LeetCode #2911 — HARD

Minimum Changes to Make K Semi-palindromes

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

The Problem

Problem Statement

Given a string s and an integer k, partition s into k substrings such that the letter changes needed to make each substring a semi-palindrome are minimized.

Return the minimum number of letter changes required.

A semi-palindrome is a special type of string that can be divided into palindromes based on a repeating pattern. To check if a string is a semi-palindrome:

Choose a positive divisor d of the string's length. d can range from 1 up to, but not including, the string's length. For a string of length 1, it does not have a valid divisor as per this definition, since the only divisor is its length, which is not allowed.
For a given divisor d, divide the string into groups where each group contains characters from the string that follow a repeating pattern of length d. Specifically, the first group consists of characters at positions 1, 1 + d, 1 + 2d, and so on; the second group includes characters at positions 2, 2 + d, 2 + 2d, etc.
The string is considered a semi-palindrome if each of these groups forms a palindrome.

Consider the string "abcabc":

The length of "abcabc" is 6. Valid divisors are 1, 2, and 3.
For d = 1: The entire string "abcabc" forms one group. Not a palindrome.
For d = 2:
- Group 1 (positions 1, 3, 5): "acb"
- Group 2 (positions 2, 4, 6): "bac"
- Neither group forms a palindrome.
For d = 3:
- Group 1 (positions 1, 4): "aa"
- Group 2 (positions 2, 5): "bb"
- Group 3 (positions 3, 6): "cc"
- All groups form palindromes. Therefore, "abcabc" is a semi-palindrome.

Example 1:

Input: s = "abcac", k = 2

Output: 1

Explanation: Divide s into "ab" and "cac". "cac" is already semi-palindrome. Change "ab" to "aa", it becomes semi-palindrome with d = 1.

Example 2:

Input: s = "abcdef", k = 2

Output: 2

Explanation: Divide s into substrings "abc" and "def". Each needs one change to become semi-palindrome.

Example 3:

Input: s = "aabbaa", k = 3

Output: 0

Explanation: Divide s into substrings "aa", "bb" and "aa". All are already semi-palindromes.

Constraints:

2 <= s.length <= 200
1 <= k <= s.length / 2
s contains only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s and an integer k, partition s into k substrings such that the letter changes needed to make each substring a semi-palindrome are minimized. Return the minimum number of letter changes required. A semi-palindrome is a special type of string that can be divided into palindromes based on a repeating pattern. To check if a string is a semi-palindrome: Choose a positive divisor d of the string's length. d can range from 1 up to, but not including, the string's length. For a string of length 1, it does not have a valid divisor as per this definition, since the only divisor is its length, which is not allowed. For a given divisor d, divide the string into groups where each group contains characters from the string that follow a repeating pattern of length d. Specifically, the first group consists of characters at positions 1, 1 + d, 1 + 2d, and so on; the second group

Baseline thinking

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

Pattern signal: Two Pointers · Dynamic Programming

Example 1

"abcac"
2

Core Insight

What unlocks the optimal approach

Define <code>dp[i][j]</code> as the minimum count of letter changes needed to split the suffix of string <code>s</code> starting from <code>s[i]</code> into <code>j</code> valid parts.
We have <code>dp[i][j] = min(dp[x + 1][j - 1] + v[i][x])</code>. Here <code>v[i][x]</code> is the minimum number of letter changes to change substring <code>s[i..x]</code> into semi-palindrome.
<code>v[i][j]</code> can be calculated separately by <b>brute-force</b>. We can create a table of <code>v[i][j]</code> independently to improve the complexity. Also note that semi-palindrome’s length is at least <code>2</code>.

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 #2911: Minimum Changes to Make K Semi-palindromes
class Solution {
    public int minimumChanges(String s, int k) {
        int n = s.length();
        int[][] g = new int[n + 1][n + 1];
        int[][] f = new int[n + 1][k + 1];
        final int inf = 1 << 30;
        for (int i = 0; i <= n; ++i) {
            Arrays.fill(g[i], inf);
            Arrays.fill(f[i], inf);
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                int m = j - i + 1;
                for (int d = 1; d < m; ++d) {
                    if (m % d == 0) {
                        int cnt = 0;
                        for (int l = 0; l < m; ++l) {
                            int r = (m / d - 1 - l / d) * d + l % d;
                            if (l >= r) {
                                break;
                            }
                            if (s.charAt(i - 1 + l) != s.charAt(i - 1 + r)) {
                                ++cnt;
                            }
                        }
                        g[i][j] = Math.min(g[i][j], cnt);
                    }
                }
            }
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                for (int h = 0; h < i - 1; ++h) {
                    f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h + 1][i]);
                }
            }
        }
        return f[n][k];
    }
}

