Hi Codeforces! ↵
This is my first blog here, so please excuse any rookie formatting or style issues.↵
I’ve been learning Dynamic Programming recently, and I’m currently trying to deepen my understanding of top-down → bottom-up conversion — not just for this problem, but conceptually. I am assuming that my top down code is conceptually sound and correctly implemented, but **I am getting WA for the iterative / bottom-up version** ↵
↵
**Problem**↵
↵
The problem is [Maximum Tip Calculator (GFG)](https://www.geeksforgeeks.org/problems/maximum-tip-calculator2631/1)↵
↵
You’re given n orders and two delivery boys — A and B.↵
Each order i gives:↵
↵
arr[i] → tip if A takes it↵
↵
brr[i] → tip if B takes it↵
↵
A can take at most x orders, B can take at most y.↵
We want to maximize the total tip.↵
↵
I already solved it using the greedy approach (sorting by |arr[i] — brr[i]|),↵
but I wanted to approach it using Dynamic Programming.↵
↵
My Memoized (Top-Down) Solution:↵
--------------------------------↵
~~~~~↵
long long getMaxTip(int idx, int aCount, int x, int y, vector<int> &arr, vector<int> &brr, vector<vector<long long> > &memo){↵
// No item left to process, means no profit↵
if(idx >= arr.size()){↵
return 0;↵
}↵
↵
if(memo[idx][aCount] != -1){↵
return memo[idx][aCount];↵
}↵
↵
long long aProfit = 0, bProfit = 0;↵
↵
//A takes the order↵
if(aCount < x){↵
aProfit = 1LL * arr[idx] + getMaxTip(idx + 1, aCount + 1, x, y, arr, brr, memo);↵
}↵
↵
//B takes the order↵
if(idx - aCount < y){↵
bProfit = 1LL * brr[idx] + getMaxTip(idx + 1, aCount, x, y, arr, brr, memo);↵
}↵
return memo[idx][aCount] = max(aProfit, bProfit);↵
}↵
↵
//State meaning:↵
↵
//memo[idx][aCount] = max total profit achievable after processing 'idx' total orders and A has already taken aCount orders. (Though it stores the answer for the lower half of the tree (idx -> n - 1))↵
~~~~~↵
↵
↵
↵
Bottom-Up Conversion↵
--------------------↵
↵
Now, I tried converting this to an iterative version , O(N^x) space and O(N*x) time— keeping the exact same state definition and recurrence structure.↵
↵
Reasoning for the code:↵
↵
In memoized recursion, subproblems come from deeper calls (idx onwards).↵
↵
So in bottom-up, we should fill from idx = n-1 → 0.↵
↵
Each state {idx, aCount} depends on {idx + 1, aCount} and {idx + 1, aCount + 1}.↵
↵
~~~~~↵
long long getMaxTipDP2D(int x, int y, vector<int> &arr, vector<int> &brr){↵
int n = arr.size();↵
// state {idx, aCount} = max profit made if we have processed (n - idx - 1) items and A has already taken aCount orders;↵
vector<vector<long long> > dp(n + 1, vector<long long>(x + 1, -1));↵
↵
//base case(from memo code, when idx == n, the profit = 0)↵
for(int j = 0; j <= x; j++){↵
dp[n][j] = 0;↵
}↵
↵
for(int idx = n - 1; idx >= 0; idx--){↵
int totalItemsProcessed = n - idx - 1;↵
for(int aCount = 0; aCount <= min(x, totalItemsProcessed); aCount++){↵
//current state to fill => {idx, aCount}↵
↵
long long aProfit = 0, bProfit = 0;↵
// A can take order iff range[0, min(totalItems,x))↵
// B can take order iff range[0, y)↵
//Boundary case check↵
if(aCount < x){↵
aProfit = 1LL * arr[idx] + dp[idx + 1][aCount + 1];↵
}↵
↵
↵
//Total Items proccessed till now - orders A have taken already↵
if(totalItemsProcessed - aCount < y){↵
bProfit = 1LL * brr[idx] + dp[idx + 1][aCount]; ↵
}↵
↵
//Answer of current state↵
dp[idx][aCount] = max(aProfit, bProfit);↵
}↵
}↵
long long res = 0;↵
for(int j = 0; j <= x; j++){↵
res = max(res, dp[0][j]);↵
}↵
return res;↵
}↵
↵
//State meaning (bottom-up):↵
↵
//dp[idx][aCount] = max profit made from processing (n - idx - 1) items if A has already taken aCount orders.↵
↵
↵
~~~~~↵
↵
↵
↵
↵
❓ My Question↵
-----------↵
↵
