Part 2: The Coder's Mind - Building the Solution
From understanding to implementation in Java
The coder sits before an empty IDE, fingers poised over the keyboard. The algorithm is understood. Now comes the translation from thought to code...
The Coder's Opening Thoughts
CODER'S MIND: "Alright, I understand the DP. Now I need to make the computer understand it. Let me start with the signature..."
public int maxPoints(int[][] points) {
// Computer needs to know dimensions first
int m = points.length; // rows - how deep do I go?
int n = points[0].length; // columns - how wide at each level?
INTERNAL MONOLOGUE: "Good. m and n are my boundaries. The computer needs these constants—no magic numbers floating around."
Creating the DP Table
CODER'S MIND: "I need storage for my chronicles. A 2D array—rows represent stages, columns represent positions at each stage."
// dp[r][c] = max points accumulated from row 0 to row r, ending at column c
long[][] dp = new long[m][n];
PAUSE: "Wait, why long? Let me think... points can be up to 10^5, and we have up to 10^5 cells. Worst case sum: 10^10. That exceeds int's 2^31 ≈ 2×10^9. Need long."
INTERNAL DEBATE:
- "Could I use int? No, test cases might fail on large inputs."
- "Memory concern? m×n ≤ 10^5, so at most 10^5 longs. That's fine."
DECISION: long[][] it is. Safety first.
The Base Case
CODER'S MIND: "Before I loop, I need the foundation. Row 0 has no history, no penalties."
// Base case: first row is just the values themselves
for (int c = 0; c < n; c++) {
dp[0][c] = points[0][c];
}
VISUALIZATION IN MIND:
dp[0][0] = points[0][0]
dp[0][1] = points[0][1]
dp[0][2] = points[0][2]
...
INTERNAL CHECK: "Do I need a loop? Yes, there are n columns. Could I skip? No, each position needs initialization. Moving on."
The First Loop: Iterating Rows
CODER'S MIND: "Now the real work. I need to fill each row based on the previous row."
// Process each row starting from row 1
for (int r = 1; r < m; r++) {
THINKING ALOUD:
- "Why start at 1? Because row 0 is already done—it's the base case."
- "Why
r < m? Because there are m rows total (0 to m-1)." - "This outer loop represents TIME STAGES. Each iteration is one step forward in my journey."
MENTAL MODEL:
r=1: Process row 1 using row 0
r=2: Process row 2 using row 1
r=3: Process row 3 using row 2
...
The Second Loop: Iterating Current Columns
CODER'S MIND: "At each row, I need to compute the value for EVERY column. Each column is a possible ending position at this stage."
// For each column in current row
for (int c = 0; c < n; c++) {
INTERNAL MONOLOGUE:
- "Why every column? Because I don't know yet which will be optimal for future rows."
- "I need to compute
dp[r][c]for ALL c, because row r+1 will need to choose from them." - "This is like computing all possible futures, not just one greedy path."
VISUALIZATION:
At r=1:
Compute dp[1][0]
Compute dp[1][1]
Compute dp[1][2]
...
Compute dp[1][n-1]
"I'm filling an entire ROW at once, not just one cell."
The Third Loop: The Heart of DP
CODER'S MIND: "Here's the critical part. For THIS cell (r, c), where could I have come from? Answer: ANY column in the previous row."
// Try coming from every possible previous column
for (int prev_c = 0; prev_c < n; prev_c++) {
Deep Dive Into Thinking
Why This Loop Exists
"Let me verbalize what the computer will do:
- I'm at position (r, c) - this is WHERE I want to end up
- I need to figure out the BEST WAY to get here
- The 'best way' depends on WHERE I was in row r-1
- I could have been at prev_c = 0, or prev_c = 1, or prev_c = 2, etc.
- For EACH possibility, I calculate: 'If I was THERE, and I move HERE, what's my total score?'
