Microsoft Coding Questions

Practice with 25+ Microsoft placement paper coding questions covering DSA problems, algorithms, and system design concepts. These questions are representative of what you’ll encounter in Microsoft’s online assessment and technical interviews.

What’s Included:

25+ Coding Problems: Easy, Medium, and Hard level questions with solutions
Multiple Language Solutions: C++, Java, Python, and C# solutions
Time Complexity Analysis: Every solution includes complexity analysis
Microsoft-Specific Patterns: Focus on arrays, trees, graphs, dynamic programming

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Microsoft OA coding section overview

Microsoft Online Assessment Breakdown:

Section	Questions	Time	Difficulty	Focus Areas
Coding Test	3-4	90 min	Medium-Hard	Arrays, trees, graphs, DP, system design

Languages Allowed: C, C++, Java, Python, C#

Coding problems by difficulty

Two Sum

Problem: Find indices of two numbers that add up to target.

Example:

Input: nums = [2, 7, 11, 15], target = 9
Output: [0, 1]

Solution (C++):

vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> map;
    for (int i = 0; i < nums.size(); i++) {
        int complement = target - nums[i];
        if (map.find(complement) != map.end()) {
            return {map[complement], i};
        }
        map[nums[i]] = i;
    }
    return {};
}

Solution (Python):

def two_sum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        if target - num in seen:
            return [seen[target - num], i]
        seen[num] = i
    return []

Time Complexity: O(n) | Space Complexity: O(n)

Valid Parentheses

Problem: Check if string with brackets is valid.

Example:

Input: "()[]{}"
Output: true

Solution (C++):

bool isValid(string s) {
    stack<char> st;
    unordered_map<char, char> pairs = {{')', '('}, {'}', '{'}, {']', '['}};

    for (char c : s) {
        if (pairs.count(c)) {
            if (st.empty() || st.top() != pairs[c]) return false;
            st.pop();
        } else {
            st.push(c);
        }
    }
    return st.empty();
}

Solution (Python):

def is_valid(s):
    stack = []
    mapping = {')': '(', '}': '{', ']': '['}
    for char in s:
        if char in mapping:
            if not stack or stack.pop() != mapping[char]:
                return False
        else:
            stack.append(char)
    return len(stack) == 0

Time Complexity: O(n) | Space Complexity: O(n)

Reverse Linked List

Problem: Reverse a singly linked list.

Solution (C++):

ListNode* reverseList(ListNode* head) {
    ListNode* prev = nullptr;
    ListNode* curr = head;

    while (curr) {
        ListNode* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

Solution (Python):

def reverse_list(head):
    prev = None
    curr = head
    while curr:
        next_temp = curr.next
        curr.next = prev
        prev = curr
        curr = next_temp
    return prev

Time Complexity: O(n) | Space Complexity: O(1)

Merge Two Sorted Lists

Problem: Merge two sorted linked lists.

Solution (C++):

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    ListNode dummy(0);
    ListNode* curr = &dummy;

    while (l1 && l2) {
        if (l1->val < l2->val) {
            curr->next = l1;
            l1 = l1->next;
        } else {
            curr->next = l2;
            l2 = l2->next;
        }
        curr = curr->next;
    }
    curr->next = l1 ? l1 : l2;
    return dummy.next;
}

Time Complexity: O(n+m) | Space Complexity: O(1)

Maximum Depth of Binary Tree

Problem: Find maximum depth of binary tree.

Solution (C++):

int maxDepth(TreeNode* root) {
    if (!root) return 0;
    return 1 + max(maxDepth(root->left), maxDepth(root->right));
}

Solution (Python):

def max_depth(root):
    if not root:
        return 0
    return 1 + max(max_depth(root.left), max_depth(root.right))

Time Complexity: O(n) | Space Complexity: O(h)

Longest Substring Without Repeating

Problem: Find length of longest substring without repeating characters.

Example:

Input: "abcabcbb"
Output: 3 (substring "abc")

Solution (C++):

int lengthOfLongestSubstring(string s) {
    unordered_map<char, int> seen;
    int maxLen = 0, left = 0;

    for (int right = 0; right < s.length(); right++) {
        if (seen.count(s[right]) && seen[s[right]] >= left) {
            left = seen[s[right]] + 1;
        }
        seen[s[right]] = right;
        maxLen = max(maxLen, right - left + 1);
    }
    return maxLen;
}

Solution (Python):

def length_of_longest_substring(s):
    seen = {}
    max_len = left = 0
    for right, char in enumerate(s):
        if char in seen and seen[char] >= left:
            left = seen[char] + 1
        seen[char] = right
        max_len = max(max_len, right - left + 1)
    return max_len

Time Complexity: O(n) | Space Complexity: O(min(m,n))

Binary Tree Maximum Path Sum

Problem: Find the maximum path sum in a binary tree.

