Palindrome Partitioning DP Minimum Cuts in DSA Javascript

Mental Model

We want to split a string into parts where each part is a palindrome, using the fewest cuts possible. We check all possible splits and remember the best results to avoid repeating work.

Analogy: Imagine cutting a ribbon into pieces so that each piece is a perfect mirror image of itself. You want to make as few cuts as possible to get all mirror pieces.

s = "a b a c"
Indices: 0 1 2 3
Cuts: between indices
Initial: a b a c
No cuts: a b a c (not all palindrome)
Possible cuts: a | b a | c

Dry Run Walkthrough

Input: string: "aab"

Goal: Find the minimum number of cuts to partition "aab" into palindromic substrings

Step 1: Initialize dp array with max cuts (worst case: cut before every character)

dp = [0, 1, 2]
Palindrome table initialized with false

Why: Start with the worst case so we can improve it

Step 2: Check substring s[0..0] = 'a' is palindrome, update palindrome table

palindrome[0][0] = true
No cut needed, dp[0] = 0

Why: Single characters are always palindromes

Step 3: Check substring s[0..1] = 'aa' is palindrome, update palindrome table and dp

palindrome[0][1] = true
Update dp[1] = 0 (no cut needed since 'aa' is palindrome)

Why: If substring is palindrome, no cut needed before it

Step 4: Check substring s[1..1] = 'a' palindrome, update palindrome table

palindrome[1][1] = true
No change to dp[1]

Why: Single character palindrome

Step 5: Check substring s[0..2] = 'aab' not palindrome, dp unchanged

palindrome[0][2] = false
Check cuts at j=1 and j=2

Why: Cannot treat whole string as palindrome

Step 6: Check substring s[1..2] = 'ab' not palindrome

palindrome[1][2] = false

Why: Not palindrome, no dp update

Step 7: Check substring s[2..2] = 'b' palindrome, update palindrome table

palindrome[2][2] = true
Update dp[2] = min(dp[2], dp[1] + 1) = min(2, 0 + 1) = 1

Why: Cut before 'b' to get palindrome partitions

Result:

dp = [0, 0, 1]
Minimum cuts needed = 1
Partitions: "aa" | "b"

Annotated Code

DSA Javascript

class PalindromePartition {
  constructor(s) {
    this.s = s;
    this.n = s.length;
    this.dp = new Array(this.n).fill(0);
    this.palindrome = Array.from({ length: this.n }, () => new Array(this.n).fill(false));
  }

  minCuts() {
    if (this.n === 0) return 0;
    for (let i = 0; i < this.n; i++) {
      let minCut = i; // max cuts: cut before every char
      for (let j = 0; j <= i; j++) {
        if (this.s[j] === this.s[i] && (i - j < 2 || this.palindrome[j + 1][i - 1])) {
          this.palindrome[j][i] = true; // mark palindrome
          minCut = j === 0 ? 0 : Math.min(minCut, this.dp[j - 1] + 1); // update min cuts
        }
      }
      this.dp[i] = minCut;
    }
    return this.dp[this.n - 1];
  }
}

// Driver code
const input = "aab";
const partition = new PalindromePartition(input);
console.log(partition.minCuts());

let minCut = i; // max cuts: cut before every char

Initialize minCut to worst case for position i

if (this.s[j] === this.s[i] && (i - j < 2 || this.palindrome[j + 1][i - 1])) {

Check if substring s[j..i] is palindrome by comparing ends and inner substring

this.palindrome[j][i] = true; // mark palindrome

Mark substring s[j..i] as palindrome

minCut = j === 0 ? 0 : Math.min(minCut, this.dp[j - 1] + 1);

Update minCut using previous dp value plus one cut if needed

this.dp[i] = minCut;

Store minimum cuts needed for substring s[0..i]

Output

1

Complexity Analysis

Time: O(n^2) because we check all substrings and update dp accordingly

Space: O(n^2) for the palindrome table and O(n) for the dp array

vs Alternative: Naive approach tries all partitions recursively with exponential time; DP reduces it to polynomial time by remembering palindrome checks and minimum cuts

Edge Cases

Empty string

Returns 0 cuts since no partitions needed

DSA Javascript

if (this.n === 0) return 0;

String with all identical characters, e.g. "aaaa"

Returns 0 cuts because whole string is palindrome

DSA Javascript

minCut = j === 0 ? 0 : Math.min(minCut, this.dp[j - 1] + 1);

String with no palindrome longer than 1, e.g. "abc"

Returns n-1 cuts, cutting before every character

DSA Javascript

let minCut = i;

When to Use This Pattern

When asked to partition a string into palindromes with minimum cuts, use DP with palindrome substring checks to optimize repeated work.

Common Mistakes

Mistake: Not checking if inner substring is palindrome before marking larger substring
Fix: Add condition to check palindrome[j+1][i-1] before marking palindrome[j][i]

Mistake: Initializing dp array incorrectly causing wrong minimum cuts
Fix: Initialize dp[i] with i (max cuts) before updating

Summary

Finds the fewest cuts needed to split a string into palindrome parts.

Use when you want to minimize cuts for palindrome partitions in a string.

Remember to use dynamic programming to check palindromes and store minimum cuts efficiently.

Practice

1. What does the term minimum cuts mean in the context of palindrome partitioning?

easy

A. The number of characters to remove to make the string a palindrome

B. The longest palindrome substring in the string

C. The total number of palindromic substrings in the string

D. The smallest number of cuts needed to split the string into palindromic substrings

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning splits a string into substrings where each substring is a palindrome.

