Minimum Length of String After Operations | Leetcode 3223

This video explains Minimum Length of String After Operations using the most optimal count approach.
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[sailendrachettri]

After this intuition 6:04 its so much easy to code .. i used unordered map to solve the problem
Thank you so much sir :)

[tahamidhossain]

Excellent explanations, easy to understand and follow through. Please consider adding Python3 solution too.

[christofrank]

Top Class! Finding this pattern is really awesome! Great Explanation!

[albin_joby]

class Solution:
def minimumLength(self, s: str) -> int:
counter = Counter(s)
for letter in counter:
if counter[letter] % 2 == 0:
counter[letter] = 2
else:
counter[letter] = 1

res = 0
for letter in counter:
res += counter[letter]

return res

[wierdo]

I too solved the problem, but with a different approach when a character hits the freqency of 3 then it is capable of having two guys on both the directions so we -2 from the string length and thats out answer


code :


class Solution {
public int minimumLength(String s) {

char arr[]=s.toCharArray();

HashMap<Character, Integer>mp=new HashMap<>();

int ans=s.length();

for(int i=0;i<arr.length;i++)
{
int freq=0;

if(mp.containsKey(arr[i]))
{
freq=mp.get(arr[i]);
}

if(freq+1==3)
{
ans-=2;
mp.put(arr[i], 1);
}
else
{
mp.put(arr[i], freq+1);
}
}

return ans;

}
}

[cheedallatejasrinivas]

I tried but reached TLE and done almost 693 / 702 test case. This explanation is very useful class Solution {

public boolean check(HashMap<Character, Integer> map){
for (int count : map.values()) {
if (count >= 3) {
return true;
}
}
return false;
}
public int minimumLength(String s) {
StringBuilder sb = new StringBuilder(s);
int n = s.length();
HashMap<Character, Integer> map = new HashMap<>();
for(int i=0;i<n;i++){
char ch = s.charAt(i);
map.put(ch, map.getOrDefault(ch, 0)+1);
}
while(check(map)){
for(char key : map.keySet()){
int val = map.get(key);
if(val >= 3){
int left_index = sb.indexOf(key+"");
int right_index = sb.lastIndexOf(key+"");
if(left_index != -1 && right_index != -1 && left_index != right_index){
sb.delete(left_index, left_index+1);
sb.delete(right_index-1, right_index);
}
map.put(key, map.get(key) - 2);
}
}
}
return sb.length();
}
}

[chocolateandmath]

How do you upload a new problem every single day??? Please tell us how to be consistent with Leetcode along with a job!

[mr.haristotle]

Hi sir. I don't understand how are you analysing the constraints at the start of every video. It's very confusing sir, analysing the time constraints