Palindrome Partitioning III

Palindrome Partitioning III - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: s = "abc", k = 2

› Output: 1

💡 Note: Change one character to create palindromes: "a" | "bb" (change 'c' to 'b') or "aa" | "c" (change 'b' to 'a'). Minimum cost is 1.

Example 2 — Already Palindromic

$ Input: s = "aabaa", k = 3

› Output: 0

💡 Note: Can partition as "a" | "aba" | "a" with no changes needed since each part is already palindromic.

Example 3 — Single Part

$ Input: s = "leetcode", k = 1

› Output: 4

💡 Note: Need to make entire string palindromic. Comparing from both ends: l↔e(1 change), e↔d(1 change), e↔o(1 change), t↔c(1 change). Total: 4 changes minimum.

Constraints

1 ≤ k ≤ s.length ≤ 100
s consists of lowercase English letters only

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to precompute palindrome conversion costs for all substrings, then use dynamic programming to find optimal partitioning. Best approach is Bottom-Up DP with Time: O(n³), Space: O(n²).

Common Approaches

✓ Bottom-Up Dynamic Programming

⏱️ Time: O(n³ + n²k) Space: O(n² + nk)

Create a 2D DP table where dp[i][j] represents minimum cost to partition first i characters into j parts. Fill the table bottom-up.

Brute Force Recursion

⏱️ Time: O(k^n) Space: O(n)

For each position, try making a cut and recursively solve the remaining string with k-1 parts. Calculate the cost to make each substring palindromic.

Memoization (Top-Down DP)

⏱️ Time: O(n³ + n²k) Space: O(n² + nk)

Same as brute force but store results of (position, parts_remaining) states in a memo table to avoid recalculating the same subproblems.

Bottom-Up Dynamic Programming — Algorithm Steps

Precompute palindrome costs for all substrings
Initialize DP table with base cases
Fill table using optimal substructure
Return dp[n][k]

Visualization

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Step-by-Step Walkthrough

Initialize Table

Set up DP table with base cases

Fill States

Use optimal substructure to fill table

Extract Answer

Return dp[n][k] as final result

Code -

solution.c — C

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

#define MAX_N 101

int min(int a, int b) {
    return a < b ? a : b;
}

int solution(char* s, int k) {
    int n = strlen(s);
    
    // Precompute palindrome costs
    static int cost[MAX_N][MAX_N];
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            cost[i][j] = 0;
            int left = i, right = j;
            while (left < right) {
                if (s[left] != s[right]) {
                    cost[i][j]++;
                }
                left++;
                right--;
            }
        }
    }
    
    // DP table
    static int dp[MAX_N][MAX_N];
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= k; j++) {
            dp[i][j] = INT_MAX;
        }
    }
    dp[0][0] = 0;
    
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= min(i, k); j++) {
            // Try all possible last segments
            for (int prev = j - 1; prev < i; prev++) {
                if (dp[prev][j - 1] != INT_MAX) {
                    dp[i][j] = min(dp[i][j], dp[prev][j - 1] + cost[prev][i - 1]);
                }
            }
        }
    }
    
    return dp[n][k];
}

int main() {
    char s[MAX_N];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    printf("%d\n", solution(s, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³ + n²k)

O(n³) to precompute palindrome costs, O(n²k) to fill DP table with O(n) transitions

⚠ Quadratic Growth

Space Complexity

O(n² + nk)

O(n²) for palindrome cost matrix, O(nk) for DP table

⚠ Quadratic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Bottom-Up Dynamic Programming — Algorithm Steps

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Code -

Time & Space Complexity

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