Object.keys(freq).sort((a, b) => a - b);
+
+  // Step 3: Init a count of numbers smaller than the active number
+  let count = 0;
+
+  // Step 4: Init a map to track number of values smaller for each number
+  const smaller = {};
+
+  // Step 5: Iterate over the sorted list
+  for (let num of sorted) {
+    // Set count for active number — BEFORE adding own frequency,
+    // so duplicates only see values strictly below them
+    smaller[num] = count;
+
+    // Update the count by frequency
+    count += freq[num];
+  }
+
+  // Step 6: Use original list and find number of smaller values than it
+  return nums.map((n) => smaller[n]);
+}
+// Narration reminders:
+// - "record before adding" is WHY [7,7,7,7] => [0,0,0,0]
+// - Object.keys returns strings; (a, b) => a - b coerces numerically
+// - quick brute-force alternative if short on time:
+//   nums.map((n) => nums.filter((m) => m < n).length)
+
+// problem2 = LC 451 — Sort Characters By Frequency (+ alpha tiebreak)
+// (your file: Medium/Hash Table/451...js — same, tiebreak included)
+// RULE: "rebuild the string most-frequent chars first; equal counts
+//        break alphabetically."
+// DISCRIMINATOR: "bookkeeper" — e:3, then k:2/o:2 tie -> k before o
+// = alphabetical, NOT input order (o appeared first in the input!).
+function problem2(s) {
+  // count the frequency of each character
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  // sort the characters by frequency, ties alphabetical
+  return s
+    .split("")
+    .sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
+    .join("");
+}
+// Note vs your repo file: identical. localeCompare orders lowercase
+// before uppercase ("bbaA"), which is what the drill examples use.
+// LC 451 proper accepts any tie order — the tiebreak is the
+// interview twist Jim's source described.
+
+// problem3 = LC 1636 — Sort Array by Increasing Frequency
+// (your file: Easy/Array/1636...js — this IS your solution)
+// RULE: "sort by frequency ascending; ties by VALUE DESCENDING."
+// DISCRIMINATOR: [2,3,1,3,2] -> 2 and 3 both appear twice, 3 first.
+function problem3(nums) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+  return nums.sort((a, b) => freq[a] - freq[b] || b - a);
+}
+// The idiom to say out loud: "primary key OR tiebreak — when the
+// frequency difference is 0 (falsy), JS falls through to b - a."
+
+// problem4 = LC 387 — First Unique Character
+// (your file: Easy/Hash Table/387...js — this IS your solution)
+// RULE: "index of the first character appearing exactly once; -1 if none."
+// SHAPE TELL: output is a NUMBER, not an array.
+function problem4(s) {
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  for (let i = 0; i < s.length; i++) {
+    if (freq[s[i]] === 1) {
+      return i;
+    }
+  }
+  return -1;
+}
+// Say it: "two passes — I can't know a char is unique until I've
+// seen the whole string."
+
+// problem5 = LC 242 — Valid Anagram
+// (not in your repo yet — written in your exact style)
+// RULE: "true iff both strings have the same characters with the
+//        same counts."
+// DISCRIMINATOR: ("aacc", "ccac") -> same char SET {a,c}, different
+// counts -> false. Kills set-equality.
+function problem5(s, t) {
+  if (s.length !== t.length) return false;
+
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  // walk t, spending counts down; a missing/exhausted char fails
+  for (let c of t) {
+    if (!freq[c]) return false;
+    freq[c]--;
+  }
+  return true;
+}
+// The length guard up front is what lets count-down work without a
+// final "all zeros" pass.
+
+// problem6 = LC 347 — Top K Frequent Elements
+// (your file: Medium/Array/347...js — this IS your solution)
+// RULE: "return the k values that appear most often, most frequent
+//        first."
+// SHAPE TELL: second argument k controls output length.
+// EDGE: [3,0,1,0] k=1 => [0] — value 0 is falsy but valid.
+function problem6(nums, k) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  return Object.keys(freq)
+    .map(Number) // Object.keys gave us strings — convert back
+    .sort((a, b) => freq[b] - freq[a])
+    .slice(0, k);
+}
+// The .map(Number) is the classic gotcha to mention: without it you
+// return ["1","2"] instead of [1,2].
