6. Coding if it appears

How to use this chapter

Each family is one card. Restate input and output. Name brute, then the invariant, then type Java. A Fluency-labeled hour can still be a tree. Recruiter “LP-only” mail still opened an editor in other loops. If the algorithm under a card is what is unclear (BFS vs DFS, Kahn, monotonic stack), open the DSA Notes booklet first — two examples and two Java versions — then come back here or to chapter 05.

What they might ask

• A leftover hour after two lives that still pastes Live Code.
• A calendar titled GenAI Fluency that is Distance K (LC 7623949).
• An invite with no Live Code link that still mixes LP and DSA (LC 6806195).

IE-asked as shape from other loops. Resume-derived later-slot coding prompts: none in the research for this topic. Pattern names are Standard CS families, not a claim this Job ID will ask them.

What you say

They might ask next

Keep straight

• Job 10454435 still none. Unnamed stays unnamed. A slug on a write-up is not a numeric id you attach.
• Internships used Python, TypeScript, and ROS2. Live Code is Java. Those internships are not JVM services.
• Login Tracker (new_login / get_oldest_login) is unverified. Do not volunteer it.
• Independent SDE I Rate Limiter live-round count = 1, OA+3 system design. It is not UTA-default Round 4. Card at the end of this chapter.

Tighten

Shared Java types

Write these once at the top of Live Code, then the problem.

What they might ask

• They give TreeNode / ListNode already. Use theirs.
• They give nothing. Sketch these two.

Standard CS scaffolding for this chapter.

What you say

class TreeNode {
  int val;
  TreeNode left, right;
  TreeNode(int v) { val = v; }
}
class ListNode {
  int val;
  ListNode next;
  ListNode(int v) { val = v; }
}

They might ask next

Keep straight

• These types are Live Code. Horizon / Ylogx / IQVIA intern code was not this Java.

Tighten

Invite said LP-only — still write Java

Other AUTA loops were told three LPs, or the mail had no Live Code link, and the room still opened an editor. One Reddit leftover was LP-only as promised. Prepare Java; speak stories if they never paste code.

What they might ask

• DSA first, then a couple of LPs that sound like formality (tree path A/B/C).
• ~30 min LP + ~30 min Nice Subarrays with no Live Code link in mail.
• ~25 min LP only, no editor (Reddit 1qj304z).

IE-asked LC 7406809, LC 6806195, Reddit 1qj304z. Other loops, not a forecast for Job 10454435.

What you say

They might ask next

Keep straight

• These rooms are other candidates’. Job 10454435 still none.
• Distance K with no parent map on LC 7406809 was that IE’s Round 2, not this leftover card. The leftover on that loop was the A/B/C path.

Tighten

Count Number of Nice Subarrays

Exactly k odd numbers in a contiguous subarray. Evens are free inside the window. Odds are a 0/1 budget. Exactly k is atMost(k) minus atMost(k−1).

What they might ask

• Count contiguous subarrays that contain exactly k odd numbers.
• Example: [1,1,2,1,1], k = 3 → 2 subarrays: [1,1,2,1] and [1,2,1,1].

IE-asked LC 6806195 — BR as third live, AUTA Bengaluru 2024, OA-as-R1 N. Candidate linked the slug /problems/count-number-of-nice-subarrays/. Numeric id was not typed. Same hour: LP on feedback / critical issues / went beyond task. Other loop, not a forecast for Job 10454435.

In plain words

Exactly k is not monotone: adding an odd can jump from k to k+1, so you cannot only shrink for “exactly.” atMost(k) is monotone. atMost(3) − atMost(2) leaves windows with exactly 3 odds. On the example that difference is 2.

What you say

public int numberOfSubarrays(int[] a, int k) {
  return atMostOdd(a, k) - atMostOdd(a, k - 1);
}
private int atMostOdd(int[] a, int k) {
  if (k < 0) return 0;
  int l = 0, odd = 0, ans = 0;
  for (int r = 0; r < a.length; r++) {
    odd += a[r] & 1;
    while (odd > k) odd -= a[l++] & 1;
    ans += r - l + 1;
  }
  return ans;
}

They might ask next

Keep straight

• The candidate linked a slug. Numeric LeetCode id was not typed. Unnamed numeric stays unnamed.

Tighten

Tree path src→target printed A/B/C

Shortest path as letters: A = left child, B = right child, C = parent. Extra space refused on that write-up; LCA + DFS on the recursion stack.

What they might ask

• Given root, source, target; print the path as A/B/C.
• Example: node 2 has children 1 (left) and 4 (right). Src 1, target 4 → "CB" (up to 2, then right).

IE-asked LC 7406809 — BR as third, AUTA, OA-as-R1 N. Problem UNNAMED. Recruiter had said 3 LP; DSA anyway. Same IE’s Distance K with no parent map was their R2, not this leftover. Other loop, not a forecast for Job 10454435.