// Accepted solution for LeetCode #2911: Minimum Changes to Make K Semi-palindromes
func minimumChanges(s string, k int) int {
	n := len(s)
	g := make([][]int, n+1)
	f := make([][]int, n+1)
	const inf int = 1 << 30
	for i := range g {
		g[i] = make([]int, n+1)
		f[i] = make([]int, k+1)
		for j := range g[i] {
			g[i][j] = inf
		}
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	f[0][0] = 0
	for i := 1; i <= n; i++ {
		for j := i; j <= n; j++ {
			m := j - i + 1
			for d := 1; d < m; d++ {
				if m%d == 0 {
					cnt := 0
					for l := 0; l < m; l++ {
						r := (m/d-1-l/d)*d + l%d
						if l >= r {
							break
						}
						if s[i-1+l] != s[i-1+r] {
							cnt++
						}
					}
					g[i][j] = min(g[i][j], cnt)
				}
			}
		}
	}
	for i := 1; i <= n; i++ {
		for j := 1; j <= k; j++ {
			for h := 0; h < i-1; h++ {
				f[i][j] = min(f[i][j], f[h][j-1]+g[h+1][i])
			}
		}
	}
	return f[n][k]
}

# Accepted solution for LeetCode #2911: Minimum Changes to Make K Semi-palindromes
class Solution:
    def minimumChanges(self, s: str, k: int) -> int:
        n = len(s)
        g = [[inf] * (n + 1) for _ in range(n + 1)]
        for i in range(1, n + 1):
            for j in range(i, n + 1):
                m = j - i + 1
                for d in range(1, m):
                    if m % d == 0:
                        cnt = 0
                        for l in range(m):
                            r = (m // d - 1 - l // d) * d + l % d
                            if l >= r:
                                break
                            if s[i - 1 + l] != s[i - 1 + r]:
                                cnt += 1
                        g[i][j] = min(g[i][j], cnt)

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

// Accepted solution for LeetCode #2911: Minimum Changes to Make K Semi-palindromes
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2911: Minimum Changes to Make K Semi-palindromes
// class Solution {
//     public int minimumChanges(String s, int k) {
//         int n = s.length();
//         int[][] g = new int[n + 1][n + 1];
//         int[][] f = new int[n + 1][k + 1];
//         final int inf = 1 << 30;
//         for (int i = 0; i <= n; ++i) {
//             Arrays.fill(g[i], inf);
//             Arrays.fill(f[i], inf);
//         }
//         for (int i = 1; i <= n; ++i) {
//             for (int j = i; j <= n; ++j) {
//                 int m = j - i + 1;
//                 for (int d = 1; d < m; ++d) {
//                     if (m % d == 0) {
//                         int cnt = 0;
//                         for (int l = 0; l < m; ++l) {
//                             int r = (m / d - 1 - l / d) * d + l % d;
//                             if (l >= r) {
//                                 break;
//                             }
//                             if (s.charAt(i - 1 + l) != s.charAt(i - 1 + r)) {
//                                 ++cnt;
//                             }
//                         }
//                         g[i][j] = Math.min(g[i][j], cnt);
//                     }
//                 }
//             }
//         }
//         f[0][0] = 0;
//         for (int i = 1; i <= n; ++i) {
//             for (int j = 1; j <= k; ++j) {
//                 for (int h = 0; h < i - 1; ++h) {
//                     f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h + 1][i]);
//                 }
//             }
//         }
//         return f[n][k];
//     }
// }

// Accepted solution for LeetCode #2911: Minimum Changes to Make K Semi-palindromes
function minimumChanges(s: string, k: number): number {
    const n = s.length;
    const g = Array.from({ length: n + 1 }, () => Array.from({ length: n + 1 }, () => Infinity));
    const f = Array.from({ length: n + 1 }, () => Array.from({ length: k + 1 }, () => Infinity));
    f[0][0] = 0;
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= n; ++j) {
            const m = j - i + 1;
            for (let d = 1; d < m; ++d) {
                if (m % d === 0) {
                    let cnt = 0;
                    for (let l = 0; l < m; ++l) {
                        const r = (((m / d) | 0) - 1 - ((l / d) | 0)) * d + (l % d);
                        if (l >= r) {
                            break;
                        }
                        if (s[i - 1 + l] !== s[i - 1 + r]) {
                            ++cnt;
                        }
                    }
                    g[i][j] = Math.min(g[i][j], cnt);
                }
            }
        }
    }
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= k; ++j) {
            for (let h = 0; h < i - 1; ++h) {
                f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h + 1][i]);
            }
        }
    }
    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)

Space

O(1)

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.