Palindrome Partitioning DP Minimum Cuts in DSA Javascript - Time & Space Complexity
We want to know how the time needed grows when we find the minimum cuts to split a string into palindromes.
How does the work increase as the string gets longer?
Analyze the time complexity of the following code snippet.
function minCut(s) {
const n = s.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 minCuts = end;
for (let start = 0; start <= end; start++) {
if (s[start] === s[end] && (end - start < 2 || palindrome[start + 1][end - 1])) {
palindrome[start][end] = true;
minCuts = start === 0 ? 0 : Math.min(minCuts, dp[start - 1] + 1);
}
}
dp[end] = minCuts;
}
return dp[n - 1];
}
This code finds the minimum number of cuts needed to split the string into palindrome substrings using dynamic programming.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: Nested loops checking all substring pairs (start to end) and palindrome checks.
- How many times: Outer loop runs n times; inner loop runs up to n times for each outer loop, so roughly n x n = n² times.
As the string length grows, the number of substring checks grows roughly with the square of the length.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 100 checks |
| 100 | About 10,000 checks |
| 1000 | About 1,000,000 checks |
Pattern observation: Doubling the input size roughly quadruples the work because of the nested loops.
Time Complexity: O(n³)
This means the time needed grows roughly with the cube of the string length.
[X] Wrong: "The algorithm runs in linear time because it just scans the string twice."
[OK] Correct: The nested loops check all substring pairs, which means the work grows much faster than just scanning once or twice.
Understanding this time complexity helps you explain how dynamic programming reduces repeated work and why nested loops matter in string problems.
"What if we precompute palindrome substrings separately before the main loop? How would that affect the time complexity?"
Practice
minimum cuts mean in the context of palindrome partitioning?Solution
Step 1: Understand palindrome partitioning
Palindrome partitioning splits a string into substrings where each substring is a palindrome.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.Final Answer:
The smallest number of cuts needed to split the string into palindromic substrings -> Option DQuick Check:
Minimum cuts = smallest cuts for palindrome parts [OK]
- Confusing minimum cuts with longest palindrome substring
- Thinking minimum cuts count characters removed
- Mixing total palindrome substrings with minimum cuts
isPalindrome for a string s of length n in JavaScript?Solution
Step 1: Understand 2D array initialization in JavaScript
UsingArray.fromwith a mapping function creates independent inner arrays.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 becausemapon uninitialized array does not work as expected.Final Answer:
const isPalindrome = Array.from({length: n}, () => Array(n).fill(false)); -> Option AQuick Check:
Use Array.from with function for independent inner arrays [OK]
- Using fill with array directly causing shared references
- Using map on uninitialized arrays
- Not initializing inner arrays separately
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));Solution
Step 1: Identify palindromic substrings in "aab"
"a", "a", "b" are palindromes; "aa" is palindrome; "aab" is not.Step 2: Calculate minimum cuts
Minimum cuts needed: split "aab" as "aa" | "b" with 1 cut.Final Answer:
1 -> Option BQuick Check:
Minimum cuts for "aab" = 1 [OK]
- Assuming no cuts needed for non-palindrome string
- Counting total palindromic substrings instead of cuts
- Misindexing palindrome checks
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;
}Solution
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.Step 2: Understand impact on palindrome checks
This causes incorrect palindrome marking because all rows share the same inner array, corrupting data.Final Answer:
isPalindrome initialization shares inner arrays causing incorrect updates -> Option CQuick Check:
Shared inner arrays cause bugs in 2D boolean array [OK]
- Ignoring shared reference problem in 2D arrays
- Thinking loop start index causes bug
- Overlooking missing return value as main error
s = "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromic substrings using dynamic programming?Solution
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.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.Final Answer:
3 -> Option AQuick Check:
Minimum cuts for complex palindrome string = 3 [OK]
- Assuming minimum cuts equals number of palindromic substrings
- Miscounting cuts by off-by-one errors
- Not considering overlapping palindrome substrings
Ready to prove it?
Pass the test and this pattern moves unknown