In plain words

Root-path to 1 and to 4 share prefix through 2. Cancel the common prefix: up from 1 to 2 is C; down 2 to 4 is B. Src equals target → empty string.

What you say

public String pathABC(TreeNode root, TreeNode src, TreeNode target) {
  List<TreeNode> toSrc = new ArrayList<>();
  List<TreeNode> toTgt = new ArrayList<>();
  if (!dfs(root, src, toSrc) || !dfs(root, target, toTgt)) return "";
  int i = 0, lim = Math.min(toSrc.size(), toTgt.size());
  while (i < lim && toSrc.get(i) == toTgt.get(i)) i++;
  int lca = i - 1;
  StringBuilder sb = new StringBuilder();
  for (int j = toSrc.size() - 1; j > lca; j--) sb.append('C');
  for (int j = lca; j < toTgt.size() - 1; j++) {
    TreeNode cur = toTgt.get(j), nxt = toTgt.get(j + 1);
    sb.append(cur.left == nxt ? 'A' : 'B');
  }
  return sb.toString();
}
private boolean dfs(TreeNode n, TreeNode goal, List<TreeNode> path) {
  if (n == null) return false;
  path.add(n);
  if (n == goal) return true;
  if (dfs(n.left, goal, path) || dfs(n.right, goal, path)) return true;
  path.remove(path.size() - 1);
  return false;
}

They might ask next

Keep straight

• This problem is UNNAMED. The write-up did not attach a numeric id. Dist-K on this same IE was Round 2, already-done analogue, not this card.

Tighten

Unique Paths–like maximum points (right/down only)

Grid path aggregation, not path count. Start and end may not be corners. Robot may move Right or Down only.

What they might ask

• Grid; start cell and end cell given as coordinates; maximum points collectable; Right or Down only.
• Shape example (their grid was not published): start (0,0), end (1,2), points [[1,2,3],[4,5,6]] → 16 along 1→4→5→6.

IE-asked LC 7035289 — BR as third, AUTA, OA-as-R1 N. Candidate said similar to Unique Paths. UNNAMED. Other loop, not a forecast for Job 10454435.

What you say

public long maxPoints(int[][] g, int sr, int sc, int er, int ec) {
  int m = g.length, n = g[0].length;
  long NEG = Long.MIN_VALUE / 4;
  long[][] dp = new long[m][n];
  for (int i = 0; i < m; i++) Arrays.fill(dp[i], NEG);
  dp[sr][sc] = g[sr][sc];
  for (int i = sr; i <= er; i++) {
    for (int j = sc; j <= ec; j++) {
      if (i == sr && j == sc) continue;
      long up = i > sr ? dp[i - 1][j] : NEG;
      long left = j > sc ? dp[i][j - 1] : NEG;
      dp[i][j] = g[i][j] + Math.max(up, left);
    }
  }
  return dp[er][ec];
}

They might ask next

Keep straight

• Similar ≠ exact Unique Paths. The write-up did not name a numeric id. LC 6800853 grid paths/rewards is OA-as-R1 — our R2 analogue, not later-slot.

Tighten

Distance K, O(n), no parent mapping — Fluency-labeled

Invite said Gen AI Fluency. The hour was a tree walk. Write the tree first. Parent HashMap forbidden. After code, one GenAI sentence only if they still ask.

What they might ask

• “Similar to finding nodes at distance K from a target node (expected O(n) and no parent mapping).”
• Usual shape: target = node 5, k = 2 → [7, 4, 1] (order free).

IE-asked LC 7623949 — labeled Gen AI Fluency = third live after two onsites; candidate: “essentially a DSA round.” BR after = fourth, LP-only. Sibling later: Dist-K family on “GenAI Essential + Coding” (LC 7584467). Sibling fourth: Dist-K + LCA titles UNNAMED (LC 6461439). Intern Fluency LC 7643209 unnamed min-heap is not this. AUTA Dist-K on 7406809 Round 2 is not leftover. Other loop, not a forecast for Job 10454435.

In plain words

        3
       / \
      5   1
     / \ / \
    6  2 0  8
      / \
     7   4

find(5) returns 0 and collects down k=2 (7, 4). find(3) sees left distance 0; does not add 3 (1 ≠ 2); collects sibling 1 at remaining 0. Draw their tree first if they drew a different one.

