Why Palindrome Partitioning DP Minimum Cuts in DSA Typescript?

The Big Idea

What if you could instantly find the perfect way to split a word into palindromes without guessing?

The Scenario

Imagine you have a long word and you want to split it into smaller parts so that each part reads the same forwards and backwards (a palindrome). Doing this by hand means checking every possible way to cut the word, which can take forever if the word is long.

The Problem

Manually checking all ways to split the word is slow and confusing. You might miss some cuts or repeat checks, making it easy to make mistakes and waste time.

The Solution

Using a smart step-by-step method called dynamic programming, we can remember which parts are palindromes and find the fewest cuts needed quickly without repeating work.

Before vs After

✗ Before

function minCuts(word: string): number {
  // Check all splits manually
  // Very slow for long words
  return 0; // placeholder
}

✓ After

function minCuts(word: string): number {
  const n = word.length;
  const dp = new Array(n).fill(0);
  const palindrome = Array.from({ length: n }, () => new Array(n).fill(false));

  for (let end = 0; end < n; end++) {
    let minCut = end; // max cuts
    for (let start = 0; start <= end; start++) {
      if (word[start] === word[end] && (end - start < 2 || palindrome[start + 1][end - 1])) {
        palindrome[start][end] = true;
        minCut = start === 0 ? 0 : Math.min(minCut, dp[start - 1] + 1);
      }
    }
    dp[end] = minCut;
  }

  return dp[n - 1];
}

What It Enables

This method lets you quickly find the smallest number of cuts to split any word into palindrome parts, even if the word is very long.

Real Life Example

Think of checking if a sentence can be split into palindromic phrases for a puzzle game or text analysis tool that needs to find symmetrical patterns fast.

Key Takeaways

Manual checking is slow and error-prone for palindrome splits.

Dynamic programming remembers palindrome parts to avoid repeated work.

It finds the minimum cuts needed efficiently for any word length.

Practice

1. What does the "Palindrome Partitioning DP Minimum Cuts" problem aim to find for a given string s?

easy

A. The minimum number of cuts needed to partition s into palindromic substrings

B. The maximum length of a palindrome substring in s

C. The total number of palindrome substrings in s

D. The longest palindrome substring in s

Solution

1. Step 1: Understand the problem goal

   The problem asks for the fewest cuts needed to split the string into parts where each part is a palindrome.

2. Step 2: Clarify what minimum cuts mean

   Minimum cuts means the smallest number of splits to make all parts palindromes.

3. Final Answer:

   The minimum number of cuts needed to partition s into palindromic substrings -> Option A

4. Quick Check:

   Minimum cuts = minimum splits for palindrome parts [OK]

Hint: Focus on splitting string into palindrome parts with fewest cuts [OK]

Common Mistakes:

• Confusing minimum cuts with longest palindrome substring
• Counting total palindrome substrings instead of cuts
• Thinking about maximum palindrome length

2. Which of the following TypeScript function signatures correctly represents a function to compute minimum palindrome partition cuts for a string s?

easy

A. function minCut(s: string): string[]

B. function minCut(s: number): string

C. function minCut(s: string): number

D. function minCut(s: boolean): number

Solution

1. Step 1: Identify input and output types

   Input is a string s, output is a number representing minimum cuts.

2. Step 2: Match correct signature

   Only function minCut(s: string): number matches input string and numeric output.

3. Final Answer:

   function minCut(s: string): number -> Option C

4. Quick Check:

   Input string, output number for cuts [OK]

Hint: Input string, output number for minimum cuts [OK]

Common Mistakes:

• Using wrong input type like number or boolean
• Returning array or string instead of number
• Mixing input and output types

3. What is the output of the following TypeScript code for input "aab"?

function minCut(s: string): number {
  const n = s.length;
  const dp = new Array(n).fill(0);
  const palindrome = Array.from({ length: n }, () => new Array(n).fill(false));

  for (let i = 0; i < n; i++) {
    let minCuts = i;
    for (let j = 0; j <= i; j++) {
      if (s[j] === s[i] && (i - j < 2 || palindrome[j + 1][i - 1])) {
        palindrome[j][i] = true;
        minCuts = j === 0 ? 0 : Math.min(minCuts, dp[j - 1] + 1);
      }
    }
    dp[i] = minCuts;
  }
  return dp[n - 1];
}
console.log(minCut("aab"));

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Trace palindrome checks for "aab"

   Substrings: "a" (palindrome), "aa" (palindrome), "b" (palindrome). Minimum cuts for "aab" is 1 (cut between "aa" and "b").

2. Step 2: Check dp array updates

   dp[0]=0 ("a"), dp[1]=0 ("aa"), dp[2]=1 (cut before "b").

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Check palindrome substrings and update dp for minimum cuts [OK]

Common Mistakes:

• Assuming no cuts needed for "aab"
• Counting total palindromes instead of cuts
• Incorrect dp index updates

4. Identify the bug in this TypeScript code snippet for minimum palindrome partition cuts:

function minCut(s: string): number {
  const n = s.length;
  const dp = new Array(n).fill(0);
  const palindrome = Array.from({ length: n }, () => new Array(n).fill(false));

  for (let i = 0; i < n; i++) {
    dp[i] = i;
    for (let j = 0; j <= i; j++) {
      if (s[j] === s[i] && (i - j < 2 || palindrome[j + 1][i - 1])) {
        palindrome[j][i] = true;
        dp[i] = Math.min(dp[i], dp[j - 1] + 1);
      }
    }
  }
  return dp[n - 1];
}

medium

A. dp[i] should be initialized to 0 instead of i

B. Inner loop should run from j = 0 to j <= i

C. palindrome array initialization is incorrect

D. dp[j] should be dp[j - 1] when updating dp[i]

Solution

1. Step 1: Analyze dp update logic

   When substring s[j..i] is palindrome, dp[i] should consider dp[j-1] + 1 cuts, not dp[j] + 1.

2. Step 2: Identify correct index for dp

   dp[j-1] represents cuts before substring start; dp[j] is incorrect and causes off-by-one error.

3. Final Answer:

   dp[j] should be dp[j - 1] when updating dp[i] -> Option D

4. Quick Check:

   Use dp[j-1] for correct minimum cuts [OK]

Hint: Use dp[j-1] when substring starts at j to avoid off-by-one error [OK]

Common Mistakes:

• Using dp[j] instead of dp[j-1]
• Incorrect loop bounds causing missed palindromes
• Wrong initialization of dp array

5. Given the string "abccbc", what is the minimum number of cuts needed to partition it into palindromic substrings using the DP approach?

hard

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Identify palindromic substrings in "abccbc"

   Palindromes include "a", "b", "cc", "bcb", "c". Partitioning as "a" | "bccb" | "c" is possible with 2 cuts.

2. Step 2: Calculate minimum cuts

   Minimum cuts = 2 because splitting into "a" | "bccb" | "c" yields all palindromes with 2 cuts.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Minimum cuts for "abccbc" = 2 [OK]

Hint: Look for largest palindrome substrings to minimize cuts [OK]

Common Mistakes:

• Assuming 0 cuts because string is not fully palindrome
• Counting each character as separate palindrome causing more cuts
• Missing palindrome "bccb" in middle