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Home AI News Mastering LeetCode Weekly Contest 272: Problem Solutions

Mastering LeetCode Weekly Contest 272: Problem Solutions

Updated on Nov 02,2025

Table of Contents

This article dives deep into the solutions for the initial problems from LeetCode Weekly Contest 272, sponsored by Amazon. We'll dissect each problem, offering clear strategies and code examples to help you improve your problem-solving skills and enhance your coding prowess. Aimed at intermediate coders and beginners alike, this guide breaks down complex problems into manageable steps, making it easier to grasp the underlying concepts and code implementation.

Key Points

Problem-solving strategies for beginner to intermediate level LeetCode problems.

Efficient code implementation in C++ for various coding challenges.

Understanding palindrome detection and manipulation.

Using dynamic programming to optimize solutions.

Approaching array manipulation problems with optimized algorithms.

Unpacking LeetCode Weekly Contest 272

2108. Find First Palindromic String in the Array

The first problem of the contest, Find First Palindromic String in the Array, presents a straightforward challenge: given an array of strings, identify and return the first palindromic string encountered. A palindromic string is one that reads the same forwards and backward. If no such string exists, the function should return an empty string.

To tackle this, you can iterate through the array and apply a palindrome check on each string. This check can be performed by comparing the string to its reversed version. This method ensures that you can efficiently determine whether a string is a palindrome and retrieve the first one found.

Strategies for Palindrome Detection:

  • Two-Pointer Approach: This method involves using two pointers, one starting from the beginning and the other from the end of the string. The characters at these pointers are compared, and if they differ, the string is not a palindrome. This continues until the pointers meet in the middle.
  • Reverse and Compare: A more direct approach is to reverse the string and then compare it with the original. If both are identical, the string is a palindrome. C++ provides functionalities like std::reverse to facilitate this.

Example C++ Code:

class Solution {
public:
    string firstPalindrome(vector<string>& words) {
        for (string word : words) {
            string reversedWord = word;
            reverse(reversedWord.begin(), reversedWord.end());
            if (word == reversedWord) {
                return word;
            }
        }
        return "";
    }
};

Keywords: Palindrome, String manipulation, Array iteration, Two-pointer approach, Reverse and compare

2109. Adding Spaces to a String

The second problem, Adding Spaces to a String, requires you to insert spaces into a given string at specified indices.

You are provided with a string s and an array spaces that contains the indices where spaces should be inserted. Your task is to modify the original string by adding spaces at these locations.

Strategies for Space Insertion:

  • Iterative Approach: Iterate through the original string, and before each character, check if the current index is present in the spaces array. If it is, insert a space before adding the character to the modified string. This approach ensures that all spaces are inserted correctly.
  • Two-Pointer Technique: Use two pointers, one to iterate through the spaces array and another to iterate through the original string. When the pointer in the spaces array matches the current index in the string, insert a space.

Example C++ Code:

class Solution {
public:
    string addSpaces(string s, vector<int>& spaces) {
        string result = "";
        int spaceIndex = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (spaceIndex < spaces.size() && i == spaces[spaceIndex]) {
                result += ' ';
                spaceIndex++;
            }
            result += s[i];
        }
        return result;
    }
};

Keywords: String insertion, Index manipulation, Array traversal, String building

2110. Number of Smooth Descent Periods of a Stock

The third problem, Number of Smooth Descent Periods of a Stock, challenges you to calculate the number of smooth descent periods in a stock's price history. A smooth descent period is defined as a contiguous sequence of days where each day’s price is exactly one less than the previous day’s price.

This requires identifying consecutive days where the stock price decreases by exactly one. The first day of any such period is always included in the count, as it initiates the descent.

Strategies for Counting Descent Periods:

  • Dynamic Programming: Maintain a counter that increments as long as the price difference between consecutive days is one. When the difference changes, apply the formula n * (n + 1) / 2 to the counter, which calculates the number of descent periods ending at that point.
  • Iterative Counting: Iterate through the prices array, tracking the length of current descent periods. Update the total count of smooth descent periods based on the length of these periods.

Example C++ Code:

class Solution {
public:
    long long getDescentPeriods(vector<int>& prices) {
        long long total = 0, count = 0;
        for (int i = 1; i < prices.size(); ++i) {
            if (prices[i] == prices[i - 1] - 1) {
                count++;
            } else {
                total += (count * (count + 1)) / 2;
                count = 0;
            }
        }
        total += (count * (count + 1)) / 2;
        return total + prices.size();
    }
};

Keywords: Stock prices, Contiguous sequence, Dynamic programming, Price difference calculation, Algorithmic trading

2111. Minimum Operations to Make the Array K-Increasing

The fourth problem, Minimum Operations to Make the Array K-Increasing, involves determining the fewest operations to make an array K-increasing. An array is K-increasing if, for every index i, the element at i - k is less than or equal to the element at i.

To solve this problem efficiently, you will need to identify the longest increasing subsequence within specific sub-arrays defined by intervals of k. The operation count is the difference between the total number of elements and the length of the longest increasing subsequence.

Strategies for K-Increasing Conversion:

  • Dynamic Subsequence Counting: Iterate over each of the k sub-arrays and compute the Longest Increasing Subsequence (LIS) for each. The total number of changes equals the original sub-array size minus its LIS.
  • Binary Search Optimization: Optimize LIS computation with binary search. This can reduce time complexity from O(n^2) to O(n log n) for larger arrays.