2. Step 2: Define minimum cuts

   Minimum cuts means the least number of cuts needed to divide the string so that every part is a palindrome.

3. Final Answer:

   The smallest number of cuts needed to split the string into palindromic substrings -> Option D

4. Quick Check:

   Minimum cuts = smallest cuts for palindrome parts [OK]

Hint: Minimum cuts split string into palindrome parts with fewest cuts [OK]

Common Mistakes:

• Confusing minimum cuts with longest palindrome substring
• Thinking minimum cuts count characters removed
• Mixing total palindrome substrings with minimum cuts

2. Which of the following is the correct way to initialize a 2D boolean array isPalindrome for a string s of length n in JavaScript?

easy

A. const isPalindrome = Array.from({length: n}, () => Array(n).fill(false));

B. const isPalindrome = Array(n).fill(Array(n).fill(false));

C. const isPalindrome = new Array(n).fill(new Array(n).fill(false));

D. const isPalindrome = Array(n).map(Array(n).fill(false));

Solution

1. Step 1: Understand 2D array initialization in JavaScript

   Using Array.from with a mapping function creates independent inner arrays.

2. Step 2: Compare options

   const isPalindrome = Array.from({length: n}, () => Array(n).fill(false)); correctly creates separate inner arrays filled with false. Options A and C create references to the same inner array, causing bugs. const isPalindrome = Array(n).map(() => Array(n).fill(false)); fails because map on uninitialized array does not work as expected.

3. Final Answer:

   const isPalindrome = Array.from({length: n}, () => Array(n).fill(false)); -> Option A

4. Quick Check:

   Use Array.from with function for independent inner arrays [OK]

Hint: Use Array.from with function to avoid shared references [OK]

Common Mistakes:

• Using fill with array directly causing shared references
• Using map on uninitialized arrays
• Not initializing inner arrays separately

3. What is the output of the following code snippet for s = "aab"?

function minCut(s) {
  const n = s.length;
  const isPalindrome = Array.from({length: n}, () => Array(n).fill(false));
  for (let i = 0; i < n; i++) {
    isPalindrome[i][i] = true;
  }
  for (let length = 2; length <= n; length++) {
    for (let start = 0; start <= n - length; start++) {
      let end = start + length - 1;
      if (s[start] === s[end]) {
        if (length === 2) {
          isPalindrome[start][end] = true;
        } else {
          isPalindrome[start][end] = isPalindrome[start + 1][end - 1];
        }
      }
    }
  }
  const cuts = Array(n).fill(0);
  for (let i = 0; i < n; i++) {
    let minCut = i;
    for (let j = 0; j <= i; j++) {
      if (isPalindrome[j][i]) {
        minCut = j === 0 ? 0 : Math.min(minCut, cuts[j - 1] + 1);
      }
    }
    cuts[i] = minCut;
  }
  return cuts[n - 1];
}
console.log(minCut(s));

medium

A. 0

B. 1

C. 2

D. 3

Solution

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

   "a", "a", "b" are palindromes; "aa" is palindrome; "aab" is not.

2. Step 2: Calculate minimum cuts

   Minimum cuts needed: split "aab" as "aa" | "b" with 1 cut.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Check palindrome substrings, then find fewest cuts [OK]

Common Mistakes:

• Assuming no cuts needed for non-palindrome string
• Counting total palindromic substrings instead of cuts
• Misindexing palindrome checks

4. The following code attempts to compute minimum cuts for palindrome partitioning but has a bug. What is the error?

function minCut(s) {
  const n = s.length;
  const isPalindrome = Array(n).fill(Array(n).fill(false));
  for (let i = 0; i < n; i++) {
    isPalindrome[i][i] = true;
  }
  // ... rest of palindrome logic ...
  return 0;
}

medium

A. The function does not return the minimum cuts value

B. Loop for setting isPalindrome[i][i] should start from 1, not 0

C. isPalindrome initialization shares inner arrays causing incorrect updates

D. The string length n is not calculated correctly

Solution

1. Step 1: Analyze isPalindrome array initialization

   Using fill with an array creates references to the same inner array, so updating one row updates all rows.

2. Step 2: Understand impact on palindrome checks

   This causes incorrect palindrome marking because all rows share the same inner array, corrupting data.

3. Final Answer:

   isPalindrome initialization shares inner arrays causing incorrect updates -> Option C

4. Quick Check:

   Shared inner arrays cause bugs in 2D boolean array [OK]

Hint: Avoid fill with arrays; use Array.from with function [OK]

Common Mistakes:

• Ignoring shared reference problem in 2D arrays
• Thinking loop start index causes bug
• Overlooking missing return value as main error

5. Given the string s = "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromic substrings using dynamic programming?

hard

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: Understand palindrome partitions for "ababbbabbababa"

   This string can be partitioned into palindromes with minimum cuts by checking all substrings and using DP to find minimal cuts.

2. Step 2: Use known result or DP approach

   The minimum cuts for this string is 3, as it can be split into palindromic parts like "a|babbbab|b|ababa" or similar partitions.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Minimum cuts for complex palindrome string = 3 [OK]

Hint: Use DP palindrome checks and cuts array to find minimum cuts [OK]

Common Mistakes:

• Assuming minimum cuts equals number of palindromic substrings
• Miscounting cuts by off-by-one errors
• Not considering overlapping palindrome substrings