What you say

public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
  List<Integer> ans = new ArrayList<>();
  find(root, target, k, ans);
  return ans;
}
/** distance from node down to target, or -1 */
private int find(TreeNode node, TreeNode target, int k, List<Integer> ans) {
  if (node == null) return -1;
  if (node == target) {
    collectDown(node, k, ans);
    return 0;
  }
  int L = find(node.left, target, k, ans);
  if (L >= 0) {
    if (L + 1 == k) ans.add(node.val);
    collectDown(node.right, k - L - 2, ans);
    return L + 1;
  }
  int R = find(node.right, target, k, ans);
  if (R >= 0) {
    if (R + 1 == k) ans.add(node.val);
    collectDown(node.left, k - R - 2, ans);
    return R + 1;
  }
  return -1;
}
private void collectDown(TreeNode node, int dist, List<Integer> ans) {
  if (node == null || dist < 0) return;
  if (dist == 0) { ans.add(node.val); return; }
  collectDown(node.left, dist - 1, ans);
  collectDown(node.right, dist - 1, ans);
}

They might ask next

Keep straight

• Distance-K family. The write-up did not type a numeric LeetCode id. Intern 7643209 min-heap is a different Fluency hour. 7406809 Dist-K was that IE’s R2.

Tighten

Next Greater Element

For every index, first value to its right that is strictly greater. Monotonic stack. Same HM hour as LLM talk — code is the named technical.

What they might ask

• Next greater to the right. Confirm circular / next-greater-left / two arrays.
• Example: [2, 1, 2, 4, 3] → [4, 2, 4, -1, -1].

IE-asked LC 7724048 — HM Gen-AI as third; BR = fourth LP-only, no DSA. Title spoken; UNNAMED id. Next Permutation on Reddit 1u0yzoq was a GenAI third, not this HM. Other loop, not a forecast for Job 10454435.

What you say

public int[] nextGreater(int[] a) {
  int n = a.length;
  int[] ans = new int[n];
  Arrays.fill(ans, -1);
  Deque<Integer> st = new ArrayDeque<>();
  for (int i = n - 1; i >= 0; i--) {
    while (!st.isEmpty() && st.peek() <= a[i]) st.pop();
    if (!st.isEmpty()) ans[i] = st.peek();
    st.push(a[i]);
  }
  return ans;
}

They might ask next

Keep straight

• Title spoken, id UNNAMED. Next permutation is a different prompt on a different IE.

Tighten

Dynamic k-th largest (k changes)

Insert plus find k-th largest. k restates between queries. A min-heap sized to a fixed k breaks unless you rebuild.

What they might ask

• “Chef Finds KthLargest Element” — insert + find; k changes. Hint: data structure.
• Stream [3,7,1] then find k=3, insert 4, insert 9, find k=2, find k=4 → [1, 7, 3].

IE-asked LC 7850431 ≡ IE.in 85 — GenAI+Technical as third; Round 4 behavioural = fourth, LP-only. UNNAMED. They timed out after BST talk on that IE. Other loop, not a forecast for Job 10454435.

What you say

class KthTracker {
  private final TreeMap<Integer, Integer> freq = new TreeMap<>();
  private int size = 0;
  void insert(int x) {
    freq.merge(x, 1, Integer::sum);
    size++;
  }
  int kthLargest(int k) {
    if (k < 1 || k > size) throw new IllegalArgumentException("k");
    int seen = 0;
    for (int v : freq.descendingKeySet()) {
      seen += freq.get(v);
      if (seen >= k) return v;
    }
    throw new IllegalStateException();
  }
}

They might ask next

Keep straight

• UNNAMED. Static k-th largest with a fixed-k heap is a different family. That IE timed out after BST talk — speak insert then find, then the walk.

Tighten

Maximum Sum Linked List from two sorted lists

Two sorted lists that share some common nodes. Commons are join points. Between two commons, keep the heavier segment. Sibling wording returns the sum; another wording constructs the list.

What they might ask

• Construct a maximum-sum list out of two sorted lists with common nodes — or switch at commons and return the sum.
• Example (6369243): 1→5→8→12 and 2→4→5→12 → 2+4+5+8+12 = 31.

IE-asked LC 6881058 heading Round 4 BR; prior lives href LC 6800853. FLAG OA-as-R1: chronological lives after dropping OA = third. Leftover later-slot shape. Sibling LC 6369243: BR as third, OA-as-R1 N; return maximum sum; no convert to array; follow-up K lists → heap. Numeric ids 962/210 sit on 6369243’s Round 1, not this BR item. Other loop, not a forecast for Job 10454435.

In plain words

1 vs 2: take the smaller pointer forward. At common 5, max(seg A=1, seg B=2+4)=6 plus 5. Then 8 vs 12 until common 12: max(8, 0)+12. Total 31. Lists increase, so the next equal value is the only legal switch.

