Algorithmic Problem Solving in Java — The LeetCode Playbook

8. Pattern-Based Practice Problems

Each problem targets one pattern. Try first, then reveal.

1. Two Pointers: given a sorted int[], remove duplicates in-place and return the new length.
   Two Pointers: sorted array থেকে duplicates সরিয়ে নতুন length ফেরান।
   ✨ Show Answer

   Main.java

   class Main {
       static int dedupe(int[] a) {
           if (a.length == 0) return 0;
           int slow = 0;
           for (int fast = 1; fast < a.length; fast++) {
               if (a[fast] != a[slow]) a[++slow] = a[fast];
           }
           return slow + 1;
       }
       public static void main(String[] args) {
           int[] a = {1,1,2,2,3,4,4,5};
           int n = dedupe(a);
           for (int i = 0; i < n; i++) System.out.print(a[i] + " ");
       }
   }

   Run Copy

2. Sliding Window: longest substring with at most 2 distinct characters.
   Sliding Window: সবচেয়ে দীর্ঘ substring যাতে ২টির বেশি distinct character নেই।
   ✨ Show Answer

   Main.java

   import java.util.*;
   
   class Main {
       static int longest2Distinct(String s) {
           Map<Character, Integer> cnt = new HashMap<>();
           int l = 0, best = 0;
           for (int r = 0; r < s.length(); r++) {
               cnt.merge(s.charAt(r), 1, Integer::sum);
               while (cnt.size() > 2) {
                   char c = s.charAt(l++);
                   if (cnt.merge(c, -1, Integer::sum) == 0) cnt.remove(c);
               }
               best = Math.max(best, r - l + 1);
           }
           return best;
       }
       public static void main(String[] args) {
           System.out.println(longest2Distinct("eceba"));  // 3
           System.out.println(longest2Distinct("ccaabbb")); // 5
       }
   }

   Run Copy

3. BFS: count islands in a 2D grid of 0/1. An island is a connected region of 1s (4-directional).
   BFS: 0/1 grid-এ কতগুলো island আছে গণনা করুন (4-direction)।
   ✨ Show Answer

   Main.java

   import java.util.*;
   
   class Main {
       static int numIslands(int[][] g) {
           int rows = g.length, cols = g[0].length, count = 0;
           int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1}};
           for (int i = 0; i < rows; i++)
               for (int j = 0; j < cols; j++)
                   if (g[i][j] == 1) {
                       count++;
                       Deque<int[]> q = new ArrayDeque<>();
                       q.add(new int[]{i, j}); g[i][j] = 0;
                       while (!q.isEmpty()) {
                           int[] c = q.poll();
                           for (int[] d : dirs) {
                               int ni = c[0] + d[0], nj = c[1] + d[1];
                               if (ni >= 0 && nj >= 0 && ni < rows && nj < cols && g[ni][nj] == 1) {
                                   g[ni][nj] = 0;
                                   q.add(new int[]{ni, nj});
                               }
                           }
                       }
                   }
           return count;
       }
       public static void main(String[] args) {
           int[][] g = {{1,1,0,0},{0,1,0,1},{0,0,0,1}};
           System.out.println("islands = " + numIslands(g));
       }
   }

   Run Copy

4. Heap: k-th largest element in an unsorted array using a min-heap of size k.
   Heap: unsorted array-এ k-তম বৃহত্তম element — size-k min-heap দিয়ে।
   ✨ Show Answer

   Main.java

   import java.util.PriorityQueue;
   
   class Main {
       static int kthLargest(int[] nums, int k) {
           PriorityQueue<Integer> min = new PriorityQueue<>();
           for (int n : nums) {
               min.offer(n);
               if (min.size() > k) min.poll();
           }
           return min.peek();
       }
       public static void main(String[] args) {
           System.out.println(kthLargest(new int[]{3,2,1,5,6,4}, 2)); // 5
       }
   }

   Run Copy

5. DP: climb N stairs, one or two steps at a time — number of ways.
   DP: N সিঁড়ি, এক বা দুই ধাপে — মোট কতভাবে উঠা যায়?
   ✨ Show Answer

   Main.java

   class Main {
       static int climbStairs(int n) {
           if (n <= 2) return n;
           int a = 1, b = 2;
           for (int i = 3; i <= n; i++) { int c = a + b; a = b; b = c; }
           return b;
       }
       public static void main(String[] args) {
           for (int n = 1; n <= 10; n++) System.out.println(n + " -> " + climbStairs(n));
       }
   }

   Run Copy