LeetCode #131 — MEDIUM

Palindrome Partitioning

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

Problem Statement

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

Example 1:

Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]

Example 2:

Input: s = "a"
Output: [["a"]]

Constraints:

1 <= s.length <= 16
s contains only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Backtracking

Example 1

"aab"

Example 2

"a"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #131: Palindrome Partitioning
class Solution {
    private int n;
    private String s;
    private boolean[][] f;
    private List<String> t = new ArrayList<>();
    private List<List<String>> ans = new ArrayList<>();

    public List<List<String>> partition(String s) {
        n = s.length();
        f = new boolean[n][n];
        for (int i = 0; i < n; ++i) {
            Arrays.fill(f[i], true);
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = s.charAt(i) == s.charAt(j) && f[i + 1][j - 1];
            }
        }
        this.s = s;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i == s.length()) {
            ans.add(new ArrayList<>(t));
            return;
        }
        for (int j = i; j < n; ++j) {
            if (f[i][j]) {
                t.add(s.substring(i, j + 1));
                dfs(j + 1);
                t.remove(t.size() - 1);
            }
        }
    }
}

// Accepted solution for LeetCode #131: Palindrome Partitioning
func partition(s string) (ans [][]string) {
	n := len(s)
	f := make([][]bool, n)
	for i := range f {
		f[i] = make([]bool, n)
		for j := range f[i] {
			f[i][j] = true
		}
	}
	for i := n - 1; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			f[i][j] = s[i] == s[j] && f[i+1][j-1]
		}
	}
	t := []string{}
	var dfs func(int)
	dfs = func(i int) {
		if i == n {
			ans = append(ans, append([]string(nil), t...))
			return
		}
		for j := i; j < n; j++ {
			if f[i][j] {
				t = append(t, s[i:j+1])
				dfs(j + 1)
				t = t[:len(t)-1]
			}
		}
	}
	dfs(0)
	return
}

# Accepted solution for LeetCode #131: Palindrome Partitioning
class Solution:
    def partition(self, s: str) -> List[List[str]]:
        def dfs(i: int):
            if i == n:
                ans.append(t[:])
                return
            for j in range(i, n):
                if f[i][j]:
                    t.append(s[i : j + 1])
                    dfs(j + 1)
                    t.pop()

        n = len(s)
        f = [[True] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                f[i][j] = s[i] == s[j] and f[i + 1][j - 1]
        ans = []
        t = []
        dfs(0)
        return ans

// Accepted solution for LeetCode #131: Palindrome Partitioning
impl Solution {
    pub fn partition(s: String) -> Vec<Vec<String>> {
        let mut res: Vec<Vec<String>> = Vec::new();
        let mut part: Vec<String> = Vec::new();

        Self::dfs(&s, &mut res, &mut part, 0);
        res
    }

    pub fn dfs(s: &String, res: &mut Vec<Vec<String>>, part: &mut Vec<String>, i: usize) {
        if i >= s.len() {
            res.push(part.clone());
            return;
        }

        for j in i..s.len() {
            if Self::is_palindrome(s, i, j) {
                let substr: String = s[i..j + 1].to_string();
                part.push(substr);
                Self::dfs(s, res, part, j + 1);
                part.pop();
            }
        }
    }

    pub fn is_palindrome(s: &String, mut left: usize, mut right: usize) -> bool {
        let s = s.as_bytes();
        while left < right {
            if s[left] != s[right] {
                return false;
            }
            left += 1;
            right -= 1;
        }
        return true;
    }
}

// Accepted solution for LeetCode #131: Palindrome Partitioning
function partition(s: string): string[][] {
    const n = s.length;
    const f: boolean[][] = Array.from({ length: n }, () => Array(n).fill(true));
    for (let i = n - 1; i >= 0; --i) {
        for (let j = i + 1; j < n; ++j) {
            f[i][j] = s[i] === s[j] && f[i + 1][j - 1];
        }
    }
    const ans: string[][] = [];
    const t: string[] = [];
    const dfs = (i: number) => {
        if (i === n) {
            ans.push(t.slice());
            return;
        }
        for (let j = i; j < n; ++j) {
            if (f[i][j]) {
                t.push(s.slice(i, j + 1));
                dfs(j + 1);
                t.pop();
            }
        }
    };
    dfs(0);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × 2^n)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.