Palindrome Partitioning DP Minimum Cuts in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the minimum cuts array with the maximum possible cuts.

DSA Javascript

const minCuts = new Array(n).fill([1]);

Drag options to blanks, or click blank then click option'

An - 1

B0

CInfinity

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing minCuts with 0 which assumes no cuts needed initially.

Using Infinity which is not necessary here.

We initialize minCuts with n-1 because the maximum cuts needed is one less than the length of the string.

2fill in blank

medium

Complete the code to check if the substring from j to i is a palindrome.

DSA Javascript

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

Drag options to blanks, or click blank then click option'

Adp

BminCuts

CisPalindrome

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using minCuts instead of the palindrome table.

Using an undefined variable.

The dp table named isPalindrome stores whether substrings are palindromes.

3fill in blank

hard

Fix the error in updating the minimum cuts when the substring is a palindrome.

DSA Javascript

minCuts[i] = j === 0 ? 0 : Math.min(minCuts[i], [1] + 1);

Drag options to blanks, or click blank then click option'

AminCuts[j + 1]

BminCuts[j]

CminCuts[i - 1]

DminCuts[j - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using minCuts[j] which is the cuts at the start index, not before it.

Using minCuts[i - 1] which relates to the current end index.

We add 1 to the minimum cuts needed before index j, which is minCuts[j - 1].

4fill in blank

hard

Fill both blanks to correctly update the palindrome table and minimum cuts inside the nested loops.

DSA Javascript

isPalindrome[j][i] = true;
minCuts[i] = j === 0 ? 0 : Math.min(minCuts[i], [1] + [2]);

Drag options to blanks, or click blank then click option'

AminCuts[j - 1]

BminCuts[i - 1]

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using minCuts[i - 1] instead of minCuts[j - 1].

Adding 0 instead of 1 to the cuts.

We set isPalindrome[j][i] to true and update minCuts[i] by adding 1 to minCuts[j - 1].

5fill in blank

hard

Fill all three blanks to complete the palindrome partitioning minimum cuts function.

DSA Javascript

function minCut(s) {
  const n = s.length;
  const minCuts = new Array(n).fill([1]);
  const isPalindrome = Array.from({ length: n }, () => new Array(n).fill(false));
  for (let i = 0; i < n; i++) {
    for (let j = 0; j <= i; j++) {
      if (s[j] === s[i] && (i - j < 2 || isPalindrome[j + 1][i - 1])) {
        isPalindrome[j][i] = true;
        minCuts[i] = j === 0 ? 0 : Math.min(minCuts[i], minCuts[[2]] + [3]);
      }
    }
  }
  return minCuts[n - 1];
}

Drag options to blanks, or click blank then click option'

An - 1

Bj - 1

C1

Di - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing minCuts with 0.

Using wrong indices for minCuts update.

Adding 0 instead of 1 to cuts.

Initialize minCuts with n-1, update minCuts[i] using minCuts[j-1] plus 1 cut.

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