ChatGPT LIS Discussion

Fastrail08

May 27th, 2025

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Great question — you're thinking in the right direction! ✅ But let's break it down carefully.
You asked:
Can we say that we can memoize the above code with memo key as memo(level, msf)?
🔍 Quick Recap of Your Code
Your function:
int getLIS2(int level, int msf, string psf, vector<int> &v)
level: current index in the vector
msf: maximum so far in the current increasing subsequence
psf: printed string for debugging (not affecting the logic)
v: the input array
✅ What Are We Trying to Compute?
You're trying to:
Print all valid increasing subsequences, and
Return the length of the Longest Increasing Subsequence (LIS)
🔁 Can We Memoize It?
Short Answer: Not directly with memo[level][msf], because msf is an integer, and it can be any value — even INT_MIN.
Let’s understand why this is tricky.
⚠️ Why memo[level][msf] Is Problematic
msf can be any integer, including large positive or negative numbers.
You can't use msf directly as an index/key in a memo table unless you:
Normalize it to a small range (not possible in general), or
Use a map or unordered_map instead of an array.
So this won't work:
vector<vector<int>> memo(n, vector<int>(INT_MAX)); // unfeasible
✅ Correct Memoization Strategy
The standard approach is to memoize by level and the index of the previous element included, not by value.
So instead of passing msf, you pass prevIdx (index of the last element included in the subsequence).
🔁 Updated Function Signature:
int getLIS(int level, int prevIdx, vector<int>& v, vector<vector<int>>& dp)
🧠 Logic:
At each level, you either include v[level] if it's greater than v[prevIdx], or skip it.
You memoize dp[level][prevIdx + 1] because prevIdx == -1 (no element included yet) needs to be stored at index 0.
✅ Correct Memoized Code:
int getLIS(int level, int prevIdx, vector<int>& v, vector<vector<int>>& dp) {
if (level == v.size()) return 0;
if (dp[level][prevIdx + 1] != -1) return dp[level][prevIdx + 1];
int taken = 0;
if (prevIdx == -1 || v[level] > v[prevIdx]) {
taken = 1 + getLIS(level + 1, level, v, dp);
}
int notTaken = getLIS(level + 1, prevIdx, v, dp);
return dp[level][prevIdx + 1] = max(taken, notTaken);
}
🧪 Example main():
int main() {
int n;
cin >> n;
vector<int> v(n);
for(int i = 0; i < n; i++){
cin >> v[i];
}
vector<vector<int>> dp(n, vector<int>(n + 1, -1));
cout << getLIS(0, -1, v, dp) << '\n';
}
🧵 Summary:
Version Works? Why
memo(level, msf) ❌ msf can be any value, unbounded, hard to index
memo(level, prevIdx) ✅ Limited range (0 to n), easy to memoize
psf for printing ✅ Useful for output/debugging but not memoized
You said:
consider the value to be in range -10^4 to 10^4. What then?
ChatGPT said:
Excellent — this changes things.
If you know in advance that msf (maximum so far) will always be in a fixed range, say from -10⁴ to 10⁴, then yes:
✅ You can memoize using memo[level][msf], with appropriate offsetting.
⚙️ Why It Works:
Total possible msf values: 20001 (from -10000 to 10000, inclusive).
So you can safely create a 2D vector like this:
cpp
Copy
Edit
vector<vector<int>> dp(n, vector<int>(20001, -1));
You’ll also need to offset the msf by 10000 when storing in the memo:
dp[level][msf + 10000]
✅ Revised Code with msf Memoization
Here's how you can modify your original function to use msf directly, assuming the known range:
int getLIS2(int level, int msf, string psf, vector<int> &v, vector<vector<int>> &dp){
if(level >= v.size()){
cout << psf << '\n';
return 0;
}
if(dp[level][msf + 10000] != -1)
return dp[level][msf + 10000];
int sizeOfSubInc = 0, sizeOfSubExc = 0;
if(v[level] > msf){
sizeOfSubInc = getLIS2(level + 1, v[level], psf + to_string(v[level]) + " ", v, dp);
}
sizeOfSubExc = getLIS2(level + 1, msf, psf, v, dp);
return dp[level][msf + 10000] = max(sizeOfSubExc, sizeOfSubInc + 1);
}
🧪 main() Function:
int main() {
int n;
cin >> n;
vector<int> v(n);
for(int i = 0; i < n; i++){
cin >> v[i];
}
vector<vector<int>> dp(n + 1, vector<int>(20001, -1));
cout << getLIS2(0, -10000, "", v, dp) << '\n';
}
🔍 Important Notes:
Parameter Value/Range Why it works
msf [-10000, 10000] Bounded, so indexable
dp[level][msf+10000] [0...n][0...20000] Offset needed for negatives
🧵 Summary
Yes, memoizing using (level, msf) is valid if msf is bounded.
