LeetCode #1147 — HARD

Longest Chunked Palindrome Decomposition

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a string text. You should split it to k substrings (subtext₁, subtext₂, ..., subtext_k) such that:

subtext_i is a non-empty string.
The concatenation of all the substrings is equal to text (i.e., subtext₁ + subtext₂ + ... + subtext_k == text).
subtext_i == subtext_{k - i + 1} for all valid values of i (i.e., 1 <= i <= k).

Return the largest possible value of k.

Example 1:

Input: text = "ghiabcdefhelloadamhelloabcdefghi"
Output: 7
Explanation: We can split the string on "(ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi)".

Example 2:

Input: text = "merchant"
Output: 1
Explanation: We can split the string on "(merchant)".

Example 3:

Input: text = "antaprezatepzapreanta"
Output: 11
Explanation: We can split the string on "(a)(nt)(a)(pre)(za)(tep)(za)(pre)(a)(nt)(a)".

Constraints:

1 <= text.length <= 1000
text consists only of lowercase English characters.

Step 01

Brute Force Baseline

Problem summary: You are given a string text. You should split it to k substrings (subtext1, subtext2, ..., subtextk) such that: subtexti is a non-empty string. The concatenation of all the substrings is equal to text (i.e., subtext1 + subtext2 + ... + subtextk == text). subtexti == subtextk - i + 1 for all valid values of i (i.e., 1 <= i <= k). Return the largest possible value of k.

Baseline thinking

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

Pattern signal: Two Pointers · Dynamic Programming · Greedy

Example 1

"ghiabcdefhelloadamhelloabcdefghi"

Example 2

"merchant"

Example 3

"antaprezatepzapreanta"

Core Insight

What unlocks the optimal approach

Using a rolling hash, we can quickly check whether two strings are equal.
Use that as the basis of a dp.

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

Largest constraint values

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 #1147: Longest Chunked Palindrome Decomposition
class Solution {
    public int longestDecomposition(String text) {
        int ans = 0;
        for (int i = 0, j = text.length() - 1; i <= j;) {
            boolean ok = false;
            for (int k = 1; i + k - 1 < j - k + 1; ++k) {
                if (check(text, i, j - k + 1, k)) {
                    ans += 2;
                    i += k;
                    j -= k;
                    ok = true;
                    break;
                }
            }
            if (!ok) {
                ++ans;
                break;
            }
        }
        return ans;
    }

    private boolean check(String s, int i, int j, int k) {
        while (k-- > 0) {
            if (s.charAt(i++) != s.charAt(j++)) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1147: Longest Chunked Palindrome Decomposition
func longestDecomposition(text string) (ans int) {
	for i, j := 0, len(text)-1; i <= j; {
		ok := false
		for k := 1; i+k-1 < j-k+1; k++ {
			if text[i:i+k] == text[j-k+1:j+1] {
				ans += 2
				i += k
				j -= k
				ok = true
				break
			}
		}
		if !ok {
			ans++
			break
		}
	}
	return
}

# Accepted solution for LeetCode #1147: Longest Chunked Palindrome Decomposition
class Solution:
    def longestDecomposition(self, text: str) -> int:
        ans = 0
        i, j = 0, len(text) - 1
        while i <= j:
            k = 1
            ok = False
            while i + k - 1 < j - k + 1:
                if text[i : i + k] == text[j - k + 1 : j + 1]:
                    ans += 2
                    i += k
                    j -= k
                    ok = True
                    break
                k += 1
            if not ok:
                ans += 1
                break
        return ans

// Accepted solution for LeetCode #1147: Longest Chunked Palindrome Decomposition
struct Solution;

impl Solution {
    fn longest_decomposition(text: String) -> i32 {
        Self::greedy(&text)
    }

    fn greedy(s: &str) -> i32 {
        let n = s.len();
        if n == 0 {
            return 0;
        }
        for i in 1..=n / 2 {
            if s[0..i] == s[n - i..] {
                return 2 + Self::greedy(&s[i..n - i]);
            }
        }
        1
    }
}

#[test]
fn test() {
    let text = "ghiabcdefhelloadamhelloabcdefghi".to_string();
    let res = 7;
    assert_eq!(Solution::longest_decomposition(text), res);
    let text = "merchant".to_string();
    let res = 1;
    assert_eq!(Solution::longest_decomposition(text), res);
    let text = "antaprezatepzapreanta".to_string();
    let res = 11;
    assert_eq!(Solution::longest_decomposition(text), res);
    let text = "aaa".to_string();
    let res = 3;
    assert_eq!(Solution::longest_decomposition(text), res);
    let text = "elvtoelvto".to_string();
    let res = 2;
    assert_eq!(Solution::longest_decomposition(text), res);
}

// Accepted solution for LeetCode #1147: Longest Chunked Palindrome Decomposition
function longestDecomposition(text: string): number {
    let ans = 0;
    for (let i = 0, j = text.length - 1; i <= j; ) {
        let ok = false;
        for (let k = 1; i + k - 1 < j - k + 1; ++k) {
            if (text.slice(i, i + k) === text.slice(j - k + 1, j + 1)) {
                ans += 2;
                i += k;
                j -= k;
                ok = true;
                break;
            }
        }
        if (!ok) {
            ++ans;
            break;
        }
    }
    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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.