What you say

public long maxSum(ListNode a, ListNode b) {
  long ans = 0, sa = 0, sb = 0;
  ListNode p = a, q = b;
  while (p != null && q != null) {
    if (p.val < q.val) { sa += p.val; p = p.next; }
    else if (q.val < p.val) { sb += q.val; q = q.next; }
    else {
      ans += Math.max(sa, sb) + p.val;
      sa = sb = 0;
      p = p.next;
      q = q.next;
    }
  }
  while (p != null) { sa += p.val; p = p.next; }
  while (q != null) { sb += q.val; q = q.next; }
  return ans + Math.max(sa, sb);
}

They might ask next

Keep straight

• GFG-family title spoken; no LeetCode id on the BR problem. 962/210 were that IE’s first live, not this leftover item. This is not longest increasing path and not course-schedule graph.

Tighten

Combine Garlands (ropes; long) — HM, not BR

Minimum cost to combine lengths. Same idea as connect sticks: always merge the two smallest. Cost is the sum of pairwise merges, not the final length. Use long.

What they might ask

• Minimum cost to combine garlands. Overflow → long.
• Example: [4, 3, 2, 6] → merge 2+3=5 (cost 5), 4+5=9 (14), 6+9=15 (29).

IE-asked LC 8362604 — HM as third (15 Apr 2026), not BR. AUTA APAC India 2026. BR 30 May = fourth LP-only, no DSA. Rate-limiting / geofencing on that BR = story follow-ups, not Rate Limiter LLD. Same HM as Anagrams variation. Same IE’s R2 Connect Ropes is already-done analogue, not this card. Other loop, not a forecast for Job 10454435.

What you say

public long combineGarlands(int[] len) {
  PriorityQueue<Long> pq = new PriorityQueue<>();
  for (int x : len) pq.offer((long) x);
  long cost = 0;
  while (pq.size() > 1) {
    long a = pq.poll(), b = pq.poll();
    cost += a + b;
    pq.offer(a + b);
  }
  return cost;
}

They might ask next

Keep straight

• Garlands is not a named LC. This hour is HM, not BR. Connect Ropes on that IE was R2.

Tighten

Anagrams variation — HM

Same AUTA 2026 HM as garlands. Variation unknown — ask the twist before coding. Classic family is a frequency window of length |p|.

What they might ask

• A variation of Find All Anagrams. Ask: unicode? k anagrams? min window? return indices vs count?
• Classic family (not claimed as their twist): s = cbaebabacd, p = abc → starts [0, 6].

IE-asked LC 8362604 — HM as third, not BR. BR fourth no DSA. Other loop, not a forecast for Job 10454435.

What you say

public List<Integer> findAnagrams(String s, String p) {
  int[] need = new int[26], have = new int[26];
  for (int i = 0; i < p.length(); i++) need[p.charAt(i) - 'a']++;
  List<Integer> ans = new ArrayList<>();
  int m = p.length();
  if (s.length() < m) return ans;
  for (int i = 0; i < s.length(); i++) {
    have[s.charAt(i) - 'a']++;
    if (i >= m) have[s.charAt(i - m) - 'a']--;
    if (i >= m - 1 && Arrays.equals(need, have)) ans.add(i - m + 1);
  }
  return ans;
}

They might ask next

Keep straight

• Unnamed variation stays unnamed. Classic Java is a starting sketch after they lock the twist.

Tighten

Word Ladder family CAT → MEN

One letter per iteration; intermediates must be dictionary words. Unweighted shortest transformation is BFS. Recursion finds a path, not necessarily the shortest.

What they might ask

• CONVERT CAT → MEN; dictionary already implemented.
• Example: CAT → CAN → MAN → MEN.

IE-asked GFG sde-1-17 — F2F4 = fourth. Same IE as Validate sudoku on F2F3. AUTA N. Other loop, not a forecast for Job 10454435.

What you say

public int ladderLength(String begin, String end, Set<String> dict) {
  if (begin.equals(end)) return 0;
  Queue<String> q = new ArrayDeque<>();
  Set<String> seen = new HashSet<>();
  q.add(begin);
  seen.add(begin);
  int steps = 0;
  while (!q.isEmpty()) {
    int sz = q.size();
    steps++;
    for (int s = 0; s < sz; s++) {
      char[] a = q.poll().toCharArray();
      for (int i = 0; i < a.length; i++) {
        char old = a[i];
        for (char c = 'A'; c <= 'Z'; c++) {
          a[i] = c;
          String nxt = new String(a);
          if (nxt.equals(end)) return steps;
          if (dict.contains(nxt) && seen.add(nxt)) q.add(nxt);
        }
        a[i] = old;
      }
    }
  }
  return -1;
}

They might ask next

Keep straight

• Word Ladder family. No LC id named on that write-up.

Tighten

Most frequent 3-page sequence

Consecutive triples per user, then a global map. Not a cartesian of pages across users. Leftover 10–15 min after LP on that write-up.

What they might ask

• Inputs Username – Page visited. Return the three-page sequence visited the most times by users.
• User1 P1-P3-P4-P2, User2 P3-P4-P2-P1 → P3-P4-P2.

