Longest Chunked Palindrome Decomposition - Practice Coding Problems

Longest Chunked Palindrome Decomposition - Problem

Two Pointers String Dynamic Programming Greedy Rolling Hash Hash Function Hard

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.

Input & Output

Example 1 — Palindromic Chunks

$ Input: text = "ghiabcdefhelloadefghiabcdef"

› Output: 7

💡 Note: Split into: (ghi)(abcdef)(hello)(a)(def)(ghiabcdef) - wait, this doesn't work. Let me recalculate... Actually: (ghi)(abc)(def)(hello)(a)(def)(abc)(ghi) = 8 chunks, but let's verify the greedy approach gives the optimal split.

Example 2 — Single Character

$ Input: text = "aa"

› Output: 2

💡 Note: Split into (a)(a) - two matching single character chunks

Example 3 — No Splits

$ Input: text = "abc"

› Output: 1

💡 Note: No way to split into palindromic chunks, so the whole string is one chunk

Constraints

1 ≤ text.length ≤ 1000
text consists of lowercase English letters only

Visualization

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Understanding the Visualization

Input String

Given string to split into palindromic chunks

Find Matches

Identify matching prefix-suffix pairs greedily

Count Chunks

Return maximum number of valid chunks

Key Takeaway

🎯 Key Insight: Greedily matching the shortest prefix-suffix pairs maximizes the total number of chunks

Asked in

G Google 15 f Facebook 8

The key insight is to use a greedy approach: start from both ends and find the shortest matching prefix-suffix pairs. This maximizes the number of chunks since shorter chunks leave more room for additional splits. Best approach uses greedy two pointers with Time: O(n²), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Two Pointers	O(n²)	O(1)	Greedily match shortest chunks from both ends using two pointers
Memoized Recursion	O(n³)	O(n²)	Add memoization to avoid recomputing overlapping subproblems
Brute Force Recursion	O(2ⁿ)	O(n)	Try all possible splits and recursively find maximum chunks

Greedy Two Pointers — Algorithm Steps

Start pointers at both ends
Expand until prefix equals suffix
Count 2 chunks and move pointers inward
Handle remaining middle part

Visualization

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Step-by-Step Walkthrough

Start Ends

Place pointers at both ends of string

Find Match

Expand until prefix equals suffix

Move Inward

Count 2 chunks, process remaining middle

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solve(char* s, int start, int end) {
    if (start >= end) return 0;
    
    // Try to find matching prefix and suffix
    for (int length = 1; length <= (end - start) / 2; length++) {
        // Check if prefix matches suffix
        int match = 1;
        for (int i = 0; i < length; i++) {
            if (s[start + i] != s[end - length + i]) {
                match = 0;
                break;
            }
        }
        
        if (match) {
            // Found match, recurse on middle part
            return 2 + solve(s, start + length, end - length);
        }
    }
    
    // No matching pair found, whole part is one chunk
    return 1;
}

int solution(char* text) {
    return solve(text, 0, strlen(text));
}

int main() {
    char text[1001];
    fgets(text, sizeof(text), stdin);
    text[strcspn(text, "\n")] = 0;
    
    int result = solution(text);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Outer loop runs n times, inner string comparison takes O(n) time

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses a few variables for pointers

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

450 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Two Pointers — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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