Solution (C++):

class Solution {
    int maxSum = INT_MIN;
public:
    int maxPathSum(TreeNode* root) {
        maxGain(root);
        return maxSum;
    }

    int maxGain(TreeNode* node) {
        if (!node) return 0;
        int left = max(0, maxGain(node->left));
        int right = max(0, maxGain(node->right));
        maxSum = max(maxSum, node->val + left + right);
        return node->val + max(left, right);
    }
};

Solution (Python):

def max_path_sum(root):
    max_sum = float('-inf')

    def max_gain(node):
        nonlocal max_sum
        if not node:
            return 0
        left = max(0, max_gain(node.left))
        right = max(0, max_gain(node.right))
        max_sum = max(max_sum, node.val + left + right)
        return node.val + max(left, right)

    max_gain(root)
    return max_sum

Time Complexity: O(n) | Space Complexity: O(h)

3Sum

Problem: Find all unique triplets that sum to zero.

Example:

Input: [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Solution (C++):

vector<vector<int>> threeSum(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    vector<vector<int>> result;

    for (int i = 0; i < nums.size() - 2; i++) {
        if (i > 0 && nums[i] == nums[i-1]) continue;

        int left = i + 1, right = nums.size() - 1;
        while (left < right) {
            int sum = nums[i] + nums[left] + nums[right];
            if (sum == 0) {
                result.push_back({nums[i], nums[left], nums[right]});
                while (left < right && nums[left] == nums[left+1]) left++;
                while (left < right && nums[right] == nums[right-1]) right--;
                left++;
                right--;
            } else if (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

Time Complexity: O(n²) | Space Complexity: O(1)

LRU Cache

Problem: Design LRU cache with O(1) get and put operations.

Solution (C++):

class LRUCache {
    int capacity;
    list<pair<int, int>> lru;
    unordered_map<int, list<pair<int, int>>::iterator> cache;
public:
    LRUCache(int cap) : capacity(cap) {}

    int get(int key) {
        if (!cache.count(key)) return -1;
        lru.splice(lru.begin(), lru, cache[key]);
        return cache[key]->second;
    }

    void put(int key, int value) {
        if (cache.count(key)) {
            cache[key]->second = value;
            lru.splice(lru.begin(), lru, cache[key]);
        } else {
            if (cache.size() == capacity) {
                cache.erase(lru.back().first);
                lru.pop_back();
            }
            lru.emplace_front(key, value);
            cache[key] = lru.begin();
        }
    }
};

Solution (Python):

from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity):
        self.cache = OrderedDict()
        self.capacity = capacity

    def get(self, key):
        if key not in self.cache:
            return -1
        self.cache.move_to_end(key)
        return self.cache[key]

    def put(self, key, value):
        if key in self.cache:
            self.cache.move_to_end(key)
        self.cache[key] = value
        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)

Time Complexity: O(1) for both | Space Complexity: O(capacity)

Validate Binary Search Tree

Problem: Determine if binary tree is a valid BST.

Solution (C++):

bool isValidBST(TreeNode* root) {
    return validate(root, LLONG_MIN, LLONG_MAX);
}

bool validate(TreeNode* node, long long minVal, long long maxVal) {
    if (!node) return true;
    if (node->val <= minVal || node->val >= maxVal) return false;
    return validate(node->left, minVal, node->val) &&
           validate(node->right, node->val, maxVal);
}

Solution (Python):

def is_valid_bst(root):
    def validate(node, min_val, max_val):
        if not node:
            return True
        if node.val <= min_val or node.val >= max_val:
            return False
        return validate(node.left, min_val, node.val) and \
               validate(node.right, node.val, max_val)
    return validate(root, float('-inf'), float('inf'))

Time Complexity: O(n) | Space Complexity: O(h)

Number of Islands

Problem: Count number of islands in a 2D grid.

Solution (C++):

int numIslands(vector<vector<char>>& grid) {
    int count = 0;
    for (int i = 0; i < grid.size(); i++) {
        for (int j = 0; j < grid[0].size(); j++) {
            if (grid[i][j] == '1') {
                dfs(grid, i, j);
                count++;
            }
        }
    }
    return count;
}

void dfs(vector<vector<char>>& grid, int i, int j) {
    if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == '0')
        return;
    grid[i][j] = '0';
    dfs(grid, i+1, j);
    dfs(grid, i-1, j);
    dfs(grid, i, j+1);
    dfs(grid, i, j-1);
}

Time Complexity: O(m×n) | Space Complexity: O(m×n)

Course Schedule

Problem: Determine if you can finish all courses given prerequisites.

Solution (C++):

bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
    vector<vector<int>> graph(numCourses);
    vector<int> inDegree(numCourses, 0);

    for (auto& p : prerequisites) {
        graph[p[1]].push_back(p[0]);
        inDegree[p[0]]++;
    }

    queue<int> q;
    for (int i = 0; i < numCourses; i++) {
        if (inDegree[i] == 0) q.push(i);
    }

    int count = 0;
    while (!q.empty()) {
        int course = q.front();
        q.pop();
        count++;
        for (int next : graph[course]) {
            if (--inDegree[next] == 0) q.push(next);
        }
    }

    return count == numCourses;
}

Time Complexity: O(V+E) | Space Complexity: O(V+E)

Median of Two Sorted Arrays

Problem: Find median with O(log(min(m,n))) complexity.