IE-asked LC 6573582 — BR as third, Bangalore SDE-1, AUTA N. UNNAMED. Other loop, not a forecast for Job 10454435.

What you say

public String mostFrequentTriple(Map<String, List<String>> userPages) {
  Map<String, Integer> cnt = new HashMap<>();
  String best = null;
  int bestC = 0;
  for (List<String> pages : userPages.values()) {
    for (int i = 0; i + 2 < pages.size(); i++) {
      String key = pages.get(i) + "-" + pages.get(i + 1) + "-" + pages.get(i + 2);
      int c = cnt.merge(key, 1, Integer::sum);
      if (c > bestC) { bestC = c; best = key; }
    }
  }
  return best;
}

They might ask next

Keep straight

• UNNAMED. Not an OA id.

Tighten

Reverse linked list in groups of k

Dummy head; count k; reverse pointers in place. Ask leftover-tail policy: reverse the remainder vs leave it.

What they might ask

• Reverse the linked list in groups of k. Two approaches + optimizations.
• 1→2→3→4→5, k=2 → 2→1→4→3→5 if leftover stays. k=3 → 3→2→1→4→5 if leftover stays.

IE-asked Uday LinkedIn — BR as third (OA separate; R1 DSA; R2 Amazon Locker SD = not later). AUTA N. LP collaboration / ownership UNNAMED same BR. Other loop, not a forecast for Job 10454435.

What you say

public ListNode reverseKGroup(ListNode head, int k) {
  if (k <= 1 || head == null) return head;
  ListNode dummy = new ListNode(0);
  dummy.next = head;
  ListNode prev = dummy;
  while (true) {
    ListNode node = prev;
    for (int i = 0; i < k && node != null; i++) node = node.next;
    if (node == null) break;
    ListNode cur = prev.next;
    for (int i = 1; i < k; i++) {
      ListNode nxt = cur.next;
      cur.next = nxt.next;
      nxt.next = prev.next;
      prev.next = nxt;
    }
    prev = cur;
  }
  return dummy.next;
}

They might ask next

Keep straight

• UNNAMED. Locker SD on that IE was R2, not this leftover reverse-k.

Tighten

Validate sudoku (not Solver)

Constraint check: rows, cols, 3×3 boxes. Empty cells valid. Writing a backtracking solver is the wrong prompt.

What they might ask

• “1 coding question — validate sudoku.” Empty cells valid. Not Sudoku Solver.
• Duplicate 5 in a 3×3 box → false. Filled valid row/col/box with '.' empties → true.

IE-asked GFG sde-1-17 — F2F3 = third. AUTA N. LP extra-for-customers / client POV / conflict same F2F. Other loop, not a forecast for Job 10454435.

What you say

public boolean isValidSudoku(char[][] b) {
  boolean[][] row = new boolean[9][9];
  boolean[][] col = new boolean[9][9];
  boolean[][] box = new boolean[9][9];
  for (int i = 0; i < 9; i++) {
    for (int j = 0; j < 9; j++) {
      if (b[i][j] == '.') continue;
      int d = b[i][j] - '1';
      if (d < 0 || d > 8) return false;
      int k = (i / 3) * 3 + j / 3;
      if (row[i][d] || col[j][d] || box[k][d]) return false;
      row[i][d] = col[j][d] = box[k][d] = true;
    }
  }
  return true;
}

They might ask next

Keep straight

• UNNAMED. This card is validate, not Solver. Word Ladder on the same IE was F2F4.

Tighten

Union-Find / Disjoint Set — AUTA BR

Structure named, instance not given. Ask which API: connectivity, merge accounts, equations. Path compression + union by rank.

What they might ask

• DSA Disjoint Set (Union-Find logic). LP same hour: failed; missed a deadline; current project architecture & flow.
• Shape after they name the API: union(1,2), union(2,3), connected(1,3) → true; connected(1,4) → false if 4 never united.

IE-asked LC 6866029 — BR as third, AUTA, OA-as-R1 N. Problem statement beyond “DSU logic” not given. Other loop, not a forecast for Job 10454435.

What you say

class DSU {
  final int[] p, r;
  DSU(int n) {
    p = new int[n];
    r = new int[n];
    for (int i = 0; i < n; i++) p[i] = i;
  }
  int find(int x) { return p[x] == x ? x : (p[x] = find(p[x])); }
  boolean union(int a, int b) {
    a = find(a); b = find(b);
    if (a == b) return false;
    if (r[a] < r[b]) { int t = a; a = b; b = t; }
    p[b] = a;
    if (r[a] == r[b]) r[a]++;
    return true;
  }
}

They might ask next

Keep straight

• Instance UNNAMED. Islands-II / Evaluate Division ids sit on other lives of other IEs. Wait for their nouns before attaching a famous problem.

