Top 10 LeetCode Problems for Beginners: Your First Coding Interview

You open LeetCode for the first time and see 2,800 problems. The filter has 50 tags. Every Reddit thread disagrees on where to start. You close the tab.

Same story five minutes later on YouTube. Seventeen different "roadmaps." One guy recommends grinding 500 problems. Another says just do the Blind 75. A third guy is somehow selling a course on how to find the right course. You close that tab too.

Good news: your first interview draws from a tiny, predictable pool. Recruiters at most companies screening entry-level engineers pull from the same 15-ish archetypes, over and over, in slightly different costumes. Learn the archetypes and the variants solve themselves.

This list is not the easiest ten problems, or the most upvoted ten. These are the ten LeetCode problems for beginners that teach the most transferable patterns, ordered roughly by how often they appear and how much each one unlocks. Solve all ten correctly, including the gotcha edge cases, and you're ready for most first-round coding screens.

1. Two Sum (LC 1): The Hash Map Is a Lookup Accelerator

The brute-force check-every-pair solution is O(n²). It also works in a contest where the test cases are weak. Do not submit it in an interview.

The trick is trading memory for time: scan once, storing each number's index in a hash map, and check whether the complement already exists.

def twoSum(nums: list[int], target: int) -> list[int]:
    seen = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i
    return []

The underlying pattern is the complement lookup. Any time a problem asks "does X exist in what I've already processed," a hash map gets you from O(n) search cost to O(1). That reflex shows up in Three Sum, Four Sum, and dozens of subarray problems. See how hash maps achieve that O(1) if the mechanism isn't clear yet.

Worth internalizing now: the answer to roughly a third of interview problems is "put it in a hash map." It's embarrassing how often it works.

2. Valid Parentheses (LC 20): Brackets Wait. The Stack Remembers.

Push open brackets. When you hit a close bracket, pop and verify the last open matches. Stack empty at the end means valid.

def isValid(s: str) -> bool:
    stack = []
    pairs = {')': '(', ']': '[', '}': '{'}
    for char in s:
        if char in '([{':
            stack.append(char)
        elif not stack or stack[-1] != pairs[char]:
            return False
        else:
            stack.pop()
    return len(stack) == 0

The edge cases are where candidates fail: closing bracket when the stack is already empty (the not stack guard), or a non-empty stack at the end (unmatched opens). Both checks are in the code above. This pattern generalizes to expression evaluation, HTML tag matching, and any problem that involves nested matching.

3. Best Time to Buy and Sell Stock (LC 121): Track the Running Minimum

You can't buy in the future and sell in the past. This is a real constraint, not a gotcha, but interviews exist where people forget it.

So scan left to right. At each price, update the cheapest buy seen so far. The profit if you sold today is current price minus that minimum. Track the maximum across all days.

def maxProfit(prices: list[int]) -> int:
    min_price = float('inf')
    max_profit = 0
    for price in prices:
        min_price = min(min_price, price)
        max_profit = max(max_profit, price - min_price)
    return max_profit

The framing to internalize: you don't need to know the future if you're tracking the best option from the past. That reasoning recurs in "maximum difference," "jump game," and anything that looks two-dimensional but collapses to one linear scan.

4. Reverse a Linked List (LC 206): Three Pointers, Every Time

In-place reversal means flipping the next pointer at each node to point backward. You need three variables: the previous node, the current node, and a saved reference to the next node before you overwrite it. The pattern is always: save next, flip the pointer, advance both prev and curr.

def reverseList(head):
    prev = None
    curr = head
    while curr:
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node
    return prev

Internalize this one. Reverse in K-Groups, reorder list, palindrome linked list check, and a dozen other problems are this pattern nested inside something else. If you blank on it during an interview, write "save, flip, advance" in the margin and work backward from that.

5. Binary Search (LC 704): Get the Invariant Right Once

The code is short. The bugs are legendary. Java's standard library had a binary search overflow bug that sat undetected for nine years. Use lo + (hi - lo) // 2 instead of (lo + hi) // 2 to avoid integer overflow, and use lo <= hi as the loop condition with mid + 1 / mid - 1 updates to avoid infinite loops.

def search(nums: list[int], target: int) -> int:
    lo, hi = 0, len(nums) - 1
    while lo <= hi:
        mid = lo + (hi - lo) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            lo = mid + 1
        else:
            hi = mid - 1
    return -1

This also unlocks "binary search on the answer" problems, where you search a value space instead of an array. The binary search invariant is worth reading once in depth so you stop guessing which termination condition to use.

6. Climbing Stairs (LC 70): DP Hides Inside Fibonacci

To reach stair n, you had to come from stair n-1 or stair n-2. There's no other option. The number of ways to reach stair n equals the ways to reach n-1 plus the ways to reach n-2. That's Fibonacci, reframed as dynamic programming.

def climbStairs(n: int) -> int:
    if n <= 2:
        return n
    prev2, prev1 = 1, 2
    for _ in range(3, n + 1):
        curr = prev1 + prev2
        prev2, prev1 = prev1, curr
    return prev1

The concept at work is overlapping subproblems. Without memoization or the rolling-variable trick above, a naive recursion recomputes the same stairs exponentially. This "extend from the last two states" structure appears in house robber, minimum cost climbing stairs, and most 1D DP problems. If you find yourself staring at it thinking "wait, this is just Fibonacci," you're right. That's the point.

