14 Patterns to Ace Any Coding Interview Question by Fahim ul Haq
Recursive approach: solve f(c, i) with c is the remaining capacity and i is th current item index At each level, we branch with the item at index i (if enough capacity) and without it
public int knapsack(int[] profits, int[] weights, int c) {
return knapsack(profits, weights, c, 0, 0);
}
public int knapsack(int[] profits, int[] weights, int c, int i, int sum) {
if (i == profits.length || c <= 0) {
return sum;
}
// Not
int sum1 = knapsack(profits, weights, c, i + 1, sum);
// With
int sum2 = 0;
if (weights[i] <= c) {
sum2 = knapsack(profits, weights, c - weights[i], i + 1, sum + profits[i]);
}
return Math.max(sum1, sum2);
}Memoization: store a[c][i] (c is the remaining capacity, i is the current item index)
As we need to store the 0 capacity, we have to init the array this way:
int[][] a = new int[c + 1][n] // n is the number of items
Time and space complexity: O(n * c)
public int knapsack(int[] profits, int[] weights, int capacity) {
// Capacity from 1 to n
Integer[][] a = new Integer[capacity][profits.length];
return knapsack(profits, weights, capacity, 0, 0, a);
}
public int knapsack(int[] profits, int[] weights, int capacity, int i, int sum, Integer[][] a) {
if (i == profits.length || capacity == 0) {
return sum;
}
// If value already exists, return
if (a[capacity - 1][i] != null) {
return a[capacity][i];
}
// With
int sum1 = knapsack(profits, weights, capacity, i + 1, sum, a);
// Without
int sum2 = 0;
if (weights[i] <= capacity) {
sum2 = knapsack(profits, weights, capacity - weights[i], i + 1, sum + profits[i], a);
}
a[capacity - 1][i] = Math.max(sum1, sum2);
return a[capacity - 1][i];
}Two dimensional array: a[n + 1][c + 1] // n the number of items and c the max capacity
First row and first column are set to 0
a[row][col] represent the max profit with items 1..row at capacity col
remainingWeight = col - itemWeight // col: current max capacity
a[row][col] = max(a[row - 1][col], itemValue + a[row - 1][remainingWeight]) // max between item not selected and item selected + max remaining weight
If remainingWeight < 0, we can't chose the item so a[row][col] = a[row - 1][col]
Return last element of the array
public int solveKnapsack(int[] profits, int[] weights, int capacity) {
int[][] a = new int[profits.length + 1][capacity + 1];
for (int row = 1; row < profits.length + 1; row++) {
int value = profits[row - 1];
int weight = weights[row - 1];
for (int col = 1; col < capacity + 1; col++) {
int remainingWeight = col - weight;
if (remainingWeight < 0) {
a[row][col] = a[row - 1][col];
} else {
a[row][col] = Math.max(
a[row - 1][col],
value + a[row - 1][remainingWeight]
);
}
}
}
return a[profits.length][capacity];
}If we need to compute a result like "determine if a subset exists" that return a boolean, the array type is boolean[][]
As we are only interested in the previous row, we can also use an int[2][n] array
Solution for solving a problem recursively
Loop:
- apply() // Apply a change
- try() // Try a solution
- reverse() // Reverse apply
Iterate over each number of an array and swap it to its correct position
At the end, we may iterate on the array to check which number is not at its correct position
If numbers are not within the 1 to n range, we can simply drop them
Alternative: marker technique (mark a result by setting a[i] to negative for example)
Identify an optimal subproblem or substructure in the problem and determine how to reach it
Focus on what you have now (don't think about what comes next)
We may want to apply the traversal technique to have a global context for the identification part (a map of letters/positions etc.)
Given K sorted array, technique to perform a sorted traversal of all the elements of all arrays
- First, push the first element of each array in a min heap
- While min heap not empty, take min element and push the next element of the same array
We need to keep track of which structure the min element come from (tracking the array index or taking the next node if it's a linked list)
Iterate over the linked list with two pointers simultaneously either with:
- One ahead by a fixed amount
- One faster
This technique can also be applied on other problems where we need to find a cycle (f(slow) and f(f(fast)) may converge)
Simplify the problem. If solvable, generalize to the initial problem.