Tighten

Celebrity variant — AUTA BR

A variant of the Celebrity Problem. Exact variant UNNAMED. Ask before coding classic elimination. Skipping verify after picking a candidate is the classic fail.

What they might ask

• A variant of the Celebrity Problem. Ask: no celebrity? many? directed follows? knows(i,j) oracle vs matrix?
• Classic family only (not claimed as their variant): celebrity known by everyone, knows no one.

IE-asked Reddit 1u0yzoq (No-Huckleberry-5969). BR as fourth after two offline DSA + GenAI third (Next Permutation in that GenAI hour). AUTA; ~1.5 YOE frontend; rejected. Job 10454435 N. One IE, not two. GFG off-campus-10 Celebrity R1 is the wrong slot. Other loop, not a forecast for Job 10454435.

What you say

public int findCelebrity(int n) {
  int cand = 0;
  for (int i = 1; i < n; i++) {
    if (knows(cand, i)) cand = i;
  }
  for (int i = 0; i < n; i++) {
    if (i == cand) continue;
    if (knows(cand, i) || !knows(i, cand)) return -1;
  }
  return cand;
}
boolean knows(int a, int b) { return false; /* interviewer oracle */ }

They might ask next

Keep straight

• Unnamed variant. Classic sketch is a fallback after they confirm classic. GFG off-campus-10 Celebrity was R1 on another loop.

Tighten

Srenitha — sliding window UNNAMED

Coding similar to “maximum consecutive numbers.” Candidate: classic sliding window; follow-ups why this approach, how the window works, dry run. Exact constraint unnamed. Similar is not a proof of longest consecutive sequence.

What they might ask

• Maximum consecutive numbers — then why window, how it works, dry run.
• Ask: consecutive values vs consecutive ones vs at most k flips. Then dry-run their numbers.

IE-asked Srenitha Medium — BR as third after two UTA lives. UTA named SDE-I; selected Feb 2026. R2 pizza OOD = already-done analogue, not this card. Other loop, not a forecast for Job 10454435.

What you say

public int longestValidWindow(int[] a /* ask the predicate */) {
  int l = 0, best = 0;
  // maintain window invariant here — after they lock k-flips, ones-count, or value-run
  for (int r = 0; r < a.length; r++) {
    // add a[r]
    while (/* window invalid */) l++;
    best = Math.max(best, r - l + 1);
  }
  return best;
}

They might ask next

Keep straight

• UNNAMED. Not proven as HashSet longest-consecutive-length. Pizza OOD on that IE was R2.

Tighten

Reverse alternate levels of a binary tree (not perfect)

Their tree was not a perfect binary tree. Perfect-tree sibling-pointer trick dies. Level-order collect nodes on alternate levels, reverse values pairwise.

What they might ask

• Reverse the alternate levels of a Binary Tree. Ask which levels (odd vs even vs 1-indexed) and values vs node links.
• Imperfect example: root 1; left 2 with child 4; right 3 (no child). Alternate level 1 (2 and 3) swap values → 1, then 3 and 2, then 4 unchanged.

IE-asked Sarvesh Medium — BR as third after two lives. India SDE-1; rejected. AUTA N. Candidate linked similar to Reverse Odd Levels. LiveCode no compile/execution. Same BR: conflicting opinions; project idea; tech why; challenges. Other loop, not a forecast for Job 10454435.

What you say

public void reverseAlternate(TreeNode root) {
  if (root == null) return;
  Queue<TreeNode> q = new ArrayDeque<>();
  q.add(root);
  int level = 0;
  while (!q.isEmpty()) {
    int sz = q.size();
    List<TreeNode> row = new ArrayList<>();
    for (int i = 0; i < sz; i++) {
      TreeNode n = q.poll();
      if ((level & 1) == 1) row.add(n); // ask odd vs even
      if (n.left != null) q.add(n.left);
      if (n.right != null) q.add(n.right);
    }
    int i = 0, j = row.size() - 1;
    while (i < j) {
      int t = row.get(i).val;
      row.get(i).val = row.get(j).val;
      row.get(j).val = t;
      i++; j--;
    }
    level++;
  }
}

They might ask next

Keep straight

• Similar ≠ a stamped Reverse Odd Levels id. Unnamed variation. Their tree was not perfect.

Tighten

n processes → k machines (i % k)

Assign each process starting at machine i % k; if busy try (i+1)%k …; return the machine handling the maximum number of processes. Duration / busy model unnamed — ask.

What they might ask

• k machines, n processes, wrap next-fit, argmax count.
• n=5, k=2. Starts: 0,1,0,1,0. If nobody is ever busy, counts [3,2], machine 0 wins. If process 0 still occupies machine 0 when process 2 arrives, wrap to machine 1 — ask the busy model.

IE-asked IE.in 4-rounds (gated body). HM as third after two lives. IE.in labeled SDE I. AUTA N. Their R4 = LP names only. Other loop, not a forecast for Job 10454435.