7. Merge Two Sorted Lists (LC 21): The Dummy Node Eliminates Edge Cases

Compare heads of both lists. Take the smaller one. Advance that pointer. Repeat. Attach whatever's left. The dummy node trick means you never need a special case for an empty result list. Your returned head is always dummy.next.

def mergeTwoLists(list1, list2):
    dummy = ListNode(0)
    curr = dummy
    while list1 and list2:
        if list1.val <= list2.val:
            curr.next = list1
            list1 = list1.next
        else:
            curr.next = list2
            list2 = list2.next
        curr = curr.next
    curr.next = list1 or list2
    return dummy.next

You'll reuse the dummy node pattern in merge K sorted lists, add two numbers, and any problem where you're constructing a linked list on the fly. It's not a convenience. It's the pattern.

8. Maximum Subarray (LC 53): Kadane's or Nothing

At each position, ask one question: is it better to extend the current subarray, or restart fresh from this element alone? If the running sum drops below zero, restarting beats extending. Otherwise, extend.

def maxSubArray(nums: list[int]) -> int:
    max_sum = current_sum = nums[0]
    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
    return max_sum

Critical edge case: initialize to nums[0], not 0. If every number is negative, the answer is the largest negative number. Starting at 0 gives you a wrong answer silently, and you will not catch it on a happy-path test case. This algorithm was designed in under a minute at a 1984 Carnegie Mellon seminar by Joseph Kadane, who was told the problem cold and just... solved it immediately. The O(n²) instinct is wrong; the O(n) intuition is learnable.

Interviewer watching a beginner still coding FizzBuzz 30 minutes in

The interviewer energy when someone skips Two Sum but can't explain what a hash map is.

9. Invert Binary Tree (LC 226): Recursion When the Structure Repeats

Swap the left and right children. Then recursively invert each child. The base case is null: a null node inverts to null. Recursion on trees almost always follows three beats: base case, process current node, recurse on children.

def invertTree(root):
    if not root:
        return None
    root.left, root.right = invertTree(root.right), invertTree(root.left)
    return root

This structure appears in tree height, path sum, diameter, lowest common ancestor, and nearly every other tree problem. Understand why the recursion terminates here and you've understood most tree recursion. (Yes, this is the problem that famously tripped up the creator of Homebrew in a Google interview. No, that didn't stop Google from using it.)

10. Number of Islands (LC 200): The Outer Loop Is Not Optional

Scan every cell. When you find a '1', increment your count and kick off a DFS to mark the whole island as visited. The outer loop is not optional because the grid can have multiple disconnected components. Without it, you'd count the first island and stop.

def numIslands(grid: list[list[str]]) -> int:
    if not grid:
        return 0
    rows, cols = len(grid), len(grid[0])
    count = 0

    def dfs(r, c):
        if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != '1':
            return
        grid[r][c] = '0'
        for dr, dc in [(1,0),(-1,0),(0,1),(0,-1)]:
            dfs(r + dr, c + dc)

    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == '1':
                dfs(r, c)
                count += 1
    return count

The same outer-loop-plus-DFS structure works for counting provinces, number of enclaves, and most 2D graph problems. This is connected component counting. Read BFS vs DFS if you're unsure which to reach for when.

LeetCode Beginner Patterns: Ten Problems, Seven Templates

Problem	Pattern
Two Sum	Hash map as lookup
Valid Parentheses	Stack for nesting
Best Time to Buy Stock	Single-pass greedy
Reverse Linked List	Three-pointer manipulation
Binary Search	Divide and conquer invariant
Climbing Stairs	1D dynamic programming
Merge Two Sorted Lists	Dummy node + merge
Maximum Subarray	Kadane's extend-or-restart
Invert Binary Tree	Tree recursion
Number of Islands	Grid DFS + connected components

None of these patterns appears only once. The variants are slightly disguised versions of the same underlying structure. Learn to see through the disguise.

Each of These Has a Harder Cousin

Solve Two Sum cold, then try Three Sum. Automate Reverse Linked List, then try Reverse in K-Groups. Once Climbing Stairs is trivial, try House Robber. Once Number of Islands bores you, try Pacific Atlantic Water Flow.

The pattern progression is the prep. You're not grinding 2,800 problems. You're building seven mental templates and stress-testing them until the disguise stops working.

Programmer levels: good at leetcode, reads the manual, decompiles binaries

Start with the top two. Do not attempt the rabbit until you're at least 200 mediums deep.

One thing solo practice doesn't train: performing under real interview conditions. Knowing the solution in your head is different from explaining it out loud to someone who's evaluating you. SpaceComplexity runs realistic voice-based DSA mock interviews that score your reasoning and communication, not just your final answer. That's how you find out which of these ten patterns you think you know but can't yet explain.

The Gotchas That Bite in Interviews

Initialize Kadane's to nums[0], not 0. All-negative input breaks the 0 initialization silently. You will not catch this on the test cases you write yourself.
Use lo + (hi - lo) // 2 in binary search. The (lo + hi) // 2 version has a real overflow bug that sat in Java's stdlib for nine years. You will not catch this on small arrays.
The dummy node in linked list construction is the pattern, not a shortcut. Use it by default.
The outer loop in Number of Islands is mandatory. Every disconnected component needs its own DFS kick. Forgetting this is the single most common way to get a wrong answer on a grid problem.
Valid Parentheses needs two guards: empty stack on close bracket, non-empty stack at the end. One guard is not enough.