“The answer for a range depends on

• The problem involves:
• You see:
• The operation:
• You hear:

• “partition”
• “split”
• “burst”
• “merge”
• “cut”

30 MUST-DO INTERVIEW PROBLEMS

1. #
   Matrix Chain Multiplication (Basic recursion idea) — GFG
2. #
   Minimum Score Triangulation of Polygon — LC 1039
3. #
   Guess Number Higher or Lower II — LC 375
4. #
   Stone Game (basic DP view) — LC 877
5. #
   Optimal Strategy for a Game — GFG
6. #
   Boolean Parenthesization (Intro) — GFG
7. #
   Minimum Cost Tree From Leaf Values — LC 1130
8. #
   Palindrome Removal — LC 1246
9. #
   Stone Game II — LC 1140
10. #
    Zuma Game — LC 488

• Split index iteration
• Two-player interval logic
• Partition-based recursion
• Combining left + right

1. #
   Matrix Chain Multiplication (Tabulation) — GFG
2. #
   Palindrome Partitioning II — LC 132
3. #
   Burst Balloons — LC 312
4. #
   Minimum Cost to Cut a Stick — LC 1547
5. #
   Strange Printer — LC 664
6. #
   Remove Boxes — LC 546
7. #
   Stone Game III — LC 1406
8. #
   Interleaving String (range-style view) — LC 97
9. #
   Longest Palindromic Subsequence — LC 516
10. #
    Count Different Palindromic Subsequences — LC 730

• dp[l][r] formulation
• Reverse iteration filling
• Multi-branch split
• State compression challenges

1. #
   Minimum Cost to Merge Stones — LC 1000
2. #
   Egg Dropping Problem — LC 887
3. #
   Remove Boxes (Optimized 3D DP) — LC 546
4. #
   Strange Printer II — LC 1591
5. #
   Burst Balloons (Optimized Thinking) — LC 312
6. #
   Maximum Score from Performing Multiplication Operations — LC 1770
7. #
   Predict the Winner — LC 486
8. #
   Partition Array for Maximum Sum – LC 1043
9. #
   Make Array Strictly Increasing – LC 1187
10. #
    Optimal Binary Search Tree – GFG

• 2D/3D DP
• Non-trivial transitions
• Split + additional state
• Optimization beyond naive O(n³)

“What does

• Forgetting base cases (l > r, l == r)
• Double-counting partitions
• Writing recursion without memoization
• Not explaining why splitting works

1️⃣ Pattern in One Line

"The answer for a range depends on WHERE you split it — try every possible partition point and pick the best."

2️⃣ Recognize It When...

• Problem involves ranges, intervals, or subarrays
• Trigger words: "partition", "split", "burst", "merge", "cut", "minimum cost over range"
• Operation combines left + right results
• "Try all possible ways" phrasing

3️⃣ Don't Confuse With...

• Regular DP (P13) → 1D DP on elements; Interval DP on ranges [l, r]
• Greedy (P9) → greedy commits to one split; interval DP tries ALL splits
• Divide & Conquer → D&C splits once; interval DP tries every split point

4️⃣ Approach in 3 Steps

1. #
   Define dp[l][r] = best answer for subarray/range [l, r]
2. #
   Try every split point k: dp[l][r] = best over all k in [l, r-1]
3. #
   Fill in order of increasing length (small ranges first)

5️⃣ Invariant to Maintain

"dp[l][r] always contains the optimal answer for the range [l, r] — smaller ranges are always computed before larger ones."

6️⃣ Complexity to Justify

• Time: O(N³) — N² states × N split points
• Space: O(N²) — 2D DP table
• Why: Every subrange tried for every split point

7️⃣ Edge Cases to Always Check

• Base case: dp[i][i] = 0 or single element value
• l > r — invalid range, return 0 or infinity
• Length 2 ranges — simplest non-trivial case
• Modular arithmetic if answer is large
• Fill order: length 1 → length 2 → ... → length N

8️⃣ One Line to Say in Interview

"This is interval DP — dp[l][r] represents the optimal answer for range [l,r], computed by trying every split point k in O(N³) time."