Solution (C++):

double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    if (nums1.size() > nums2.size()) swap(nums1, nums2);

    int m = nums1.size(), n = nums2.size();
    int low = 0, high = m;

    while (low <= high) {
        int i = (low + high) / 2;
        int j = (m + n + 1) / 2 - i;

        int maxLeft1 = (i == 0) ? INT_MIN : nums1[i-1];
        int minRight1 = (i == m) ? INT_MAX : nums1[i];
        int maxLeft2 = (j == 0) ? INT_MIN : nums2[j-1];
        int minRight2 = (j == n) ? INT_MAX : nums2[j];

        if (maxLeft1 <= minRight2 && maxLeft2 <= minRight1) {
            if ((m + n) % 2 == 0) {
                return (max(maxLeft1, maxLeft2) + min(minRight1, minRight2)) / 2.0;
            }
            return max(maxLeft1, maxLeft2);
        }
        else if (maxLeft1 > minRight2) high = i - 1;
        else low = i + 1;
    }
    return 0.0;
}

Time Complexity: O(log(min(m,n))) | Space Complexity: O(1)

Merge k Sorted Lists

Problem: Merge k sorted linked lists.

Solution (C++):

ListNode* mergeKLists(vector<ListNode*>& lists) {
    auto cmp = [](ListNode* a, ListNode* b) { return a->val > b->val; };
    priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> pq(cmp);

    for (auto node : lists) {
        if (node) pq.push(node);
    }

    ListNode dummy(0);
    ListNode* curr = &dummy;

    while (!pq.empty()) {
        curr->next = pq.top();
        pq.pop();
        curr = curr->next;
        if (curr->next) pq.push(curr->next);
    }

    return dummy.next;
}

Time Complexity: O(N log k) | Space Complexity: O(k)

Trapping Rain Water

Problem: Calculate water trapped after raining.

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

Solution (C++):

int trap(vector<int>& height) {
    int left = 0, right = height.size() - 1;
    int leftMax = 0, rightMax = 0;
    int water = 0;

    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= leftMax) {
                leftMax = height[left];
            } else {
                water += leftMax - height[left];
            }
            left++;
        } else {
            if (height[right] >= rightMax) {
                rightMax = height[right];
            } else {
                water += rightMax - height[right];
            }
            right--;
        }
    }
    return water;
}

Time Complexity: O(n) | Space Complexity: O(1)

Word Break II

Problem: Find all possible sentences from word dictionary.

Solution (C++):

vector<string> wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> dict(wordDict.begin(), wordDict.end());
    unordered_map<int, vector<string>> memo;
    return helper(s, 0, dict, memo);
}

vector<string> helper(string& s, int start, unordered_set<string>& dict,
                     unordered_map<int, vector<string>>& memo) {
    if (memo.count(start)) return memo[start];
    if (start == s.length()) return {""};

    vector<string> result;
    for (int end = start + 1; end <= s.length(); end++) {
        string word = s.substr(start, end - start);
        if (dict.count(word)) {
            auto rest = helper(s, end, dict, memo);
            for (auto& r : rest) {
                result.push_back(word + (r.empty() ? "" : " " + r));
            }
        }
    }
    return memo[start] = result;
}

Time Complexity: O(n × 2^n) | Space Complexity: O(n × 2^n)

Serialize and Deserialize Binary Tree

Problem: Design algorithm to serialize/deserialize binary tree.

Solution (C++):

class Codec {
public:
    string serialize(TreeNode* root) {
        if (!root) return "null,";
        return to_string(root->val) + "," +
               serialize(root->left) + serialize(root->right);
    }

    TreeNode* deserialize(string data) {
        queue<string> nodes;
        string token;
        for (char c : data) {
            if (c == ',') {
                nodes.push(token);
                token.clear();
            } else {
                token += c;
            }
        }
        return build(nodes);
    }

private:
    TreeNode* build(queue<string>& nodes) {
        string val = nodes.front();
        nodes.pop();
        if (val == "null") return nullptr;
        TreeNode* node = new TreeNode(stoi(val));
        node->left = build(nodes);
        node->right = build(nodes);
        return node;
    }
};

Time Complexity: O(n) | Space Complexity: O(n)

Practice tips for Microsoft OA

Master DSA fundamentals

Arrays, strings, trees, graphs
Dynamic programming patterns
Linked lists, stacks, queues
Hash maps and sets

Focus on optimization

Microsoft values optimal solutions
Know time/space tradeoffs
Practice Big O analysis
Multiple approaches per problem

Time management

20-25 minutes per problem
Read problem carefully
Test with edge cases
Optimize if time permits

Common patterns

Two pointers
Sliding window
Binary search
BFS/DFS
Dynamic programming
Graph algorithms

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