Example C++ Code:

class Solution {
public:
    int kIncreasing(vector<int>& arr, int k) {
        int total = 0;
        int n = arr.size();
        for (int i = 0; i < k; ++i) {
            vector<int> sub;
            for (int j = i; j < n; j += k) {
                sub.push_back(arr[j]);
            }
            total += sub.size() - lis(sub);
        }
        return total;
    }

    int lis(vector<int>& nums) {
        vector<int> dp;
        for (int num : nums) {
            auto it = upper_bound(dp.begin(), dp.end(), num);
            if (it == dp.end()) {
                dp.push_back(num);
            } else {
                *it = num;
            }
        }
        return dp.size();
    }
};

Keywords: K-Increasing array, Longest increasing subsequence, Dynamic programming, Operation count optimization, Binary search algorithm

Detailed Analysis of the Problems

Core Concepts & Algorithmic Strategies

These challenges emphasize the need for efficiency, clarity, and the ability to handle array and string manipulations effectively. Understanding the nuances of palindrome detection, string manipulation, dynamic programming, and algorithmic trading will help coders perform well in similar contexts. The efficient insertion of spaces and calculation of the Longest Increasing Subsequence (LIS) demonstrate practical algorithmic techniques crucial for problem-solving in coding interviews.

Understanding dynamic programming's applications and binary search enhances algorithm efficiency and solution reliability.

These fundamental ideas are essential for anyone looking to improve their abilities to solve challenging coding questions, regardless of experience level.

Pros and Cons of Problem-Solving Approaches

👍 Pros

Optimized algorithmic skills lead to faster execution times.

Effective problem-solving techniques enhance code clarity.

Dynamic programming can handle complex data structures well.

Binary search can optimize solutions by cutting search space effectively.

String and array manipulation techniques are adaptable to different problems.

👎 Cons

Dynamic programming can be space-intensive due to memoization.

Algorithmic optimizations may increase code complexity.

Understanding dynamic programming intricacies is essential.

Binary search requires a sorted dataset to function properly.

Careful selection of algorithmic strategy is crucial to achieve optimal results.

FAQ

What is a palindromic string?
A palindromic string is a string that reads the same forwards and backward. Common examples include 'madam,' 'level,' and 'racecar.'
Why is it important to optimize algorithmic trading code?
Optimization is critical in algorithmic trading to ensure quick decision-making and efficient execution of trades, which can directly impact profitability.
How does dynamic programming optimize the 'smooth descent periods' problem?
Dynamic programming allows you to reuse previously calculated values, avoiding redundant computations and significantly improving efficiency for large datasets.

Related Questions

How can dynamic programming techniques be applied to solve array manipulation problems?
Dynamic programming is instrumental in solving array manipulation problems by dividing them into subproblems. Each subproblem is solved and stored, preventing redundant calculations and leading to more efficient solutions. Techniques like memoization and tabulation, commonly used in dynamic programming, help in managing the complexity associated with array manipulations. By storing intermediate results, dynamic programming ensures that no computation is performed more than once, optimizing both time and space complexity. Consider the problem of finding the longest increasing subsequence. A dynamic programming solution involves constructing a table that stores the lengths of the longest increasing subsequences ending at each index, which is then used to find the global optimum. To illustrate, let’s consider the Longest Increasing Subsequence (LIS) problem. This is a classic example where dynamic programming shines. Given an array of numbers, the goal is to find the length of the longest subsequence that is sorted in increasing order. Dynamic programming provides an efficient way to solve this by breaking the problem into smaller, manageable pieces. Dynamic Programming Solution for LIS: Initialization: Create an array dp of the same length as the input array nums. Each element dp[i] will store the length of the LIS ending at index i. Initialize all elements of dp to 1, because a single element is itself a subsequence of length 1. Iteration: Iterate through the nums array from the second element onwards. For each element nums[i], iterate through all preceding elements nums[j] where j < i. If nums[i] is greater than nums[j], it means we can extend the LIS ending at j by including nums[i]. Update dp[i] to be the maximum of its current value and dp[j] + 1. Result: After filling the dp array, the maximum value in dp represents the length of the LIS for the entire input array. This process allows each element to build upon the solutions of its predecessors, ensuring the most efficient route to the global solution. Here’s the C++ code to help visualize: int lis(vector<int>& nums) { int n = nums.size(); vector<int> dp(n, 1); for (int i = 1; i < n; ++i) { for (int j = 0; j < i; ++j) { if (nums[i] > nums[j]) { dp[i] = max(dp[i], dp[j] + 1); } } } return *max_element(dp.begin(), dp.end()); } Time Complexity: This dynamic programming solution has a time complexity of O(n^2), where n is the number of elements in the array. Example Walkthrough: Consider the array nums = [10, 22, 9, 33, 21, 50, 41, 60]. Let's walk through how the dp array is filled: dp[0] = 1 (LIS ending at 10 is just 10 itself) dp[1] = 2 (LIS ending at 22 is 10, 22) dp[2] = 1 (LIS ending at 9 is just 9) dp[3] = 3 (LIS ending at 33 is 10, 22, 33 or 10, 9, 33) dp[4] = 2 (LIS ending at 21 is 10, 21 or 9, 21) dp[5] = 4 (LIS ending at 50 is 10, 22, 33, 50 or 10, 22, 21, 50 etc.) dp[6] = 4 (LIS ending at 41 is 10, 22, 33, 41 or 10, 22, 21, 41 etc.) dp[7] = 5 (LIS ending at 60 is 10, 22, 33, 50, 60 or 10, 22, 41, 60 etc.) The maximum value in dp is 5, which is the length of the longest increasing subsequence. This example illustrates how dynamic programming can efficiently compute the longest increasing subsequence by breaking the problem into smaller, overlapping subproblems. Understanding these techniques is crucial for solving more complex coding challenges.
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