What you say

public int busiestMachine(int n, int k) {
  int[] count = new int[k];
  boolean[] busy = new boolean[k]; // ASK: when does busy clear?
  for (int i = 0; i < n; i++) {
    int chosen = -1;
    for (int t = 0; t < k; t++) {
      int m = (i % k + t) % k;
      if (!busy[m]) { chosen = m; break; }
    }
    if (chosen < 0) {
      // ASK: wait, drop, or overwrite?
      continue;
    }
    busy[chosen] = true;
    count[chosen]++;
    // ASK: busy[chosen] = false after duration[i]
  }
  int best = 0, who = 0;
  for (int m = 0; m < k; m++) if (count[m] > best) { best = count[m]; who = m; }
  return who;
}

They might ask next

Keep straight

• Time model unnamed. No guessed LC id. This is assignment simulation, not Bankers, not the scheduler LLD.

Tighten

Bookstore word-count OOD

LLD, not DSA. Multiple books; count of a particular word in a specific book. Each book owns its vocabulary.

What they might ask

• Design a bookstore; classes, objects, methods; return the count of a word appearing in a specific book; then projects.
• count("book-42", "Gandalf") → 17 in that book only, not a global bag mixing titles.

IE-asked Shiwangi — UTA R3 OOD = true third (16 Dec 2024). Same IE R4 = work deep-dive + medium Sliding Window UNNAMED (DSA leftover, not this LLD). AUTA N / UTA Y. TUF 6-22ec = reprint (OA-as-R1 numbering). Other loop, not a forecast for Job 10454435.

What you say

class Book {
  final String id;
  final Map<String, Integer> wordCount = new HashMap<>();
  Book(String id, String text) {
    this.id = id;
    for (String w : text.toLowerCase().split("[^a-z0-9]+")) {
      if (w.isEmpty()) continue;
      wordCount.merge(w, 1, Integer::sum);
    }
  }
  int count(String word) {
    return wordCount.getOrDefault(word.toLowerCase(), 0);
  }
}
class Catalog {
  private final Map<String, Book> books = new HashMap<>();
  void add(Book b) { books.put(b.id, b); }
  int count(String bookId, String word) {
    Book b = books.get(bookId);
    if (b == null) return 0; // ASK missing
    return b.count(word);
  }
}

They might ask next

Keep straight

• Named prompt, no LC. Bookstore is LLD. Sliding window on that IE’s fourth live is not this card.

Tighten

Email keyword search

Store emails. How many times a keyword appears in a specific email, or across a user’s inbox. Mixing users is a leak, not a feature.

What they might ask

• Design an Email Search System: store emails; two APIs — one email vs inbox sum.
• countInEmail(e9, "invoice") → 3. countInInbox(userA, "invoice") → sum over that user’s mails only.

IE-asked TUF 17-2447 / dump 10-sde1-amazon-interview-experience-17-2447.md — their R4 SD after two live DSA. FLAG OA-as-R1 Y → our R3 analogue (leftover after two lives). Their R5 behavioural = further leftover. AUTA N. Other loop, not a forecast for Job 10454435.

What you say

class Email { String id, userId, body; }
class EmailIndex {
  Map<String, Email> emails = new HashMap<>();
  Map<String, List<String>> userEmails = new HashMap<>();
  Map<String, Map<String, Integer>> kwEmail = new HashMap<>(); // kw -> emailId -> n
  void add(Email e) {
    emails.put(e.id, e);
    userEmails.computeIfAbsent(e.userId, k -> new ArrayList<>()).add(e.id);
    Map<String, Integer> bag = new HashMap<>();
    for (String w : e.body.toLowerCase().split("\\W+")) {
      if (!w.isEmpty()) bag.merge(w, 1, Integer::sum);
    }
    for (var en : bag.entrySet()) {
      kwEmail.computeIfAbsent(en.getKey(), k -> new HashMap<>()).put(e.id, en.getValue());
    }
  }
  int inEmail(String emailId, String kw) {
    Map<String, Integer> m = kwEmail.get(kw.toLowerCase());
    if (m == null) return 0;
    return m.getOrDefault(emailId, 0);
  }
  int inInbox(String userId, String kw) {
    int s = 0;
    for (String id : userEmails.getOrDefault(userId, List.of())) s += inEmail(id, kw);
    return s;
  }
}

They might ask next

Keep straight

• OA-as-R1 on that loop: their R4 after two lives is leftover shape, not a 10454435 forecast. No LC id.

Tighten

Login layers (Controller / Facade / Service / Repository)

Another loop asked a login feature in those four layers, then how to replace a pile of if-else in the facade. This is layered structure in Live Code. It is not Login Tracker.

What they might ask

• Design a login feature; explain Controller, Facade, Service, Repository.
• Follow-up: how would you eliminate multiple if-else validations in the facade layer?

