Palindrome Partitioning II
A hard Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Cut a string into pieces so that every piece reads the same forwards and backwards. Return the smallest number of cuts that achieves this; a string that is already a palindrome needs 0 cuts.
Example 1
- Input
- s = "aab"
- Output
- 1
- Why
- The palindrome partitioning ["aa","b"] needs one cut.
Example 2
- Input
- s = "a"
- Output
- 0
- Why
- A single character is already a palindrome, no cuts needed.
Example 3
- Input
- s = "ab"
- Output
- 1
- Why
- Partition ["a","b"] needs one cut.
Constraints
- 1 <= s.length <= 2000
- s consists of lowercase English letters
How to think about it
Updated 2026-09-29Every cut splits off a palindromic prefix and leaves a remaining suffix to be partitioned. If substring s[j..i] is a palindrome, the cuts needed for prefix s[0..i] is at most 1 + cuts(s[0..j-1]). Precomputing palindrome checks between all index pairs turns each partition check into an O(1) transition.
Approaches, worst first
Recursive cut branching
time O(2^n) · space O(n)
From current index i, try every split point j > i where s[i..j] is a palindrome and recurse on the rest. Explores an exponential tree of segmentations without memoization.
Two-stage dynamic programming
time O(n^2) · space O(n^2)
Precompute an n x n boolean table isPalin[i][j]. Then define dp[i] as the minimum cuts for s[0..i]. For each i, if s[0..i] is a palindrome dp[i] = 0; otherwise scan j from 1 to i and take min(dp[i], dp[j-1] + 1) whenever s[j..i] is a palindrome.
Expand around center with 1D DPWrite this one
time O(n^2) · space O(n)
Maintain 1D array dp of size n where dp[i] is min cuts for s[0..i]. For each center, expand both odd and even palindromes outward (left, right). Whenever a valid palindrome s[left..right] is found, update dp[right] = min(dp[right], (left == 0 ? 0 : dp[left - 1] + 1)).
Where people lose marks · 3
- A string that is already a palindrome requires 0 cuts, not 1. Handling the whole-string case as 0 cuts prevents off-by-one errors across the entire table.
- Allocating an n x n palindrome table can trigger memory pressure for n = 2000 in memory-constrained runtimes; expanding around centers uses linear auxiliary space.
- Seeding dp[i] with 0 instead of maximum possible cuts (i cuts).
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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