Interview Algorithms

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Insights from Leetcode problems.

Linear Searching

Finding a cycle in a linked-list

Use two runners.
Runner 1 goes two steps per iteration.
Runner 2 goes one step per iteration.
If there is a cycle, runner 2 will lap runner 1 within 2 cycles.



Print all permutations of an array.
The idea here is that you need a for-loop nested for each element of an array.
So n elements means n nested for-loops. You use recursion to nest the loops.
You can also think of this as a graph problem when you need to find every possible path with DFS.

Permute Brute-force solution

A more advanced solution would use backtracking, however it is time-consuming to implement for an interview.
Here is a quick brute-force solution.

  • Enumerate all possible n^n numbers from 000 to 999 and then filter out number with duplicate digits
class Solution {
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> indices(nums.size());
        permuteHelper(nums, indices, 0, ans);
        return ans;
    bool isValid(vector<int>& indices) {
        vector<char> check(indices.size(), false);
        for (int i = 0; i < indices.size(); i++) {
            if (check[indices[i]]) {
                return false;
            check[indices[i]] = true;
        return true;
    void permuteHelper(vector<int>& nums, vector<int>& indices,
                       int col, vector<vector<int>>& ans) {
        if (col == nums.size()) {
            if (isValid(indices))
                print(nums, indices, ans);
        for (int i = 0; i < nums.size(); i++) {
            indices[col] = i;
            permuteHelper(nums, indices, col+1, ans);
    void print(vector<int>& nums, vector<int>& indices, vector<vector<int>>& ans) {
        vector<int> new_nums(nums.size());
        for (int i = 0; i < nums.size(); i++) {
            new_nums[i] = nums[indices[i]];

Misc Tricks

Finding duplicates in an array

If you have an array of ints where each number appears times and one number appears times where , then you count the number of times each bit appears and take it mod .
The remaining bits will remain times.