Now, here’s where I’m confused conceptually.↵
↵
Even if this version might not pass large inputs, I’m curious conceptually:↵
↵
If I directly mirror the recurrence and keep the same DP state meaning (dp[idx][aCount] = answer of subproblem starting at idx),↵
Is there something fundamentally wrong with this kind of conversion?↵
Or should the state meaning itself change when moving to an iterative representation?↵
↵
#### Are my top-down and bottom-up definitions actually identical?↵
↵
In the memoized version, the subproblem is defined as:↵
↵
**“max profit after processing idx items and A has taken aCount orders”**↵
but it stores the answer for the later half (from idx → n-1).↵
↵
↵
In the iterative version, the subproblem is defined as:↵
↵
**“max profit after processing (n — idx — 1) items and A has taken aCount orders,”**↵
and it also stores the answer for the later half.↵
↵
So, my definitions are slightly different in wording — one counts how many items have been processed so far, the other counts how many remain —↵
yet they both seem to represent the same portion of the problem space / recursive tree logically.↵
↵
In other words, the definition is different, but the meaning feels the same.↵
↵
**Am I right in thinking that both are just two perspectives of the same state?**↵
Or does this verbal mismatch actually affect the correctness or iteration order of the DP?↵
↵
#### What I’m Hoping to Learn↵
I’d really appreciate if someone could explain:↵
↵
Whether my iterative logic is conceptually consistent with the recursive recurrence, and↵
↵
If not, what conceptual correction (in meaning, iteration order, or indexing) is needed to make it work cleanly when going bottom-up.↵
↵
**I’m more interested in the transition reasoning & sub problems (definition & actual meaning) than just the final optimized code** (though that’s welcome too! :)).↵
↵
↵
↵
↵
This is my first blog here, so please excuse any rookie formatting or style issues.↵
I’ve been learning Dynamic Programming recently, and I’m currently trying to deepen my understanding of top-down → bottom-up conversion — not just for this problem, but conceptually. I am assuming that my top down code is conceptually sound and correctly implemented, but **I am getting WA for the iterative / bottom-up version** ↵
↵
**Problem**↵
↵
The problem is [Maximum Tip Calculator (GFG)](https://www.geeksforgeeks.org/problems/maximum-tip-calculator2631/1)↵
↵
You’re given n orders and two delivery boys — A and B.↵
Each order i gives:↵
↵
arr[i] → tip if A takes it↵
↵
brr[i] → tip if B takes it↵
↵
A can take at most x orders, B can take at most y.↵
We want to maximize the total tip.↵
↵
I already solved it using the greedy approach (sorting by |arr[i] — brr[i]|),↵
but I wanted to approach it using Dynamic Programming.↵
↵
My Memoized (Top-Down) Solution:↵
--------------------------------↵
~~~~~↵
long long getMaxTip(int idx, int aCount, int x, int y, vector<int> &arr, vector<int> &brr, vector<vector<long long> > &memo){↵
// No item left to process, means no profit↵
if(idx >= arr.size()){↵
return 0;↵
}↵
↵
if(memo[idx][aCount] != -1){↵
return memo[idx][aCount];↵
}↵
↵
long long aProfit = 0, bProfit = 0;↵
↵
//A takes the order↵
if(aCount < x){↵
aProfit = 1LL * arr[idx] + getMaxTip(idx + 1, aCount + 1, x, y, arr, brr, memo);↵
}↵
↵
//B takes the order↵
if(idx - aCount < y){↵
bProfit = 1LL * brr[idx] + getMaxTip(idx + 1, aCount, x, y, arr, brr, memo);↵
}↵
return memo[idx][aCount] = max(aProfit, bProfit);↵
}↵
↵
//State meaning:↵
↵
//memo[idx][aCount] = max total profit achievable after processing 'idx' total orders and A has already taken aCount orders. (Though it stores the answer for the lower half of the tree (idx -> n - 1))↵
