Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Create 1898-maximum-number-of-removable-characters.js #2621

Merged
merged 5 commits into from
Sep 9, 2023
+95 −0
Merged

Create 1898-maximum-number-of-removable-characters.js #2621

Changes from all commits
Commits
Show all changes
5 commits
Select commit Hold shift + click to select a range
3ca1f99
Create 1898-maximum-number-of-removable-characters.js
aadil42 Jun 22, 2023
6bafb1f
Update 1898-maximum-number-of-removable-characters.js
aadil42 Jul 4, 2023
03154ca
Update 1898-maximum-number-of-removable-characters.js
aadil42 Jul 22, 2023
6ae6bb1
Update 1898-maximum-number-of-removable-characters.js
aadil42 Jul 22, 2023
f58bdb2
Update 1898-maximum-number-of-removable-characters.js
aadil42 Sep 7, 2023
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
95 changes: 95 additions & 0 deletions javascript/1898-maximum-number-of-removable-characters.js
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,95 @@
/**
* https://leetcode.com/problems/maximum-number-of-removable-characters/
*
* Brute force
* Time O(removable.length * s.length) | Space O(n)
* @param {string} s
* @param {string} p
* @param {number[]} removable
* @return {number}
*/
var maximumRemovals1 = function(s, p, removable) {
let k = 0;
// removable.reverse();
s = s.split('');
Copy link
Collaborator

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

This will increase space complexity

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

I updated the space complexity to n. But we already had s and p as strings we just used them in an array instead of a string. wouldn't time complexity be the same for that? Before we were using n space for string which we were getting as an argument. Now we use the same number of array elements.

In other words, since we're just converting some string of n char to be an array of n elements. Then, the space complexity is the same as before right? because we were already using space(n) for the strings. Let me know if I'm making a mistake. Thank you.

Copy link
Collaborator

@aakhtar3 aakhtar3 Jul 22, 2023 •
edited
Loading

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Sting is immutable primitive data type. You can not modify its state or space in memory.

If you split and convert to char array, it will need new space to allow the list to exist.

If you need to modify the string you can split, but if you need to read by index, you do not need to split and it will lower space complexity.

/* Time O(N + N) | Space O(N) */
for (const char of str.split('')) { ... }

/* Time O(N)     | Space O(1) */
for (const char of str) { ... }

/* Time O(N)     | Space O(1) */
for (const index in str) {
    const char = str.charAt(index);
    const char = str[index];
}

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

@aakhtar3, I updated the space complexity.

Copy link
Collaborator

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

You can reduce the space O(1) by avoiding the split
and just iterate with for (const char of str)

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

@aakhtar3, but if we don't convert it to the array then how are we going to update a specific char of the string? See line 18.
I did remove the split method on the binary-search solution and used the bitwise operation to find the mid now. Please check.

p = p.split('');

for (let i = 0; i < removable.length; i++) {
s[removable[i]] = -1;
if (isSubSet1(s, p)) {
k++;
continue;
}
return k;
}

return k;
};

// helper function.
function isSubSet1(s, p) {
let i = 0;
let j = 0;

while (i < s.length && j < p.length) {
if (s[i] === p[j]) {
i++;
j++;
} else {
i++;
}
}

return j === p.length;
}


/**
*
* Binary Search
* n = length of string, k = length of removable
* Time O(log(k)*n) | Space O(1)
* @param {string} s
* @param {string} p
* @param {number[]} removable
* @return {number}
*/
var maximumRemovals = function(s, p, removable) {

let left = 0;
let right = removable.length - 1;
let k = 0;

while (left <= right) {
const mid = (left + right) >> 1;
const hash = new Set(removable.slice(0, mid + 1));

if (isSubSet(hash, s, p)) {
k = Math.max(k, mid + 1);
left = mid + 1;
continue;
}

right = mid - 1;
}

return k;
};

// helper function.
function isSubSet(hash, s, p) {
let i = 0;
let j = 0;

while (i < s.length && j < p.length) {
if (s[i] === p[j] && !hash.has(i)) {
i++;
j++;
continue;
}

i++;
}

return j === p.length;
}