Results for
Over the past three weeks, players have been having great fun solving problems, sharing knowledge, and connecting with each other. Did you know over 15,000 solutions have already been submitted?
This is the final week to solve Cody problems and climb the leaderboard in the main round. Remember: solving just one problem in the contest problem group gives you a chance to win MathWorks T-shirts or socks.
🎉 Week 3 Winners:
Weekly Prizes for Contest Problem Group Finishers:
@Umar, @David Hill, @Takumi, @Nicolas, @WANG Zi-Xiang, @Rajvir Singh Gangar, @Roberto, @Boldizsar, @Abi, @Antonio
Weekly Prizes for Contest Problem Group Solvers:
Weekly Prizes for Tips & Tricks Articles:
This week’s prize goes to @Cephas. See the comments from our judge and problem group author @Matt Tearle:
'Some folks have posted deep dives into how to tackle specific problems in the contest set. But others have shared multiple smaller, generally useful tips. This week, I want to congratulate the cumulative contribution of Cool Coder Cephas, who has shared several of my favorite MATLAB techniques, including logical indexing, preallocation, modular arithmetic, and more. Cephas has also given some tips applying these MATLAB techniques to specific contest problems, such as using a convenient MATLAB function to vectorize the Leaderboard problem. Tip for Problem 61059 – Leaderboard for the Nedball World Cup:'
Congratulations to all Week 3 winners! Let’s carry this momentum into the final week!
If you haven't solved the problem yet, below hints guide how the algorithm should be implemented and clarify subtle rules that are easy to miss.
1. Shield is ONLY defended in HOME matches of the CURRENT holder - Even if a team beats the Shield holder in an away match, that does NOT count as a Shield defense.
2. A team defends the Shield ONLY when:
> They currently hold it.
> They are home team in that match
3. Shield transfer happens ONLY if the HOLDER plays a home match AND loses - A team may lose an away match — no effect.
4. The output ALWAYS includes the initial holder as the first row.
5. Defenses count resets for each new holder. - Every holder accumulates their own count until they lose it at home.
6. Match numbers are 1-indexed in the input, but “0” is used for initial state - The first real match is Match 1, but the output starts with Match 0.
7. Output row is created ONLY WHEN SHIELD CHANGES HANDS - This is an important hidden detail. A new row is appended, When the current holder loses a home match → Shield taken by visitor. If no loss at home occurs after that → no new row until next change.
8. The last holder’s defense count goes until the season ends - Even if they lose away later.
9. If a holder never gets a home match, defenses = 0.
10. In case the holder loses their very first home match → defenses = 0.
11. Shield changes only on HOME LOSS, not on a draw.
I hope above hints will help you in solving the problem.
Thanks and Regards,
Dev
In just two weeks, the competition has become both intense and friendly as participants race to climb the team leaderboard, especially in Team Creative, where @Mehdi Dehghan currently leads with 1400+ points, followed by @Vasilis Bellos with 900+ points.
There’s still plenty of time to participate before the contest's main round ends on December 7. Solving just one problem in the contest problem group gives you a chance to win MathWorks T-shirts or socks. Completing the entire problem group not only boosts your odds but also helps your team win.
🎉 Week 2 Winners:
Weekly Prizes for Contest Problem Group Finishers:
@Cephas, @Athi, @Bin Jiang, @Armando Longobardi, @Simone, @Maxi, @Pietro, @Hong Son, @Salvatore, @KARUPPASAMYPANDIYAN M
Weekly Prizes for Contest Problem Group Solvers:
Weekly Prizes for Tips & Tricks Articles:
This week’s prize goes to @Athi for the highly detailed post Solving Systematically The Clueless - Lord Ned in the Game Room.
Comment from the judge:
Shortly after the problem set dropped, several folks recognized that the final problem, "Clueless", was a step above the rest in difficulty. So, not surprisingly, there were a few posts in the discussion boards related to how to tackle this problem. Athi, of the Cool Coders, really dug deep into how the rules and strategies could be turned into an algorithm. There's always more than one way to tackle any difficult programming problem, so it was nice to see some discussion in the comments on different ways you can structure the array that represents your knowledge of who has which cards.
Congratulations to all Week 2 winners! Let’s keep the momentum going!