Use offsetting for negative values.
This approach is slightly less standard than using prevIdx, but totally valid with known bounds.
You said:
Can it be also memoised 1d where memo[level] denotes what is the LIS up 0 to level - 1 index
ChatGPT said:
This is a great question, and it touches on the difference between top-down recursion with memoization and bottom-up DP approaches.
Let’s carefully unpack it.
🔁 Your Idea:
You're asking:
Can we memoize in 1D with memo[level], where memo[level] stores the LIS ending just before index level?
The short answer:
🟡 Sort of — but only in specific formulations.
Let me explain the distinction clearly.
✅ 1D Memoization — Possible, but Not with Your Current Top-Down Formulation
In your current recursive function, you track:
level: current index you're trying to process
msf: maximum value of the current subsequence
So the decision to include/exclude v[level] depends on the actual value of msf
Hence, you need both level and msf to uniquely identify subproblems.
So in this top-down recursive structure, memo[level] alone is not sufficient to capture all subproblem states, unless msf is encoded somehow.
✅ When 1D DP Works: Bottom-Up Dynamic Programming
This is the classical LIS in O(n²) using 1D DP:
int LIS(vector<int>& v) {
int n = v.size();
vector<int> dp(n, 1); // dp[i] = LIS ending at index i
for(int i = 1; i < n; ++i) {
for(int j = 0; j < i; ++j) {
if(v[i] > v[j]) {
dp[i] = max(dp[i], dp[j] + 1);
}
}
}
return *max_element(dp.begin(), dp.end());
}
This uses dp[i] as:
Length of the Longest Increasing Subsequence ending at index i
This works because the state space is fully described by just the current index — no need to carry msf or prevIdx.
🧠 Why Top-Down Needs 2D, Bottom-Up Can Use 1D
Approach State Needed DP Table Dimensions
Top-Down (Recursive) level, msf or prevIdx 2D
Bottom-Up (Iterative) Just i and inner loop on j 1D
✅ In Summary:
❌ Top-down recursion cannot use memo[level] alone unless you're sure msf is fixed or encoded.
✅ Bottom-up DP formulation of LIS does use a 1D dp[] array.
✅ If you switch to the iterative version, then yes — you only need dp[level].
You said:
show me an example where memo[level] will not identify an unique sub problem during memoisation
ChatGPT said:
Excellent — let's demonstrate this with an example that shows **why memo[level] alone is not sufficient in a top-down recursive approach to LIS.
🎯 Objective:
We’ll show a case where two different recursive calls reach the same level, but with different msf values, which means memo[level] would overwrite or misrepresent subproblems.
🔢 Example Input:
vector<int> v = {3, 2, 4};
Let’s trace your original function:
int getLIS2(int level, int msf, string psf, vector<int> &v)
🌲 Call Tree (Partial)
Initial Call:
getLIS2(0, -∞, "")
→ level = 0, msf = -∞, can include 3 → call:
getLIS2(1, 3, "3 ")
→ at level = 1, msf = 3
- v[1] = 2 < 3 → can't include → call:
getLIS2(2, 3, "3 ")
→ from level = 0, try excluding 3 → call:
getLIS2(1, -∞, "")
→ at level = 1, msf = -∞
- v[1] = 2 > -∞ → can include → call:
getLIS2(2, 2, "2 ")
🔍 Key Observation:
Look at these two recursive calls:
getLIS2(2, 3, "3 ")
getLIS2(2, 2, "2 ")
They both have level = 2, but:
Call msf psf
A 3 "3 "
B 2 "2 "
They represent two different states of the recursion — with different constraints:
In A, you can include v[2] = 4 because 4 > 3
In B, you can also include 4 because 4 > 2
But what if they had other values (say v = {3, 5, 4})?
