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Insights

In-order, Pre-order, and Post-order traversals are Depth-First traversals. For a Graph, the complexity of a Depth First Traversal is O(n + m), where n is the number of nodes, and m is the number of edges. Since a Binary Tree is also a Graph, the same applies here. The complexity of each of these Depth-first traversals is O(n+m). Since the number of edges that can originate from a node is limited to 2 in the case of a Binary Tree, the maximum number of total edges in a Binary Tree is n-1, where n is the total number of nodes. The complexity then becomes O(n + n-1), which is O(n).

Top-Down DP 🆚 Bottom-Up D

For anyone who is getting confused with Top-down and bottom-up, Recursion memoization is always TOP-DOWN(and not bottom-up), as we take a bigger problem and recursively solve for the smaller subproblems. Whereas in tabular DP where we start filling the table from top left to bottom right is actually BOTTOM-UP because we compute dp values of smaller subproblems first and then using these values compute dp values of bigger problems.

Recursion + memoization is called Top-Down DP and Tabulation method is called Bottom-Up DP.

PS: Top-down and Bottom up is decided by the essence of methodology and not by whether we are filling table from top to bottom or vice versa!

Longest Common Subsequence Recursive Visualization

Print any 2D matrix

for (int[] row : matrix)
    System.out.println(Arrays.toString(row));

Reverse a string in efficient way

StringBuilder reverseStr = new StringBuilder();
reverseStr.append(str).reverse();

Initialize any matrix with any specific value ( Here it is -1)

int[][] matrix = new int[len][len];
for (int[] row : matrix)
    Arrays.fill(row, -1);

String to Character Array

char[] ch = str.toCharArray();

== 🆚 equals()

If you are working with the Character class and want to compare two char values, then use equals() method

with primitive char values, you can simply use the == equal operator

getOrDefault method in HashMap

If given key is exist in the map then it will return corresponding value otherwise it return the default given value by us here it's 500

              //(key, defaultValue)
map.getOrDefault(200, 500) + 1

map.put(str.charAt(j), map.getOrDefault(str.charAt(j), 1) + 1)

HashSet 🆚 TreeSet

In many permutations/subset generation problems question demand for unique and sorted elements in it, so for that we can use TreeSet which follow both of these properties. On the other hand, Hashset only handle unique but can't handle ordering whereas TreeSet can handle ordering of elements.

 TreeSet<String> set = new TreeSet<String>();

Convert TreeSet to List

 new ArrayList<String>(set)

❌ Remove last character of a StringBuilder

 sb.setLength(sb.length() - 1);

Convert array to list ( Array of int to List of Integer)

 List<Integer> subList =  Arrays.stream(edges[i]).boxed().collect(Collectors.toList());

Clone & Sort Array

int[][] sorted = intervals.clone();
Arrays.sort(sorted, (a,b) -> Integer.compare(a[0], b[0]));

Convert ArrayList to Array

int[][] arr = new int[list.size()][];
		return list.toArray(arr);

Short-Circuit

In Java || is a short-circuit operator, the right hand side will not be evaluated if the left hand side is true

Print HashMap

map.entrySet().forEach(entry -> {
      System.out.println(entry.getKey() + " " + entry.getValue());
});

Create Array of lists -> adjecency matrix of graph

ArrayList<Integer>[] adj = new ArrayList[numCourses];

JaydipBarvaliya / Data-Structure-Algorithms