% Recreation of Saturn photo
figure('Color', 'k', 'Position', [100, 100, 800, 800]);
ax = axes('Color', 'k', 'XColor', 'none', 'YColor', 'none', 'ZColor', 'none');
hold on;
% Create the planet sphere
[x, y, z] = sphere(150);
% Saturn colors - pale yellow/cream gradient
saturn_radius = 1;
% Create color data based on latitude for gradient effect
lat = asin(z);
color_data = rescale(lat, 0.3, 0.9);
% Plot Saturn with smooth shading
planet = surf(x*saturn_radius, y*saturn_radius, z*saturn_radius, ...
color_data, ...
'EdgeColor', 'none', ...
'FaceColor', 'interp', ...
'FaceLighting', 'gouraud', ...
'AmbientStrength', 0.3, ...
'DiffuseStrength', 0.6, ...
'SpecularStrength', 0.1);
% Use a cream/pale yellow colormap for Saturn
cream_map = [linspace(0.4, 0.95, 256)', ...
linspace(0.35, 0.9, 256)', ...
linspace(0.2, 0.7, 256)'];
colormap(cream_map);
% Create the ring system
n_points = 300;
theta = linspace(0, 2*pi, n_points);
% Define ring structure (inner radius, outer radius, brightness)
rings = [
1.2, 1.4, 0.7; % Inner ring
1.45, 1.65, 0.8; % A ring
1.7, 1.85, 0.5; % Cassini division (darker)
1.9, 2.3, 0.9; % B ring (brightest)
2.35, 2.5, 0.6; % C ring
2.55, 2.8, 0.4; % Outer rings (fainter)
];
% Create rings as patches
for i = 1:size(rings, 1)
r_inner = rings(i, 1);
r_outer = rings(i, 2);
brightness = rings(i, 3);
% Create ring coordinates
x_inner = r_inner * cos(theta);
y_inner = r_inner * sin(theta);
x_outer = r_outer * cos(theta);
y_outer = r_outer * sin(theta);
% Front side of rings
ring_x = [x_inner, fliplr(x_outer)];
ring_y = [y_inner, fliplr(y_outer)];
ring_z = zeros(size(ring_x));
% Color based on brightness
ring_color = brightness * [0.9, 0.85, 0.7];
fill3(ring_x, ring_y, ring_z, ring_color, ...
'EdgeColor', 'none', ...
'FaceAlpha', 0.7, ...
'FaceLighting', 'gouraud', ...
'AmbientStrength', 0.5);
end
% Add some texture/gaps in the rings using scatter
n_particles = 3000;
r_particles = 1.2 + rand(1, n_particles) * 1.6;
theta_particles = rand(1, n_particles) * 2 * pi;
x_particles = r_particles .* cos(theta_particles);
y_particles = r_particles .* sin(theta_particles);
z_particles = (rand(1, n_particles) - 0.5) * 0.02;
% Vary particle brightness
particle_colors = repmat([0.8, 0.75, 0.6], n_particles, 1) .* ...
(0.5 + 0.5*rand(n_particles, 1));
scatter3(x_particles, y_particles, z_particles, 1, particle_colors, ...
'filled', 'MarkerFaceAlpha', 0.3);
% Add dramatic outer halo effect - multiple layers extending far out
n_glow = 20;
for i = 1:n_glow
glow_radius = 1 + i*0.35; % Extend much farther
alpha_val = 0.08 / sqrt(i); % More visible, slower falloff
% Color gradient from cream to blue/purple at outer edges
if i <= 8
glow_color = [0.9, 0.85, 0.7]; % Warm cream/yellow
else
% Gradually shift to cooler colors
mix = (i - 8) / (n_glow - 8);
glow_color = (1-mix)*[0.9, 0.85, 0.7] + mix*[0.6, 0.65, 0.85];
end
surf(x*glow_radius, y*glow_radius, z*glow_radius, ...
ones(size(x)), ...
'EdgeColor', 'none', ...
'FaceColor', glow_color, ...
'FaceAlpha', alpha_val, ...
'FaceLighting', 'none');
end
% Add extensive glow to rings - make it much more dramatic
n_ring_glow = 12;
for i = 1:n_ring_glow
glow_scale = 1 + i*0.15; % Extend farther
alpha_ring = 0.12 / sqrt(i); % More visible
for j = 1:size(rings, 1)
r_inner = rings(j, 1) * glow_scale;
r_outer = rings(j, 2) * glow_scale;
brightness = rings(j, 3) * 0.5 / sqrt(i);
x_inner = r_inner * cos(theta);
y_inner = r_inner * sin(theta);
x_outer = r_outer * cos(theta);
y_outer = r_outer * sin(theta);
ring_x = [x_inner, fliplr(x_outer)];
ring_y = [y_inner, fliplr(y_outer)];
ring_z = zeros(size(ring_x));
% Color gradient for ring glow
if i <= 6
ring_color = brightness * [0.9, 0.85, 0.7];
else
mix = (i - 6) / (n_ring_glow - 6);
ring_color = brightness * ((1-mix)*[0.9, 0.85, 0.7] + mix*[0.65, 0.7, 0.9]);
end
fill3(ring_x, ring_y, ring_z, ring_color, ...
