Problem Solving Strategies and Tips for Competitive Programming in Java

Strategies for approaching competitive programming problems, debugging techniques, and time management tips to improve your performance in contests.

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Competitive Programming in Java with Solutions

Practice Problems and Solutions

This section provides a collection of practice problems designed to improve your problem-solving skills in competitive programming. Each problem is accompanied by a detailed solution, often with multiple approaches to enhance understanding.

Example Problem 1: Sum of Two Numbers

Given two integers, calculate their sum.

 public class SumOfTwoNumbers {
                            public static void main(String[] args) {
                                int a = 5;
                                int b = 10;
                                int sum = a + b;
                                System.out.println("The sum is: " + sum);
                            }
                        }

Example Problem 2: Find the Maximum

Given an array of integers, find the maximum element.

 public class FindMaximum {
                            public static void main(String[] args) {
                                int[] arr = {1, 5, 2, 8, 3};
                                int max = arr[0];
                                for (int i = 1; i < arr.length; i++) {
                                    if (arr[i] > max) {
                                        max = arr[i];
                                    }
                                }
                                System.out.println("The maximum element is: " + max);
                            }
                        }

Competitive Programming Problems with Solutions

This section features a collection of solved competitive programming problems sourced from various online platforms (e.g., Codeforces, LeetCode, HackerRank). Each problem includes a detailed explanation of the problem statement, the solution approach, and the corresponding Java code. We emphasize code clarity and efficiency to promote best practices in competitive programming.

LeetCode Problem: Two Sum

Given an array of integers `nums` and an integer `target`, return indices of the two numbers such that they add up to `target`.
You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

 import java.util.HashMap;
                        import java.util.Map;

                        class TwoSum {
                            public int[] twoSum(int[] nums, int target) {
                                Map<Integer, Integer> numMap = new HashMap<>();
                                for (int i = 0; i < nums.length; i++) {
                                    int complement = target - nums[i];
                                    if (numMap.containsKey(complement)) {
                                        return new int[] { numMap.get(complement), i };
                                    } else {
                                        numMap.put(nums[i], i);
                                    }
                                }
                                throw new IllegalArgumentException("No two sum solution");
                            }
                        }

Explanation: This solution uses a HashMap to store each number and its index. For each number, we check if its complement (target - number) exists in the HashMap. If it does, we return the indices of the number and its complement. This approach has a time complexity of O(n) due to the single iteration through the array and O(n) space complexity because of potential need to store all array elements in the hashmap.

Codeforces Problem: Watermelon (A)

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest one they could find, and after they weighed it, the scales showed w kilos. They hurried home, dying of thirst, and decided to divide the watermelon, however they faced a hard problem. Pete and Billy are big fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

 import java.util.Scanner;

                        public class Watermelon {
                            public static void main(String[] args) {
                                Scanner scanner = new Scanner(System.in);
                                int w = scanner.nextInt();

                                if (w > 2 && w % 2 == 0) {
                                    System.out.println("YES");
                                } else {
                                    System.out.println("NO");
                                }
                                scanner.close();
                            }
                        }

Explanation: This solution checks if the weight `w` is greater than 2 and is an even number. If both conditions are met, it prints "YES" because the watermelon can be divided into two even parts (e.g., w/2 and w/2). Otherwise, it prints "NO" because either the watermelon weighs too little or it can't be divided into two even parts.

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