Then:
A: msf = 5, v[2] = 4 → can't include 4
B: msf = 2, v[2] = 4 → can include 4
These would result in different valid subsequences and different LIS lengths.
❌ Why memo[level] Fails:
If you memoize as memo[level], then both of these states (level = 2) would share the same memo cell, even though:
Their previous choices were different
The subsequences being formed are different
The result of including or excluding v[2] depends on the value of msf, not just the level
✅ Solution:
Use memo[level][msfIndex] or memo[level][prevIdx] to capture both the position and the current state of the subsequence.
🧵 Summary
🛑 memo[level] assumes the result at that index is always the same, regardless of prior context.
✅ But LIS depends on whether or not current value can be appended to an increasing subsequence, so you must track that context (msf or prevIdx).
You said:
How would my memo[level][msf] have gone out of bound? Initially I have passed msf as INT_MIN, In recursion it won't have given any issue as I am not adding anything, just taking max at every level.
It would have given error in allocation of msf while initializing the memo table as can't allocate the array beyond 10^9 . Right? Could I have prevented it by using unordered_map<int, long long>?
ChatGPT said:
Absolutely! You're spot on with your diagnosis and reasoning — let's break it down and confirm everything clearly.
✅ What You Got Right:
1. Why memo[level][msf] Goes Out of Bounds
You're right — msf (maximum so far) can be any value in the integer range, potentially even INT_MIN (−2³¹). So trying to define:
vector<vector<int>> memo(n, vector<int>(INT_MAX));
is:
Not safe: Because msf values are not bounded to a small indexable range
Not feasible: You'd try to allocate a table with 2^31 columns, which is > 2 billion entries per row, leading to memory allocation errors
2. Why It Doesn’t Go Out of Bounds in Recursive Calls
Again, correct:
You’re not incrementing msf or modifying it in any unpredictable way — you only update it with v[level] if included.
So msf always comes from the input array or stays INT_MIN.
But the problem is not in runtime logic — it's in memory allocation.
✅ Can You Fix It with unordered_map?
Yes, and that’s exactly how you should handle unbounded or sparse states.
Instead of:
vector<vector<int>> dp(n, vector<int>(somethingHuge));
You do:
unordered_map<long long, int> dp;
Where your key is encoded like:
long long key = (long long)level * OFFSET + msf;
or use a pair:
unordered_map<pair<int, int>, int, pair_hash>
With a custom hash function if needed (for pair<int,int> as key):
struct pair_hash {
template<typename T1, typename T2>
size_t operator()(const pair<T1, T2>& p) const {
return hash<T1>()(p.first) ^ (hash<T2>()(p.second) << 1);
}
};
🧵 Summary
Statement Verdict
msf in recursive call won't go out of bound ✅ Correct
Allocating memo[level][msf] with full int range is unsafe ✅ Correct
Using unordered_map<int, long long> prevents this ✅ Correct
Can you memoize msf in map even with large input values? ✅ Yes — safely, because only valid paths are stored
You said:
how does this work?
unordered_map<pair<int, int>, int, pair_hash>
pair<int, int> for (level, msf) and the other int for storing the LIS ?
ChatGPT said:
Great, you're diving into a more elegant and readable way of handling memoization with composite keys using:
cpp
Copy
Edit
unordered_map<pair<int, int>, int, pair_hash>
Let’s break this down step-by-step:
✅ What This Declaration Means
unordered_map<pair<int, int>, int, pair_hash> dp;
This creates a hash map (dp) that:
Uses a pair<int, int> as the key — specifically, (level, msf)
Stores an int as the value — in your case, the LIS length from that subproblem
Uses a custom hash function (pair_hash) to correctly hash the pair<int, int>
👇 Why Use This Over long long Encoding?