'EdgeColor', 'none', ...
'FaceAlpha', alpha_ring, ...
'FaceLighting', 'none');
end
end
% Add diffuse glow particles for atmospheric effect
n_glow_particles = 8000;
glow_radius_particles = 1.5 + rand(1, n_glow_particles) * 5;
theta_glow = rand(1, n_glow_particles) * 2 * pi;
phi_glow = acos(2*rand(1, n_glow_particles) - 1);
x_glow = glow_radius_particles .* sin(phi_glow) .* cos(theta_glow);
y_glow = glow_radius_particles .* sin(phi_glow) .* sin(theta_glow);
z_glow = glow_radius_particles .* cos(phi_glow);
% Color particles based on distance - cooler colors farther out
particle_glow_colors = zeros(n_glow_particles, 3);
for i = 1:n_glow_particles
dist = glow_radius_particles(i);
if dist < 3
particle_glow_colors(i,:) = [0.9, 0.85, 0.7];
else
mix = (dist - 3) / 4;
particle_glow_colors(i,:) = (1-mix)*[0.9, 0.85, 0.7] + mix*[0.5, 0.6, 0.9];
end
end
scatter3(x_glow, y_glow, z_glow, rand(1, n_glow_particles)*2+0.5, ...
particle_glow_colors, 'filled', 'MarkerFaceAlpha', 0.05);
% Lighting setup
light('Position', [-3, -2, 4], 'Style', 'infinite', ...
'Color', [1, 1, 0.95]);
light('Position', [2, 3, 2], 'Style', 'infinite', ...
'Color', [0.3, 0.3, 0.4]);
% Camera and view settings
axis equal off;
view([-35, 25]); % Angle to match saturn_photo.jpg - more dramatic tilt
camva(10); % Field of view - slightly wider to show full halo
xlim([-8, 8]); % Expanded to show outer halo
ylim([-8, 8]);
zlim([-8, 8]);
% Material properties
material dull;
title('Saturn - Left click: Rotate | Right click: Pan | Scroll: Zoom', 'Color', 'w', 'FontSize', 12);
% Enable interactive camera controls
cameratoolbar('Show');
cameratoolbar('SetMode', 'orbit'); % Start in rotation mode
% Custom mouse controls
set(gcf, 'WindowButtonDownFcn', @mouseDown);
function mouseDown(src, ~)
selType = get(src, 'SelectionType');
switch selType
case 'normal' % Left click - rotate
cameratoolbar('SetMode', 'orbit');
rotate3d on;
case 'alt' % Right click - pan
cameratoolbar('SetMode', 'pan');
pan on;
end
end
In https://www.mathworks.com/matlabcentral/answers/38165-how-to-remove-decimal#comment_3345149 @Luisa asks,
@Cody Team, how can I vote or give a like in great comments?
It seems that there are not such options.
In just one week, we have hit an amazing milestone: 500+ players registered and 5000+ solutions submitted! We’ve also seen fantastic Tips & Tricks articles rolling in, making this contest a true community learning experience.
And here’s the best part: you don’t need to be a top-ranked player to win. To encourage more casual and first-time players to jump in, we’re introducing new weekly prizes starting Week 2!
New Casual Player Prizes:
- 5 extra MathWorks T-shirts or socks will be awarded every week.
- All you need to qualify is to register and solve one problem in the Contest Problem Group.
Jump in, try a few problems, and don’t be shy to ask questions in your team’s channel. You might walk away with a prize!
Week 1 Winners:
Weekly Prizes for Contest Problem Group Finishers:
@Mazhar, @Julien, @Mohammad Aryayi, @Pawel, @Mehdi Dehghan, @Christian Schröder, @Yolanda, @Dev Gupta, @Tomoaki Takagi, @Stefan Abendroth
Weekly Prizes for Tips & Tricks Articles:
We had a lot of people share useful tips (including some personal favorite MATLAB tricks). But Vasilis Bellos went *deep* into the Bridges of Nedsburg problem. Fittingly for a Creative Coder, his post was innovative and entertaining, while also cleverly sneaking in some hints on a neat solution method that wasn't advertised in the problem description.
