Suggestion: - Don't cheat in any contest. - Challenge yourself with harder problems independently. If you encounter difficulties, start with a brute force approach. If it's inefficient (e.g., Time Limit Exceeded), then consider alternative strategies such as: — Mathematical optimizations — Implementation techniques like greedy algorithms

For instance, when faced with a challenging problem like "Longest Increasing Subsequence (LIS)", you might initially attempt a brute force solution that checks all subsequences. However, this approach can become impractical for larger inputs. To optimize, you could implement a dynamic programming solution that uses memoization or tabulation to efficiently compute the LIS length in ( O(n^2) ) or ( O(n \log n) ) time complexity, respectively.

approach of the LIS problem using dynamic programming:

• Understand the Problem: Given an array of integers, find the length of the longest increasing subsequence.

• Start with Recursive Approach: Begin by defining a recursive function to explore all subsequences:

def lis_recursive(nums, prev_idx, current_idx):
    if current_idx == len(nums):
        return 0
    taken = 0
    if prev_idx < 0 or nums[current_idx] > nums[prev_idx]:
        taken = 1 + lis_recursive(nums, current_idx, current_idx + 1)
    not_taken = lis_recursive(nums, prev_idx, current_idx + 1)
    return max(taken, not_taken)

nums = [10, 22, 9, 33, 21, 50, 41, 60]
print(lis_recursive(nums, -1, 0))  # Output: 5 (for LIS [10, 22, 33, 50, 60])

• Add Memoization: Improve efficiency by memoizing results to avoid redundant calculations:

memo = {}

def lis_memo(nums, prev_idx, current_idx):
    if (prev_idx, current_idx) in memo:
        return memo[(prev_idx, current_idx)]
    if current_idx == len(nums):
        return 0
    taken = 0
    if prev_idx < 0 or nums[current_idx] > nums[prev_idx]:
        taken = 1 + lis_memo(nums, current_idx, current_idx + 1)
    not_taken = lis_memo(nums, prev_idx, current_idx + 1)
    memo[(prev_idx, current_idx)] = max(taken, not_taken)
    return memo[(prev_idx, current_idx)]

nums = [10, 22, 9, 33, 21, 50, 41, 60]
print(lis_memo(nums, -1, 0))  # Output: 5 (for LIS [10, 22, 33, 50, 60])

• Transition to Tabulation: Solve iteratively using a bottom-up approach:

def lis_tabulation(nums):
    if not nums:
        return 0
    n = len(nums)
    dp = [1] * n
    for i in range(1, n):
        for j in range(0, i):
            if nums[i] > nums[j]:
                dp[i] = max(dp[i], dp[j] + 1)
    return max(dp)

nums = [10, 22, 9, 33, 21, 50, 41, 60]
print(lis_tabulation(nums))  # Output: 5 (for LIS [10, 22, 33, 50, 60])

• Practice and Compare: Solve other dynamic programming problems using both recursive with memoization and iterative tabulation approaches to understand their strengths and when to use each.

• Refine Solutions: Continuously review and optimize your solutions for clarity and efficiency based on problem requirements.