- Then I pick the MAXIMUM"
The Mental Execution
CODER'S MIND: "Let me trace through what happens. Say r=2, c=3 (I'm computing dp[2][3])."
prev_c = 0:
Was at (1, 0)
Had dp[1][0] points
Moving to (2, 3)
Gain points[2][3]
Pay penalty abs(0 - 3) = 3
Total: dp[1][0] + points[2][3] - 3
prev_c = 1:
Was at (1, 1)
Had dp[1][1] points
Moving to (2, 3)
Gain points[2][3]
Pay penalty abs(1 - 3) = 2
Total: dp[1][1] + points[2][3] - 2
prev_c = 2:
Was at (1, 2)
Had dp[1][2] points
Moving to (2, 3)
Gain points[2][3]
Pay penalty abs(2 - 3) = 1
Total: dp[1][2] + points[2][3] - 1
...and so on for all prev_c
"This inner loop is EXHAUSTIVE SEARCH over all previous positions. It's brute force, but on a small space (just n columns)."
Why We Need This Inner Loop
CODER'S MIND: "Could I avoid this inner loop? Let me think..."
HYPOTHETICAL: "What if I only checked prev_c = c (staying in same column)?"
- ❌ "No! I'd miss better paths that involve column changes."
HYPOTHETICAL: "What if I only checked the best prev_c from row r-1?"
- ❌ "No! The 'best' prev_c for reaching (r, c) depends on the penalty, which varies by c."
- ❌ "Example: best for reaching (r, 0) might be prev_c=0 (low penalty), but best for reaching (r, 5) might be prev_c=4 or prev_c=5."
CONCLUSION: ✅ "I MUST check all prev_c. There's no shortcut without more sophisticated optimization."
Computing the Score
// Calculate score: previous total + current gain - movement penalty
long score = dp[r - 1][prev_c] + points[r][c] - Math.abs(prev_c - c);
CODER'S MIND: "Let me break down this line..."
The Three Components
COMPONENT 1: dp[r - 1][prev_c]
- "This is the HISTORY. The accumulated chronicle."
- "Not just
points[r-1][prev_c], but the TOTAL from row 0 to row r-1."
COMPONENT 2: points[r][c]
- "This is the PRESENT. What I gain right now."
- "Notice it doesn't depend on prev_c. It's constant for this cell."
COMPONENT 3: Math.abs(prev_c - c)
- "This is the COST. The penalty for movement."
- "This DOES depend on prev_c. Different starting points = different penalties."
MENTAL CHECK: "Type safety? dp is long, points is int, Math.abs returns int. The expression promotes to long. Good."
Updating the Maximum
// Keep track of the best score for this cell
dp[r][c] = Math.max(dp[r][c], score);
}
CODER'S MIND: "After trying each prev_c, I keep the best one."
IMPORTANT REALIZATION: "I'm NOT storing which prev_c was best—just the score. If I needed path reconstruction, I'd need a separate parent array."
DEFAULT VALUE CONCERN: "Wait, dp[r][c] starts at 0. Is that okay?"
ANALYSIS:
- ✓ "All points are non-negative (constraints say ≥ 0)."
- ✓ "Penalties are non-negative."
- ✓ "So worst case score would be 0 if we only had 0-valued cells."
- ✓ "Any actual path will be ≥ 0."
- ✓ "Initial 0 is safe."
Finding the Final Answer
// The answer is the maximum value in the last row
long answer = dp[m - 1][0];
for (int c = 1; c < n; c++) {
answer = Math.max(answer, dp[m - 1][c]);
}
return (int) answer;
CODER'S MIND: "The last row contains all possible ending positions. I need the best one."
WHY NOT MAX IN MIDDLE ROWS?
- "Because I haven't picked ALL rows yet. The journey isn't complete."
TYPE CONVERSION: "Casting long to int. Is this safe?"
ANALYSIS:
- "The answer is at most sum of all points minus minimum penalties."
- "Penalties reduce the total, so answer ≤ sum of all points."
- "But wait... with 10^5 cells at 10^5 each, that's 10^10."
- "That exceeds int! But the problem guarantees the answer fits in int."