Congratulations to all Week 1 winners! Prizes will be awarded after the contest ends. Let’s keep the momentum going!
Many MATLAB Cody problems involve solving congruences, modular inverses, Diophantine equations, or simplifying ratios under constraints. A powerful tool for these tasks is the Extended Euclidean Algorithm (EEA), which not only computes the greatest common divisor, gcd(a,b), but also provides integers x and y such that: a*x + b*y = gcd(a,b) - which is Bezout's identity.
Use of the Extended Euclidean Algorithm is very using in solving many different types of MATLAB Cody problems such as:
- Computing modular inverses safely, even for very large numbers
- Solving linear Diophantine equations
- Simplifing fractions or finding nteger coefficients without using symbolic tools
- Avoiding loops (EEA can be implemented recursively)
Below is a recursive implementation of the EEA.
function [g,x,y] = egcd(a,b)
% a*x + b*y = g [gcd(a,b)]
if b == 0
g = a; x = 1; y = 0;
else
[g, x1, y1] = egcd(b, mod(a,b));
x = y1;
y = x1 - floor(a/b)*y1;
end
end
Problem:
Given integers a and m, return the modular inverse of a (mod m).
If the inverse does not exist, return -1.
function inv = modInverse(a,m)
[g,x,~] = egcd(a,m);
if g ~= 1 % inverse doesn't exist
inv = -1;
else
inv = mod(x,m); % Bézout coefficient gives the inverse
end
end
%find the modular inverse of 19 (mod 5)
inv=modInverse(19,5)
Congratulations to all the Relentless Coders who have completed the problem set. I hope you weren't too busy relentlessly solving problems to enjoy the silliness I put into them.
If you've solved the whole problem set, don't forget to help out your teammates with suggestions, tips, tricks, etc. But also, just for fun, I'm curious to see which of my many in-jokes and nerdy references you noticed. Many of the problems were inspired by things in the real world, then ported over into the chaotic fantasy world of Nedland.
I guess I'll start with the obvious real-world reference: @Ned Gulley (I make no comment about his role as insane despot in any universe, real or otherwise.)
Hi Everyone!
As this is the most difficult question in problem group "Cody Contest 2025". To solve this problem, It is very important to understand all the hidden clues in the problem statement. Because everything is not directly visible.
For those who tried the problem, but were not able to solve. You might have missed any of the below hints -
- “The other players do not get to see which card has been shown, but they do know which three cards were asked for and that the player asked had one of them.” - Even when the card identity isn’t revealed (result = 0), you still gain partial knowledge — the asked player must have at least one of those three cards, meaning you can mark other players as not having all three simultaneously.
- "If it is your turn, you know the exact identity of that card" - You only know the exact shown card when result = 1, 2, or 3 — and it must be your turn. If someone else asked (even if you know result = 0), you don’t know which one was shown. So the meaning of result depends on whose turn it was, which is implicit — MATLAB code must assume that turns alternate 1→m→1, so your turn index is determined by (t-1) mod m + 1 == pnum.
- "Any leftover cards are placed face-up so that all players can see them" - These cards (commoncards) are not in anyone’s hand and cannot be in the envelope. So they’re not just visible — they’re logical constraints to eliminate from deduction.
- “It may be possible to determine the solution from less information than is given, but the information given will always be sufficient.”
- "Turn order is implied, not given explicitly" - Players take turns in order (1 to m, and back to 1).
On considering all the clues and constraints in the question, you will definitely be able to card for each category present in envelope.
I hope above clues will be useful for you.
Thank you, wishing you the success!
Regards,
Dev
Experimenting with Agentic AI
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I am an AI skeptic
0%
AI is banned at work
11%
I am happy with Conversational AI
44%
9 votes
When solving Cody problems, sometimes your solution takes too long — especially if you’re recomputing large arrays or iterative sequences every time your function is called.
The Cody work area resets between separate runs of your code, but within one Cody test suite, your function may be called multiple times in a single session.
This is where persistent variables come in handy.
A persistent variable keeps its value between function calls, but only while MATLAB is still running your function suite.
This means:
- You can cache results to avoid recomputation.
- You can accumulate data across multiple calls.
- But it resets when Cody or MATLAB restarts.