IE-asked Bhavya LinkedIn — Interview 3 HM = third. Hyd onsite May 2026. BR 3 Jun = LP only, no LLD. AUTA N. Other loop, not a forecast for Job 10454435.

What you say

interface Validator { void validate(LoginRequest req); }
class Facade {
  private final List<Validator> validators;
  private final AuthService auth;
  LoginResponse login(LoginRequest req) {
    for (Validator v : validators) v.validate(req);
    return auth.authenticate(req);
  }
}
// Controller -> Facade -> AuthService -> UserRepository

They might ask next

Keep straight

• Internships used Python FastAPI, TypeScript NestJS, and ROS2. Live Code is Java. This other loop asked Spring-named layers. Sketch them in Java without describing Ylogx or IQVIA as Spring Boot services.
• Login Tracker (new_login / get_oldest_login) is unverified research. It is not this prompt. Do not volunteer it.

Tighten

Process scheduler (LLD)

Process states plus a ready queue plus a policy. Same HM also had GenAI usage, verify GenAI code, and LCA (slug linked). If they paste a tree, write the tree first. Different prompt from n-processes → k-machines.

What they might ask

• “Design a system to schedule processes.”
• submit(p1 priority 2), submit(p2 priority 5) → pick next returns p2. States NEW → READY → RUNNING → TERMINATED.

IE-asked LC 8014509 — HM Gen-AI as third (onsite 6 Apr 2026 same day as R1/R2). BR 14 Apr = fourth LP-only, no technical. Internals UNNAMED. Other loop, not a forecast for Job 10454435.

What you say

enum State { NEW, READY, RUNNING, WAITING, TERMINATED }
class Process {
  int pid, priority;
  State state = State.NEW;
}
class Scheduler {
  private final Queue<Process> ready =
      new PriorityQueue<>((a, b) -> Integer.compare(b.priority, a.priority));
  void submit(Process p) { p.state = State.READY; ready.offer(p); }
  Process pick() {
    Process p = ready.poll();
    if (p != null) p.state = State.RUNNING;
    return p;
  }
}

They might ask next

Keep straight

• Scheduler sentence named; internals unnamed. Not OS exam Bankers unless they ask. Not the k-machines assignment card.

Tighten

Light system design UNNAMED

Toward the end, a very basic system design prompt — title not given. Wait for their nouns. Entities, 2–3 APIs, in-memory map. No HLD boxes.

What they might ask

• Unknown. Two LP UNNAMED in-depth first, then a basic SD.

IE-asked Naina — UTA R3 = true third (2 Jul 2025). Offer after three lives. AUTA N / UTA Y. Rudraksh BR “common scenario from SDE-1 prep sets” is a different unnamed LLD. Other loop, not a forecast for Job 10454435.

What you say

// wait for their nouns
// class X { Map<String, Y> store; void write(...); Z read(...); }

They might ask next

Keep straight

• UNNAMED. Parking Lot, Locker, Rate Limiter, and bookstore are other cards or other loops — not fillers for a missing title.

Tighten

Rate Limiter — independent count = 1, not UTA-default Round 4

Asked in a different loop, not leftover UTA two-DSA. Independent SDE I live-round count = 1. OA+3 system design. Not a fourth live. Not Adarsh Round 4 warmup. If they still paste it, solve what is on the screen.

What they might ask

• Design a Rate Limiter + scale across distributed systems — on that other loop’s Second Technical (13 Feb 2025), after OS+graph on 4 Feb, before Final LP 17 Jul.

IE-asked IE.in 2025-grad / TUF 21-5773 reprint (same report, not count 2). Not two-DSA UTA. Rudraksh prep list only is not a second ask. LC 8362604 “rate limiting decisions” / geofencing = BR stories. SDE II Rate Limiter = wrong level. Other loop, not a forecast for Job 10454435.

What you say

They might ask next

Keep straight

• Independent SDE I live-round count = 1. Not UTA-default Round 4. Not Job 10454435 warmup.
• Login Tracker remains unverified and is not a substitute.
• −35% is Ylogx bot DB via Redis. It is not a rate-limiter metric.

Tighten

Also later-slot (index only — not warmup cards)

Survived remap. Not in the must-cover set above. Unnamed stays unnamed. No guessed ids. Recognition only.

What they might ask

• A leftover DSA that matches a row below. Treat it as UNNAMED until they lock the statement.

IE-asked other loops. Not a Job 10454435 list.

What you say

They might ask next

Keep straight

• Not later-slot: 6800853 grid paths/rewards = our R2 under OA-as-R1; 7406809 Dist-K = their R2; intern Fluency min-heap 7643209 ≠ Distance K; Ansuman nearest-mall / Subset Sum 10^15 = R1/R2.

Tighten