+
+// ============================================================
+// THE HAMMER (all six are this skeleton):
+//   1. BUILD  -> const freq = {}; for (let x of input)
+//                freq[x] = (freq[x] || 0) + 1;
+//   2. ORDER  -> sort keys/elements by the criterion the pattern demands
+//   3. DERIVE -> compute what each key maps to (count / rank / index)
+//   4. EMIT   -> map back to input order / rebuild string / slice k
+//
+// SELF-SCORE per problem:
+//   Rule stated in one sentence, verified vs ALL examples:  /1
+//   Tiebreak/edge named BEFORE coding (the discriminator):  /1
+//   Working code, narrated while typing:                    /1
+// 15+/18 = ready. Misses tell you what to re-drill Tuesday
+// morning (max 2 reps, then stop — rest beats an 11th rep).
+// ============================================================
diff --git a/work/Blind/deduction.js b/work/Blind/deduction.js
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+++ b/work/Blind/deduction.js
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+// ============================================================
+// BLIND DEDUCTION SET — Monday evening, ONE round, then stop.
+// The 6 core interview problems, disguised exactly as they'd
+// appear Tuesday: no statement, just input => output pairs.
+// Rules: deduce the rule, SAY it in one sentence out loud,
+// state the plan (freq map? sort? tiebreak?), then implement.
+// Do NOT open blind-deduction-answers.js until finished.
+// Target: rule stated < 3 min, implemented < 6 min each.
+// ============================================================
+
+// ---- problem1 ----
+// problem1([8, 1, 2, 2, 3])   => [4, 0, 1, 1, 3]
+// problem1([6, 5, 4, 8])      => [2, 1, 0, 3]
+// problem1([7, 7, 7, 7])      => [0, 0, 0, 0]
+// problem1([3, 1, 2])         => [2, 0, 1]
+// problem1([5])               => [0]
+// problem1([4, 1, 4, 1])      => [2, 0, 2, 0]
+function problem1(arr) {
+  // your code
+}
+
+// ---- problem2 ----
+// problem2("tree")            => "eert"
+// problem2("cccaaa")          => "aaaccc"
+// problem2("Aabb")            => "bbaA"
+// problem2("z")               => "z"
+// problem2("bookkeeper")      => "eeekkoobpr"
+// problem2("mississippi")     => "iiiissssppm"
+function problem2(str) {
+  // your code
+}
+
+// ---- problem3 ----
+// problem3([1, 1, 2, 2, 2, 3])           => [3, 1, 1, 2, 2, 2]
+// problem3([2, 3, 1, 3, 2])              => [1, 3, 3, 2, 2]
+// problem3([-1, 1, -6, 4, 5, -6, 1, 4, 1]) => [5, -1, 4, 4, -6, -6, 1, 1, 1]
+// problem3([9])                          => [9]
+// problem3([5, 5, 4, 4])                 => [5, 5, 4, 4]
+function problem3(arr) {
+  // your code
+}
+
+// ---- problem4 ----
+// problem4("leetcode")        => 0
+// problem4("loveleetcode")    => 2
+// problem4("aabb")            => -1
+// problem4("x")               => 0
+// problem4("aabbc")           => 4
+// problem4("aa")              => -1
+function problem4(str) {
+  // your code
+}
+
+// ---- problem5 ----
+// problem5("anagram", "nagaram")  => true
+// problem5("rat", "car")          => false
+// problem5("a", "ab")             => false
+// problem5("", "")                => true
+// problem5("aacc", "ccac")        => false
+// problem5("listen", "silent")    => true
+function problem5(s, t) {
+  // your code
+}
+
+// ---- problem6 ----
+// problem6([1, 1, 1, 2, 2, 3], 2)        => [1, 2]
+// problem6([1], 1)                       => [1]
+// problem6([4, 4, 4, 6, 6, 7, 7, 7, 7], 2) => [7, 4]
+// problem6([5, 5, 5, 5], 1)              => [5]
+// problem6([3, 0, 1, 0], 1)              => [0]
+// problem6([2, 2, 3, 3, 1], 3)           => [2, 3, 1]
+function problem6(arr, k) {
+  // your code
+}
+
+// ---- harness: uncomment per problem after implementing ----
+// console.log(problem1([8, 1, 2, 2, 3]), problem1([4, 1, 4, 1]));
+// console.log(problem2("tree"), problem2("bookkeeper"));
+// console.log(problem3([2, 3, 1, 3, 2]), problem3([5, 5, 4, 4]));
+// console.log(problem4("loveleetcode"), problem4("aabbc"));
+// console.log(problem5("anagram", "nagaram"), problem5("aacc", "ccac"));
+// console.log(problem6([4, 4, 4, 6, 6, 7, 7, 7, 7], 2), problem6([2, 2, 3, 3, 1], 3));
diff --git a/work/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js b/work/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js
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+++ b/work/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js
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+/*
+ * 1365. How Many Numbers Are Smaller Than the Current Number
+ * Difficulty: Easy
+ * https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * Given the array nums, for each nums[i] find out how many numbers in
+ * the array are smaller than it. That is, for each nums[i] you have to
+ * count the number of valid j's such that j != i and nums[j] < nums[i].