Suppose you’re asked to find the n-th Fibonacci number efficiently — Cody may time out if you use recursion naively. Here’s how to use persistent to store computed values:
function f = fibPersistent(n)
import java.math.BigInteger
persistent F
if isempty(F)
F=[BigInteger('0'),BigInteger('1')];
for k=3:10000
F(k)=F(k-1).add(F(k-2));
end
end
% Extend the stored sequence only if needed
while length(F) <= n
F(end+1)=F(end).add(F(end-1));
end
f = char(F(n+1).toString); % since F(1) is really F(0)
end
%calling function 100 times
K=arrayfun(@(x)fibPersistent(x),randi(10000,1,100),'UniformOutput',false);
K(100)
The fzero function can handle extremely messy equations — even those mixing exponentials, trigonometric, and logarithmic terms — provided the function is continuous near the root and you give a reasonable starting point or interval.
It’s ideal for cases like:
- Solving energy balance equations
- Finding intersection points of nonlinear models
- Determining parameters from experimental data
Example: Solving for Equilibrium Temperature in a Heat Radiation-Conduction Model
Suppose a spacecraft component exchanges heat via conduction and radiation with its environment. At steady state, the power generated internally equals the heat lost:
Given constants:
= 25 W- k = 0.5 W/K
- ϵ = 0.8
- σ = 5.67e−8 W/m²K⁴
- A = 0.1 m²
= 250 K
Find the steady-state temperature, T.
% Given constants
Qgen = 25;
k = 0.5;
eps = 0.8;
sigma = 5.67e-8;
A = 0.1;
Tinf = 250;
% Define the energy balance equation (set equal to zero)
f = @(T) Qgen - (k*(T - Tinf) + eps*sigma*A*(T.^4 - Tinf^4));
% Plot for a sense of where the root lies before implementing
fplot(f, [250 300]); grid on
xlabel('Temperature (K)'); ylabel('f(T)')
title('Energy Balance: Root corresponds to steady-state temperature')
% Use fzero with an interval that brackets the root
T_eq = fzero(f, [250 300]);
fprintf('Steady-state temperature: %.2f K\n', T_eq);
It’s exciting to dive into a new dataset full of unfamiliar variables but it can also be overwhelming if you’re not sure where to start. Recently, I discovered some new interactive features in MATLAB live scripts that make it much easier to get an overview of your data. With just a few clicks, you can display sparklines and summary statistics using table variables, sort and filter variables, and even have MATLAB generate the corresponding code for reproducibility.
The Graphics and App Building blog published an article that walks through these features showing how to explore, clean, and analyze data—all without writing any code.
If you’re interested in streamlining your exploratory data analysis or want to see what’s new in live scripts, you might find it helpful:
If you’ve tried these features or have your own tips for quick data exploration in MATLAB, I’d love to hear your thoughts!
I set my 3D matrix up with the players in the 3rd dimension. I set up the matrix with: 1) player does not hold the card (-1), player holds the card (1), and unknown holding the card (0). I moved through the turns (-1 and 1) that are fixed first. Then cycled through the conditional turns (0) while checking the cards of each player using the hints provided until it was solved. The key for me in solving several of the tests (11, 17, and 19) was looking at the 1's and 0's being held by each player.
sum(cardState==1,3);%any zeros in this 2D matrix indicate possible cards in the solution
sum(cardState==0,3)>0;%the ones in this 2D matrix indicate the only unknown positions
sum(cardState==1,3)|sum(cardState==0,3)>0;%oring the two together could provide valuable information
Some MATLAB Cody problems prohibit loops (for, while) or conditionals (if, switch, while), forcing creative solutions.
One elegant trick is to use nested functions and recursion to achieve the same logic — while staying within the rules.
Example: Recursive Summation Without Loops or Conditionals
Suppose loops and conditionals are banned, but you need to compute the sum of numbers from 1 to n. This is a simple example and obvisously n*(n+1)/2 would be preferred.
function s = sumRecursive(n)
zero=@(x)0;
s = helper(n); % call nested recursive function
function out = helper(k)
L={zero,@helper};
out = k+L{(k>0)+1}(k-1);
end
end
sumRecursive(10)
- The helper function calls itself until the base case is reached.
- Logical indexing into a cell array (k>0) act as an 'if' replacement.
- MATLAB allows nested functions to share variables and functions (zero), so you can keep state across calls.
Tips:
- Replace 'if' with logical indexing into a cell array.
- Replace for/while with recursion.
- Nested functions are local and can access outer variables, avoiding global state.
