Блог пользователя nisanduuu

Please help me to solve this problem
Автор nisanduuu, история, 2 года назад, По-английски

You are given an string s of length n. And there is a eraser and it deletes neighboring characters of a particular letter when it place on a specific position. For an example 'aaaaba' when we place eraser on 1st 'a'(index 0), string becomes 'ba' (since we deleted all the letters of the neighboring 'a's). Task is to delete the entire string in minimum operations. 

Test 1
Input -> 6 aabcbb
Output -> 3

Constraints
1<=n<=1000
  • Проголосовать: нравится
  • 0
  • Проголосовать: не нравится

»
2 года назад, скрыть # |
← Rev. 7  
Проголосовать: нравится +3 Проголосовать: не нравится 
cout << (n+2)/3 << endl;

Assuming it can erase both neighbours at same time. So elimininate 3 in one go.

Edit: This is wrong. By neighbours, it means all consecutive neighbours that are same, not just adjacent.

→ Ответить
»
2 года назад, скрыть # |
 
Проголосовать: нравится 0 Проголосовать: не нравится

just do exactly what eraser do:

int i=0,ans=0;
while(i<n){
   int j=i+1;
   while(j<n&&s[j]==s[i])j++;
   ans++;
   i=j;
}
cout<<ans<<endl;

I think this will work

→ Ответить
»
2 года назад, скрыть # |
← Rev. 3  
Проголосовать: нравится 0 Проголосовать: не нравится
  1. We may remove instantly all successive duplicates of elements, e. i., aaaaba => aba, as 1 eraser removes all equal elements.
  2. We can think of such dp:
for i = n..1
for j = i..n
    if i == j dp[i][j] = 1 // we need exactly 1 eraser for 1 element
    else
              dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + s[i] != s[j] ? 1 : 0

What are we doing here is we calculate the optimal answer for the ranges (i + 1, j) and (i, j - 1) and then combine them adding 1 eraser if edge elements are different.

This should work, I hope it helps.

→ Ответить
»
2 года назад, скрыть # |
← Rev. 3  
Проголосовать: нравится 0 Проголосовать: не нравится

This problem can be solved using range DP.

$$$ \text{ }\\ dp[i][i] = 1 \\

dp[i][j] = \min \begin{cases}

dp[i+1][j] + 1, \\ dp[i][j-1] + 1, \\ dp[i+1][j], & \text{if } a[i] = a[i+1]\\ dp[i][j-1], & \text{if } a[j] = a[j-1]\\ \min(dp[i+1][j], dp[i][j-1]), & \text{if } a[i] = a[j] \end{cases} $$$

The solution comes from the fact that you do not need to perform extra operations on characters that are same to the next one.

→ Ответить