Cleaner and more readable
Avoids manual encoding and offset logic
Lets the compiler and hash function take care of collisions and uniqueness
✅ How to Define pair_hash
The STL doesn't define a hash function for pair by default, so you provide one:
struct pair_hash {
template <class T1, class T2>
size_t operator()(const pair<T1, T2>& p) const {
return hash<T1>()(p.first) ^ (hash<T2>()(p.second) << 1);
}
};
This:
Hashes p.first and p.second independently
Combines them using XOR and shifting (to reduce collision)
You said:
Could I have done my some changes on my memo approach to reduce the caching to 1 D array.
ChatGPT said:
Fantastic question — you're asking:
Could I make some changes to reduce my memoization approach (which uses (level, msf)) from a 2D cache to 1D?
✅ Short Answer:
Yes — but only under a different interpretation of the problem, specifically if you switch to:
“What’s the LIS ending at each index i?”
In that form, yes — you can use 1D caching, and it's the classical bottom-up dynamic programming approach. But within your current top-down recursive structure with (level, msf), you cannot reduce to 1D without losing essential information.
🔍 Why Your Current Memo Needs 2D
State in your recursion:
getLIS(level, msf)
Here, level tells you where you are, and msf tells you what you've built so far.
🟡 level alone is not enough to determine a unique subproblem — because the result depends on what the previous maximum was (msf).
So you need both level and msf — hence, a 2D structure.
✅ When Can You Reduce to 1D?
If you restructure the problem to match the classic LIS DP:
dp[i] = length of LIS ending at index i
This is a bottom-up tabulation approach:
vector<int> dp(n, 1); // Base case: each element alone is length 1
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (arr[i] > arr[j])
dp[i] = max(dp[i], dp[j] + 1);
}
}
int lis = *max_element(dp.begin(), dp.end());
✅ This Works with 1D Because:
All information needed is encoded in i and previous dp[j].
You don't need to track the actual previous value (msf), only compare arr[i] > arr[j].
You said:
I want a 1 D array with O(N) space in memo code?
ChatGPT said:
Great — you're now targeting:
🧠 A Top-Down Recursive LIS implementation with only a 1D O(N) memo array, not O(N²) or using 2D or map.
That’s absolutely possible, if you slightly reframe your problem.
✅ What You Want:
Recursive LIS
Top-down (with memoization)
Only 1D dp[n] space
So: dp[i] = LIS starting from index i (or ending at i)
🔧 Strategy: Memoize LIS starting from index i
This shifts the problem slightly:
int lisFrom(int i, vector<int>& v, vector<int>& dp) {
if (dp[i] != -1) return dp[i];
int maxLen = 1; // At least the element itself
for (int j = i + 1; j < v.size(); ++j) {
if (v[j] > v[i]) {
maxLen = max(maxLen, 1 + lisFrom(j, v, dp));
}
}
return dp[i] = maxLen;
}
Then call it for all positions in main():
🧠 Explanation:
lisFrom(i) means: "What's the LIS starting at index i?"
Only dp[i] is needed to memoize it — that’s 1D O(N) space
Total time is still O(N²) due to nested recursion
🔄 Optional Improvement (Path Recovery)
If you want to print the actual LIS path too, we can maintain a next[i] array alongside.
Let me know if you'd like that enhancement!
✅ Summary:
Goal Achieved? Method
Top-down(recursive LIS) ✅ Done
Memoized ✅ With dp[i]
1D space, O(N) ✅ Uses dp[n] only
You said:
The O(N^2) time and O(N) space top down DP seemed less intuitive than my memo DP with O(N^2) space
ChatGPT said:
Totally understandable! The top-down O(N²) DP with O(N) space approach (the dp[i] = length of LIS starting at i) can feel a bit less intuitive at first compared to the classic memo DP that tracks (level, msf) with a 2D structure. Here's why, and some tips on how to think about them:
Your original memo DP with (level, msf):
Explicitly tracks the maximum element so far (msf), i.e., the last included element value.