What a fantastic start to Cody Contest 2025! In just 2 days, over 300 players joined the fun, and we already have our first contest group finishers. A big shoutout to the first finisher from each team:
- Team Creative Coders: @Mehdi Dehghan
- Team Cool Coders: @Pawel
- Team Relentless Coders: @David Hill
- 🏆 First finisher overall: Mehdi Dehghan
Other group finishers: @Bin Jiang (Relentless), @Mazhar (Creative), @Vasilis Bellos (Creative), @Stefan Abendroth (Creative), @Armando Longobardi (Cool), @Cephas (Cool)
Kudos to all group finishers! 🎉
Reminder to finishers: The goal of Cody Contest is learning together. Share hints (not full solutions) to help your teammates complete the problem group. The winning team will be the one with the most group finishers — teamwork matters!
To all players: Don’t be shy about asking for help! When you do, show your work — include your code, error messages, and any details needed for others to reproduce your results.
Keep solving, keep sharing, and most importantly — have fun!
Many MATLAB Cody problems involve recognizing integer sequences.
If a sequence looks familiar but you can’t quite place it, the On-Line Encyclopedia of Integer Sequences (OEIS) can be your best friend.
OEIS will often identify the sequence, provide a formula, recurrence relation, or even direct MATLAB-compatible pseudocode.
Example: Recognizing a Cody Sequence
Suppose you encounter this sequence in a Cody problem:
1, 1, 2, 3, 5, 8, 13, 21, ...
Entering it on OEIS yields A000045 – The Fibonacci Numbers, defined by:
F(n) = F(n-1) + F(n-2), with F(1)=1, F(2)=1
You can then directly implement it in MATLAB:
function F = fibSeq(n)
F = zeros(1,n);
F(1:2) = 1;
for k = 3:n
F(k) = F(k-1) + F(k-2);
end
end
fibSeq(15)
When solving MATLAB Cody problems involving very large integers (e.g., factorials, Fibonacci numbers, or modular arithmetic), you might exceed MATLAB’s built-in numeric limits.
To overcome this, you can use Java’s java.math.BigInteger directly within MATLAB — it’s fast, exact, and often accepted by Cody if you convert the final result to a numeric or string form.
Below is an example of using it to find large factorials.
function s = bigFactorial(n)
import java.math.BigInteger
f = BigInteger('1');
for k = 2:n
f = f.multiply(BigInteger(num2str(k)));
end
s = char(f.toString); % Return as string to avoid overflow
end
bigFactorial(100)
If you have solved a Cody problem before, you have likely seen the Scratch Pad text field below the Solution text field. It provides a quick way to get feedback on your solution before submitting it. Since submitting a solution takes you to a new page, any time a wrong solution is submitted, you have to navigate back to the problem page to try it again.
Instead, I use the Scratch Pad to test my solution repeatedly before submitting. That way, I get to a working solution faster without having to potentially go back and forth many times between the problem page and the wrong-solution page.
Here is my approach:
- Write a tentative solution.
- Copy a test case from the test suite into the Scratch Pad.
- Click the Run Function button—this is immediately below the Scratch Pad and above the Output panel and Submit buttons.
- If the solution does not work, modify the solution code, sometimes putting in disp() lines and/or removing semicolons to trace what the code is doing. Repeat until the solution passes.
- If the solution does work, repeat steps 2 through 4.
- Once there are no more test cases to copy and paste, clean up the code, if necessary (delete disp lines, reinstate all semicolons to suppress output). Click the Run Function button once more, just to make sure I did not break the solution while cleaning it up. Then, click the Submit button.
For problems with large test suites, you may find it useful to copy and paste in multiple test cases per iteration.
Hopefully you find this useful.
The main round of Cody Contest 2025 kicks off today! Whether you’re a beginner or a seasoned solver, now’s your time to shine.
Here’s how to join the fun:
- Pick your team — choose one that matches your coding personality.
- Solve Cody problems — gain points and climb the leaderboard.
- Finish the Contest Problem Group — help your team win and unlock chances for weekly prizes by finishing the Cody Contest 2025 problem group.
- Share Tips & Tricks — post your insights to win a coveted MathWorks Yeti Bottle.
- Bonus Round — 2 players from each team will be invited to a fun live code-along event!
- Watch Party – join the big watch event to see how top players tackle Cody problems
Contest Timeline:
- Main Round: Nov 10 – Dec 7, 2025
- Bonus Round: Dec 8 – Dec 19, 2025
Big prizes await — MathWorks swag, Amazon gift cards, and shiny virtual badges!
We look forward to seeing you in the contest — learn, compete, and have fun!