- ✅ "Trust the problem constraints. Cast is safe."
The Complete Code
public class Solution {
public int maxPoints(int[][] points) {
// Get dimensions
int m = points.length; // number of rows
int n = points[0].length; // number of columns
// dp[r][c] = max points from row 0 to row r, ending at column c
// Using long to prevent overflow during computation
long[][] dp = new long[m][n];
// Base case: first row has no previous row, no penalties
for (int c = 0; c < n; c++) {
dp[0][c] = points[0][c];
}
// Fill each subsequent row
for (int r = 1; r < m; r++) { // For each row (stage)
for (int c = 0; c < n; c++) { // For each column (current position)
// Try all possible previous columns
for (int prev_c = 0; prev_c < n; prev_c++) { // For each previous position
// Calculate: previous_total + current_gain - movement_penalty
long score = dp[r - 1][prev_c] + points[r][c] - Math.abs(prev_c - c);
dp[r][c] = Math.max(dp[r][c], score);
}
}
}
// Find maximum in last row
long answer = dp[m - 1][0];
for (int c = 1; c < n; c++) {
answer = Math.max(answer, dp[m - 1][c]);
}
return (int) answer;
}
}
The Nested Loop Hierarchy: A Coder's Perspective
Loop 1 (Outer): The Timeline
for (int r = 1; r < m; r++)
- PURPOSE: "March through time. Process stages sequentially."
- MENTAL MODEL: "I'm moving DOWN the matrix, one row at a time."
- INVARIANT: "After this loop iteration, dp[r][] is fully computed."*
Loop 2 (Middle): The Possibilities at Current Stage
for (int c = 0; c < n; c++)
- PURPOSE: "At this stage (row r), compute value for EVERY possible position."
- MENTAL MODEL: "I'm moving ACROSS the row, filling each cell."
- INVARIANT: "After this loop iteration, dp[r][c] is fully computed."
- WHY NECESSARY: "Future rows will need to choose from ALL columns here, not just one."
Loop 3 (Inner): The Search for Optimal Transition
for (int prev_c = 0; prev_c < n; prev_c++)
- PURPOSE: "For THIS specific cell (r,c), try ALL ways to get here from previous row."
- MENTAL MODEL: "I'm looking BACKWARDS at row r-1, considering every column as a potential source."
- INVARIANT: "After this loop iteration, dp[r][c] contains the MAXIMUM over all possible transitions."
"This is where the 'dynamic' in dynamic programming happens. I'm synthesizing the answer for (r,c) from multiple subproblems."
The Coder's Final Thoughts
Complexity Analysis
TIME COMPLEXITY:
Three nested loops:
- Outer: O(m)
- Middle: O(n)
- Inner: O(n)
Total: O(m × n × n) = O(m × n²)
SPACE COMPLEXITY:
dp array: O(m × n)
Could optimize to O(n) by using only two rows (current and previous)
BOTTLENECK: "That inner loop. It runs n times for each of m×n cells. That's the killer when n is large."
OPTIMIZATION TEASER: "There must be a way to avoid trying all n values of prev_c. Maybe precompute something? That's for later..."
The Meta-Process: Coder's Checklist
When implementing DP, the coder thinks:
- ✅ Dimensions: Get m and n first
- ✅ Storage: Create dp array with correct types
- ✅ Base case: Initialize first row (or column, depending on problem)
- ✅ Loop structure: Outer (stages) → Middle (current choices) → Inner (previous choices)
- ✅ Transition: Calculate score from each subproblem
- ✅ Aggregation: Take max (or min, depending on problem)
- ✅ Answer extraction: Find result in final stage
- ✅ Edge cases: Check for 0 dimensions, single row/column
- ✅ Type safety: Ensure no overflow
Final Reflection
CODER'S WISDOM:
"The code mirrors the thought. Three levels of iteration = three levels of decision-making. The computer will execute this millions of times, but each execution follows the same logical path I traced once."
The coder hits compile. The code runs. The tests pass.
"Understanding first, implementation second. That's how it should always be."