Very direct and clear: you decide for each element if you can include it based on msf.
But this requires caching based on the value of msf, which is tricky (huge range → 2D map or complex structures).
So, harder to memoize efficiently.
The 1D top-down DP (dp[i]):
Tracks LIS length starting at index i (or equivalently, LIS ending at index i in the bottom-up).
No need to store or compare values explicitly, because you just check if next elements are bigger.
Much easier to memoize: a simple 1D array where dp[i] caches the result.
But conceptually a bit more indirect — you think in terms of subproblems starting at i, not maintaining an explicit max-so-far value.
How to get more intuition for the 1D DP?
Try this mindset:
“What is the longest increasing subsequence starting from position i?”
If you can find that for all i, the answer is just the max of those results.
You check all j > i, and if v[j] > v[i], you can move from i to j.
You don't carry around msf, because i itself guarantees the current "max" (since v[i] is the last included element at that subproblem).
Visualization:
v = [10, 9, 2, 5, 3, 7]
dp[i] = length of LIS starting at i
- dp[5] = 1 (only 7)
- dp[4] = 1 + max(dp[j] where v[j] > v[4]) = 1 + max(dp[5]) = 2
- dp[3] = 1 + max(dp[4], dp[5]) = 1 + max(2,1) = 3
- dp[2] = 1 + max(dp[3], dp[4], dp[5]) where v[j] > 2 = 1 + max(3,2,1) = 4
- ...
Summary
Approach Intuitiveness Memo Size
Memo (level, msf) (value) High Large (maps)
Memo dp[i] (start from i) Medium O(N)
If you want, I can help walk through the 1D DP with a step-by-step example or visualization — sometimes seeing it unfold is what makes it click.
Want me to do that?
You said:
yes
ChatGPT said:
Awesome! Let’s walk through the 1D top-down DP approach to LIS with a concrete example and visualization. This will help you see why it works and how memoization fits naturally.
Step-by-step: LIS with 1D DP dp[i] = length of LIS starting at i
Input:
v = [10, 9, 2, 5, 3, 7]
Goal:
For each index i, find the length of the longest increasing subsequence starting at i.
1️⃣ Base idea:
If you start at i, you must include v[i].
Then look forward for any j > i where v[j] > v[i] — you can extend the sequence by going to j.
Among all such js, pick the one that leads to the longest subsequence.
If no such j exists, the LIS starting at i is just length 1 (the element itself).
2️⃣ Initialization:
vector<int> dp(n, -1);
dp[i] = -1 means "not calculated yet."
When computed, dp[i] will store length of LIS starting at i.
3️⃣ Recursive formula:
int lisFrom(int i) {
if (dp[i] != -1) return dp[i];
int maxLength = 1; // minimum LIS length starting at i (the element itself)
for (int j = i + 1; j < n; j++) {
if (v[j] > v[i]) {
maxLength = max(maxLength, 1 + lisFrom(j));
}
}
return dp[i] = maxLength;
}
4️⃣ Walkthrough with example:
Index Value Calculations dp[i] result
5 7 No next elements >7 1
4 3 Next bigger: 7 at index 5 → 1 + dp[5] = 2 2
3 5 Next bigger: 7 at index 5 → 1 + dp[5] = 2 2
2 2 Next bigger: 5 at 3 (dp=2), 3 at 4 (dp=2), 7 at 5 (dp=1) → max is 1+2=3 3
1 9 Next bigger: 10 at 0 is no (index 0 < 1), no next bigger → 1 1
0 10 No bigger next → 1 1
But wait — index 1’s value is 9 and 0’s value is 10, so next bigger after 9? Actually, no next bigger after 9 because 10 is before 9.
5️⃣ Final answer:
The LIS length is max of all dp[i]:
max(dp) = max(1, 1, 3, 2, 2, 1) = 3
The LIS length is 3 (which is [2, 3, 7] or [2, 5, 7]).