Example: sort the array first
Range of elements in a specific window size
Two pointers left and right:
- Move right while condition is valid
- Move left if condition is not valid
Technique to find all the possible permutations or combinations
Start with an empty set, for each element of the input, add them to all the existing subsets to create new subsets
Example:
- Given [1, 5, 3]
- => [] // Start
- => [], [1]
- => [], [1], [5], [1,5]
- => [], [1], [5], [1,5], [3], [1,3], [1,5,3]
For each level, we iterate from 0 to size // size is the fixed size of the list
List<List<Integer>> findSubsets(int[] a) {
List<List<Integer>> subsets = new ArrayList<>();
// Add subset []
subsets.add(new ArrayList<>());
for (int n : a) {
// Fix the current size
int size = subsets.size();
for (int i = 0; i < size; i++) {
// Copy subset
ArrayList<Integer> newSubset = new ArrayList<>(subsets.get(i));
// Add element
newSubset.add(n);
subsets.add(newSubset);
}
}
return subsets;
}Runner technique
Subsets technique or recursion + backtracking
Binary search
Sliding window technique
Example:
- Given an array, find the average of all subarrays of size ‘K’ in it
Sliding window technique
Example:
- Longest substring with K distinct characters
- Longest substring without repeating characters
Two heaps technique
Runner technique
Two pointers technique
Example:
- Given a sorted array and a target sum, find a pair in the array whose sum is equal to the given target
- Given an array of unsorted numbers, find all unique triplets in it that add up to zero
- Comparing strings containing backspaces
Cyclic sort technique
Traversal technique
Iterate with two pointers, one over the starts, another one over the ends
Handle the element with the lowest value first and generate an event
Example: how many rooms for n meetings => meeting started, meeting started, meeting ended etc.
Top K elements technique
Greedy technique
K-way merge technique
Technique - Scheduling problem with n tasks where each task can have constraints to be completed before others
Topological sort technique
Heap data structure
Possibly two heaps technique
Finding the K biggest elements:
- Min heap
- Add k elements
- Then iterate over the remaining elements, if current > min => remove min, add current
Finding the k smallest elements:
- Max heap
- Add k elements
- Then iterate over the remaining elements, if current < max => remove max, add current
If there is an edge from U to V, then U <= V
Possible only if the graph is a DAG
Algo:
- Create a graph representation (adjacency list) and an in degree counter (Map<Integer, Integer>)
- Zero them for each vertex
- Fill the adjacency list and the in degree counter for each edge
- Add in a queue each vertex whose in degree count is 0 (source vertex with no parent)
- While the queue is not empty, poll a vertex from it then decrement the in degree of its children (no removal)
To check if there is a cycle, we must compare the size of the produced array to the number of vertices
List<Integer> sort(int vertices, int[][] edges) {
if (vertices == 0) {
return Collections.EMPTY_LIST;
}
List<Integer> sorted = new ArrayList<>(vertices);
// Adjacency list graph
Map<Integer, List<Integer>> graph = new HashMap<>();
// Count of incoming edges for each vertex
Map<Integer, Integer> inDegree = new HashMap<>();
for (int i = 0; i < vertices; i++) {
inDegree.put(i, 0);
graph.put(i, new LinkedList<>());
}
// Init graph and inDegree
for (int[] edge : edges) {
int parent = edge[0];
int child = edge[1];
graph.get(parent).add(child);
inDegree.put(child, inDegree.get(child) + 1);
}
// Create a source queue and add each source (a vertex whose inDegree count is 0)
Queue<Integer> sources = new LinkedList<>();
for (Map.Entry<Integer, Integer> entry : inDegree.entrySet()) {
if (entry.getValue() == 0) {
sources.add(entry.getKey());
}
}
while (!sources.isEmpty()) {
int vertex = sources.poll();
sorted.add(vertex);
// For each vertex, we will decrease the inDegree count of its children
List<Integer> children = graph.get(vertex);
for (int child : children) {
inDegree.put(child, inDegree.get(child) - 1);
if (inDegree.get(child) == 0) {
sources.add(child);
}
}
}
// Topological sort is not possible as the graph has a cycle
if (sorted.size() != vertices) {
return new ArrayList<>();
}
return sorted;
}Traverse the input and generate another data structure or optional events
Start the problem from this new state
Keep two heaps:
- A max heap for the first half
- Then a min heap for the second half
May be required to balance them to have at most a difference in terms of size of 1
Two pointers iterating through the data structure in tandem until one or both pointers hit a certain condition
Often useful when structure is sorted. If not sorted, we may want to sort it first.
Most of the times (not always): first pointer is at the start, the second pointer is at the end
The two pointers can also be on two different ds, still iterating in tandem (e.g. comparing strings containing backspaces)
Time complexity is linear
What if we need to iterate backwards on a singly linked list in constant space without mutating the input?
Reverse the liked list (or a subpart only), implement the algo then reverse it again to the initial state