🧠 Solving LeetCode Until I Become Top 1% — Day `46`

July 17, 2025

This content originally appeared on DEV Community and was authored by Md. Rishat Talukder

Problem: Maximum Length of a Subsequence With Modular Sum Zero

Difficulty: #Medium
Tags: #DP, #Array, #Greedy, #Math

Problem Summary

Given an array nums and an integer k, you need to find the length of the longest subsequence where every pair of adjacent elements (a, b) satisfies:
(a + b) % k == 0.

My Thought Process

Naive Brainstorming (a.k.a “The idea was there… kinda”):
- I figured it had something to do with remainders because of % k.
- I was thinking in terms of pairing mods somehow — so the core idea managed to show up to class.
- But when it came time to actually write working code? Hello darkness, my old friend — tabulation DP strikes again.
Why I Couldn’t Implement It:
- Because it’s tabulation and my brain just refuses to store 2D DP tables in memory like a normal human.
- Every time I try to update a dp[i][j] I end up wondering if I’m accidentally simulating the collapse of spacetime.
What Actually Works:
- Create a 2D DP table dp[prev][curr], where each cell stores the max length of a subsequence ending with modulo pair (prev, curr).
- Update by flipping: If current number is num % k, then for each prev in [0, k-1], dp[prev][num] = dp[num][prev] + 1 Because (a + b) % k == 0 ⇨ a % k == (-b) % k.

Code Implementation (Python)

from typing import List

class Solution:
    def maximumLength(self, nums: List[int], k: int) -> int:
        dp = [[0] * k for _ in range(k)]  # dp[prev_mod][curr_mod]
        res = 0
        for num in nums:
            num %= k
            for prev in range(k):
                dp[prev][num] = dp[num][prev] + 1
                res = max(res, dp[prev][num])
        return res

This solution works because it leverages the modular symmetry. You basically abuse the 2D array to remember the last good match, and mutate it like a boss.

Time & Space Complexity

Time: O(n * k)
We iterate over all n elements and try pairing each with k mod classes.
Space: O(k^2)
The 2D DP table holds best subsequence lengths for each modulo pair.

Key Takeaways

Even if your tabulation brain is weak, recognizing a problem as modular + pair-based is already half the battle.
When % k is involved, think in terms of remainders, not actual numbers.
Don’t be ashamed of struggling with DP tables. They’re evil, and you’re brave for trying.