+ *
+ * Return the answer in an array.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: nums = [8,1,2,2,3]
+ * Output: [4,0,1,1,3]
+ * Explanation:
+ * For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and
+ * 3).
+ * For nums[1]=1 does not exist any smaller number than it.
+ * For nums[2]=2 there exist one smaller number than it (1).
+ * For nums[3]=2 there exist one smaller number than it (1).
+ * For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
+ *
+ * Example 2:
+ *
+ * Input: nums = [6,5,4,8]
+ * Output: [2,1,0,3]
+ *
+ * Example 3:
+ *
+ * Input: nums = [7,7,7,7]
+ * Output: [0,0,0,0]
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 2 <= nums.length <= 500
+ *
+ * 	• 0 <= nums[i] <= 100
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+
+var smallerNumbersThanCurrent = function (nums) {
+  // Step 1: Begin by initializing a [Frequency Map]()
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  // Step 2: Sort the numbers by ascending order
+  const sorted = Object.keys(freq).sort((a, b) => a - b);
+
+  // Step 3: Init a count of numbers smaller than the active number
+  let count = 0;
+
+  // Step 4: Init a map to track number of values smaller for each number
+  const smaller = {};
+
+  // Step 5: Iterate over the sorted list
+  for (let num of sorted) {
+    // Set count for active number
+    smaller[num] = count;
+
+    // Update the count by frequency
+    count += freq[num];
+  }
+
+  // Step 6: Use original list and find number of smaller values than it
+  return nums.map((n) => smaller[n]);
+};
diff --git a/work/Easy/Array/1636.sort-array-by-increasing-frequency.js b/work/Easy/Array/1636.sort-array-by-increasing-frequency.js
new file mode 100644
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+++ b/work/Easy/Array/1636.sort-array-by-increasing-frequency.js
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+/*
+ * 1636. Sort Array by Increasing Frequency
+ * Difficulty: Easy
+ * https://leetcode.com/problems/sort-array-by-increasing-frequency/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * Given an array of integers nums, sort the array in increasing order
+ * based on the frequency of the values. If multiple values have the same
+ * frequency, sort them in decreasing order.
+ *
+ * Return the sorted array.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: nums = [1,1,2,2,2,3]
+ * Output: [3,1,1,2,2,2]
+ * Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and
+ * '2' has a frequency of 3.
+ *
+ * Example 2:
+ *
+ * Input: nums = [2,3,1,3,2]
+ * Output: [1,3,3,2,2]
+ * Explanation: '2' and '3' both have a frequency of 2, so they are
+ * sorted in decreasing order.
+ *
+ * Example 3:
+ *
+ * Input: nums = [-1,1,-6,4,5,-6,1,4,1]
+ * Output: [5,-1,4,4,-6,-6,1,1,1]
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 1 <= nums.length <= 100
+ *
+ * 	• -100 <= nums[i] <= 100
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var frequencySort = function (nums) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+  return nums.sort((a, b) => freq[a] - freq[b] || b - a);
+};
diff --git a/work/Easy/Hash Table/387.first-unique-character-in-a-string.js b/work/Easy/Hash Table/387.first-unique-character-in-a-string.js
new file mode 100644
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+++ b/work/Easy/Hash Table/387.first-unique-character-in-a-string.js	
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+/*
+ * 387. First Unique Character in a String
+ * Difficulty: Easy
+ * https://leetcode.com/problems/first-unique-character-in-a-string/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * Given a string s, find the first non-repeating character in it and
+ * return its index. If it does not exist, return -1.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: s = "leetcode"
+ *
+ * Output: 0
+ *
+ * Explanation:
+ *
+ * The character 'l' at index 0 is the first character that does not
+ * occur at any other index.