6️⃣ Visualization as recursion tree (partial):
lisFrom(0) = 1
lisFrom(1) = 1
lisFrom(2) = 1 + max(lisFrom(3), lisFrom(4), lisFrom(5))
= 1 + max(2, 2, 1)
= 3
lisFrom(3) = 1 + lisFrom(5) = 1 + 1 = 2
lisFrom(4) = 1 + lisFrom(5) = 1 + 1 = 2
lisFrom(5) = 1
7️⃣ Memoization effect:
Once dp[i] is computed, it’s stored. If lisFrom(i) is called again, it instantly returns the cached result — no repeated recomputation!
You said:
how can map.reserve(1 << 20) or something similar can be used to speed up the execution
ChatGPT said:
Excellent question! When using unordered_map in C++, especially in performance-critical or competitive programming settings, calling .reserve() like:
unordered_map<KeyType, ValueType> map;
map.reserve(1 << 20);
can make a noticeable difference in execution time. Here's a breakdown of why it works, when to use it, and how it helps:
🚀 Why map.reserve() speeds up execution
🔄 unordered_map uses hash tables internally
It stores key-value pairs in buckets.
When the number of elements exceeds the current capacity * load factor, it rehashes — meaning it resizes and re-inserts all elements into a new table.
Rehashing is expensive (time & memory) and can happen multiple times if many elements are inserted without reserving space.
💡 reserve(n) tells the map:
"I will insert around n elements — allocate bucket space ahead of time."
“I plan to insert about 1 million entries, so allocate space now to avoid costly rehashing later.”
This avoids costly rehashing during insertions.
🧠 Example: Speeding up memoized DP
In your memoized LIS:
unordered_map<pair<int, int>, int, pair_hash> dp;
If you expect about 1 million states (e.g., N ≈ 10^5, small number of msf values), you can safely reserve:
dp.reserve(1 << 20); // ≈ 1 million slots
This avoids:
Multiple rehashing operations
Random memory allocations
Poor cache locality due to resizing
Result: ~10–50% speedup is not uncommon in practice.
✅ When to use .reserve()
Situation Should use .reserve()?
Inserting many elements (> 1000) ✅
Memoization with large state spaces ✅
Competitive programming ✅✅✅
Small inputs (e.g. < 100 entries) ❌ (won’t help much)
⚠️ Notes and Tips
reserve() is not a hard limit — you can still insert more; it just pre-allocates.
You can also use .max_load_factor() to control when rehashing happens.
Example:
dp.max_load_factor(0.7); // Default is 1.0
dp.reserve(1 << 20);
If you're using unordered_map<pair<int,int>,...> often, and speed is really critical:
Consider switching to gp_hash_table from GNU pbds (even faster).
Or roll your own struct HashPair with bit-mixing for better hash distribution.
✅ In Short:
Calling:
map.reserve(1 << 20);
prevents internal rehashing, keeps performance smooth and predictable, and is a best practice for any non-trivial use of unordered_map.
You said:
best hashing for pair as a key
ChatGPT said:
Great question — using std::pair as a key in an unordered_map requires a custom hash function, because the standard library does not provide a hash for pair by default.
Here’s how to define and use an efficient, reliable, and portable custom hash for pair<int, int>.
✅ Best Hashing for pair<int, int> (C++17+ friendly)
🔧 Recommended Approach: struct pair_hash
struct pair_hash {
size_t operator()(const pair<int, int>& p) const {
return hash<int>()(p.first) ^ (hash<int>()(p.second) << 1);
}
};
This works well and is often used in competitive programming and practical systems.
Example usage:
unordered_map<pair<int, int>, int, pair_hash> dp;
dp.reserve(1 << 20); // Reserve space to avoid rehashing
🧠 Why this hash works:
Uses std::hash<int> for each part.
^ (x << 1) mixes the second value, reducing collisions.
This is based on boost::hash_combine pattern.
🧪 If you want stronger collision resistance (e.g., in adversarial settings):
Use a hash function like SplitMix64, which is very robust:
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(pair<int, int> p) const {
uint64_t hash1 = splitmix64(p.first);
uint64_t hash2 = splitmix64(p.second);
return hash1 ^ (hash2 << 1);
}
};
This is particularly useful in codeforces/online judges to prevent TLEs due to hash collisions in unordered_map.

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