LeetCode #1278 — HARD

Palindrome Partitioning III

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 s containing lowercase letters and an integer k. You need to :

First, change some characters of s to other lowercase English letters.
Then divide s into k non-empty disjoint substrings such that each substring is a palindrome.

Return the minimal number of characters that you need to change to divide the string.

Example 1:

Input: s = "abc", k = 2
Output: 1
Explanation: You can split the string into "ab" and "c", and change 1 character in "ab" to make it palindrome.

Example 2:

Input: s = "aabbc", k = 3
Output: 0
Explanation: You can split the string into "aa", "bb" and "c", all of them are palindrome.

Example 3:

Input: s = "leetcode", k = 8
Output: 0

Constraints:

1 <= k <= s.length <= 100.
s only contains lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s containing lowercase letters and an integer k. You need to : First, change some characters of s to other lowercase English letters. Then divide s into k non-empty disjoint substrings such that each substring is a palindrome. Return the minimal number of characters that you need to change to divide the string.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

"abc"
2

Example 2

"aabbc"
3

Example 3

"leetcode"
8

Core Insight

What unlocks the optimal approach

For each substring calculate the minimum number of steps to make it palindrome and store it in a table.
Create a dp(pos, cnt) which means the minimum number of characters changed for the suffix of s starting on pos splitting the suffix on cnt chunks.

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 #1278: Palindrome Partitioning III
class Solution {
    public int palindromePartition(String s, int k) {
        int n = s.length();
        int[][] g = new int[n][n];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                g[i][j] = s.charAt(i) != s.charAt(j) ? 1 : 0;
                if (i + 1 < j) {
                    g[i][j] += g[i + 1][j - 1];
                }
            }
        }
        int[][] f = new int[n + 1][k + 1];
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= Math.min(i, k); ++j) {
                if (j == 1) {
                    f[i][j] = g[0][i - 1];
                } else {
                    f[i][j] = 10000;
                    for (int h = j - 1; h < i; ++h) {
                        f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
                    }
                }
            }
        }
        return f[n][k];
    }
}

// Accepted solution for LeetCode #1278: Palindrome Partitioning III
func palindromePartition(s string, k int) int {
	n := len(s)
	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, n)
	}
	for i := n - 1; i >= 0; i-- {
		for j := 1; j < n; j++ {
			if s[i] != s[j] {
				g[i][j] = 1
			}
			if i+1 < j {
				g[i][j] += g[i+1][j-1]
			}
		}
	}
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
	}
	for i := 1; i <= n; i++ {
		for j := 1; j <= min(i, k); j++ {
			if j == 1 {
				f[i][j] = g[0][i-1]
			} else {
				f[i][j] = 100000
				for h := j - 1; h < i; h++ {
					f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
				}
			}
		}
	}
	return f[n][k]
}

# Accepted solution for LeetCode #1278: Palindrome Partitioning III
class Solution:
    def palindromePartition(self, s: str, k: int) -> int:
        n = len(s)
        g = [[0] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                g[i][j] = int(s[i] != s[j])
                if i + 1 < j:
                    g[i][j] += g[i + 1][j - 1]

        f = [[0] * (k + 1) for _ in range(n + 1)]
        for i in range(1, n + 1):
            for j in range(1, min(i, k) + 1):
                if j == 1:
                    f[i][j] = g[0][i - 1]
                else:
                    f[i][j] = inf
                    for h in range(j - 1, i):
                        f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1])
        return f[n][k]

// Accepted solution for LeetCode #1278: Palindrome Partitioning III
impl Solution {
    pub fn palindrome_partition(s: String, k: i32) -> i32 {
        let n = s.len();
        let s: Vec<char> = s.chars().collect();
        let mut g = vec![vec![0; n]; n];

        for i in (0..n).rev() {
            for j in i + 1..n {
                g[i][j] = if s[i] != s[j] { 1 } else { 0 };
                if i + 1 < j {
                    g[i][j] += g[i + 1][j - 1];
                }
            }
        }

        let mut f = vec![vec![0; (k + 1) as usize]; n + 1];
        let inf = i32::MAX;

        for i in 1..=n {
            for j in 1..=i.min(k as usize) {
                if j == 1 {
                    f[i][j] = g[0][i - 1];
                } else {
                    f[i][j] = inf;
                    for h in (j - 1)..i {
                        f[i][j] = f[i][j].min(f[h][j - 1] + g[h][i - 1]);
                    }
                }
            }
        }

        f[n][k as usize]
    }
}

// Accepted solution for LeetCode #1278: Palindrome Partitioning III
function palindromePartition(s: string, k: number): number {
    const n = s.length;
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    for (let i = n - 1; i >= 0; i--) {
        for (let j = i + 1; j < n; j++) {
            g[i][j] = s[i] !== s[j] ? 1 : 0;
            if (i + 1 < j) {
                g[i][j] += g[i + 1][j - 1];
            }
        }
    }
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
    for (let i = 1; i <= n; i++) {
        for (let j = 1; j <= Math.min(i, k); j++) {
            if (j === 1) {
                f[i][j] = g[0][i - 1];
            } else {
                f[i][j] = 1 << 30;
                for (let h = j - 1; h < i; h++) {
                    f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
                }
            }
        }
    }
    return f[n][k];
}

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 × k)

Space

O(n × (n + k)

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.