+ *
+ * Example 2:
+ *
+ * Input: s = "loveleetcode"
+ *
+ * Output: 2
+ *
+ * Example 3:
+ *
+ * Input: s = "aabb"
+ *
+ * Output: -1
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 1 <= s.length <= 10^5
+ *
+ * 	• s consists of only lowercase English letters.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var firstUniqChar = function (s) {
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  for (let i = 0; i < s.length; i++) {
+    if (freq[s[i]] === 1) {
+      return i;
+    }
+  }
+  return -1;
+};
diff --git a/work/Medium/Array/347.top-k-frequent-elements.js b/work/Medium/Array/347.top-k-frequent-elements.js
new file mode 100644
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--- /dev/null
+++ b/work/Medium/Array/347.top-k-frequent-elements.js
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+/*
+ * 347. Top K Frequent Elements
+ * Difficulty: Medium
+ * https://leetcode.com/problems/top-k-frequent-elements/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * Given an integer array nums and an integer k, return the k most
+ * frequent elements. You may return the answer in any order.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: nums = [1,1,1,2,2,3], k = 2
+ *
+ * Output: [1,2]
+ *
+ * Example 2:
+ *
+ * Input: nums = [1], k = 1
+ *
+ * Output: [1]
+ *
+ * Example 3:
+ *
+ * Input: nums = [1,2,1,2,1,2,3,1,3,2], k = 2
+ *
+ * Output: [1,2]
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 1 <= nums.length <= 10^5
+ *
+ * 	• -10^4 <= nums[i] <= 10^4
+ *
+ * 	• k is in the range [1, the number of unique elements in the array].
+ *
+ * 	• It is guaranteed that the answer is unique.
+ *
+ *
+ *
+ * Follow up: Your algorithm's time complexity must be better than O(n
+ * log n), where n is the array's size.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var topKFrequent = function (nums, k) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  return Object.keys(freq)
+    .map(Number)
+    .sort((a, b) => freq[b] - freq[a])
+    .slice(0, k);
+};
diff --git a/work/Medium/Hash Table/451.sort-characters-by-frequency.js b/work/Medium/Hash Table/451.sort-characters-by-frequency.js
new file mode 100644
index 0000000..0357d5c
--- /dev/null
+++ b/work/Medium/Hash Table/451.sort-characters-by-frequency.js	
@@ -0,0 +1,64 @@
+/*
+ * 451. Sort Characters By Frequency
+ * Difficulty: Medium
+ * https://leetcode.com/problems/sort-characters-by-frequency/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * Given a string s, sort it in decreasing order based on the frequency
+ * of the characters. The frequency of a character is the number of times
+ * it appears in the string.
+ *
+ * Return the sorted string. If there are multiple answers, return any
+ * of them.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: s = "tree"
+ * Output: "eert"
+ * Explanation: 'e' appears twice while 'r' and 't' both appear once.
+ * So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also
+ * a valid answer.
+ *
+ * Example 2:
+ *
+ * Input: s = "cccaaa"
+ * Output: "aaaccc"
+ * Explanation: Both 'c' and 'a' appear three times, so both "cccaaa"
+ * and "aaaccc" are valid answers.
+ * Note that "cacaca" is incorrect, as the same characters must be
+ * together.
+ *
+ * Example 3:
+ *
+ * Input: s = "Aabb"
+ * Output: "bbAa"
+ * Explanation: "bbaA" is also a valid answer, but "Aabb" is incorrect.
+ * Note that 'A' and 'a' are treated as two different characters.
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 1 <= s.length <= 5 * 10^5
+ *
+ * 	• s consists of uppercase and lowercase English letters and digits.
+ */
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var frequencySort = function (s) {
+  // count the frequency of each character
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  // sort the characters by frequency
+  return s
+    .split("")
+    .sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
+    .join("");
+};