~~~~~↵
↵
↵
↵
Bottom-Up Conversion↵
--------------------↵
↵
Now, I tried converting this to an iterative version , O(N^x) space and O(N*x) time— keeping the exact same state definition and recurrence structure.↵
↵
Reasoning for the code:↵
↵
In memoized recursion, subproblems come from deeper calls (idx onwards).↵
↵
So in bottom-up, we should fill from idx = n-1 → 0.↵
↵
Each state {idx, aCount} depends on {idx + 1, aCount} and {idx + 1, aCount + 1}.↵
↵
~~~~~↵
long long getMaxTipDP2D(int x, int y, vector<int> &arr, vector<int> &brr){↵
int n = arr.size();↵
// state {idx, aCount} = max profit made if we have processed (n - idx - 1) items and A has already taken aCount orders;↵
vector<vector<long long> > dp(n + 1, vector<long long>(x + 1, -1));↵
↵
//base case(from memo code, when idx == n, the profit = 0)↵
for(int j = 0; j <= x; j++){↵
dp[n][j] = 0;↵
}↵
↵
for(int idx = n - 1; idx >= 0; idx--){↵
int totalItemsProcessed = n - idx - 1;↵
for(int aCount = 0; aCount <= min(x, totalItemsProcessed); aCount++){↵
//current state to fill => {idx, aCount}↵
↵
long long aProfit = 0, bProfit = 0;↵
// A can take order iff range[0, min(totalItems,x))↵
// B can take order iff range[0, y)↵
//Boundary case check↵
if(aCount < x){↵
aProfit = 1LL * arr[idx] + dp[idx + 1][aCount + 1];↵
}↵
↵
↵
//Total Items proccessed till now - orders A have taken already↵
if(totalItemsProcessed - aCount < y){↵
bProfit = 1LL * brr[idx] + dp[idx + 1][aCount]; ↵
}↵
↵
//Answer of current state↵
dp[idx][aCount] = max(aProfit, bProfit);↵
}↵
}↵
long long res = 0;↵
for(int j = 0; j <= x; j++){↵
res = max(res, dp[0][j]);↵
}↵
return res;↵
}↵
↵
//State meaning (bottom-up):↵
↵
//dp[idx][aCount] = max profit made from processing (n - idx - 1) items if A has already taken aCount orders.↵
↵
↵
~~~~~↵
↵
↵
↵
↵
❓ My Question↵
-----------↵
↵
Now, here’s where I’m confused conceptually.↵
↵
Even if this version might not pass large inputs, I’m curious conceptually:↵
↵
If I directly mirror the recurrence and keep the same DP state meaning (dp[idx][aCount] = answer of subproblem starting at idx),↵
Is there something fundamentally wrong with this kind of conversion?↵
Or should the state meaning itself change when moving to an iterative representation?↵
↵
#### Are my top-down and bottom-up definitions actually identical?↵
↵
In the memoized version, the subproblem is defined as:↵
↵
**“max profit after processing idx items and A has taken aCount orders”**↵
but it stores the answer for the later half (from idx → n-1).↵
↵
↵
In the iterative version, the subproblem is defined as:↵
↵
**“max profit after processing (n — idx — 1) items and A has taken aCount orders,”**↵
and it also stores the answer for the later half.↵
↵
So, my definitions are slightly different in wording — one counts how many items have been processed so far, the other counts how many remain —↵
yet they both seem to represent the same portion of the problem space / recursive tree logically.↵
↵
In other words, the definition is different, but the meaning feels the same.↵
↵
**Am I right in thinking that both are just two perspectives of the same state?**↵
Or does this verbal mismatch actually affect the correctness or iteration order of the DP?↵
↵
#### What I’m Hoping to Learn↵
I’d really appreciate if someone could explain:↵
↵
Whether my iterative logic is conceptually consistent with the recursive recurrence, and↵
↵
If not, what conceptual correction (in meaning, iteration order, or indexing) is needed to make it work cleanly when going bottom-up.↵
↵
**I’m more interested in the transition reasoning & sub problems (definition & actual meaning) than just the final optimized code** (though that’s welcome too! :)).↵
↵
↵
↵
↵
