2024. maximum difficulty of the exam - sliding window

2024. The biggest trouble of the exam

A teacher is giving a play by n A test consisting of three judgment questions , The answer to each question is true （ use ‘T’ Express ） perhaps false （ use ‘F’ Express ）. Teachers want to increase students' uncertainty about their answers , The method is Maximize Yes Continuous same The number of results .（ That is, continuous true Or continuous false）.

Give you a string answerKey , among answerKey[i] It's No i The correct result of a question . in addition to , Give you an integer k , Indicates the maximum number of times you can do the following ：

 In every operation , Change the correct answer to the question to  'T'  perhaps  'F' （ Also is to  answerKey[i]  Change it to  'T'  perhaps  'F' ）.

Please return no more than k In case of one operation , Maximum continuity ‘T’ perhaps ‘F’ Number of .

Example 1：

Input ：answerKey = “TTFF”, k = 2
Output ：4
explain ： We can put two ‘F’ All become ‘T’ , obtain answerKey = “TTTT” .
There are four consecutive ‘T’ .

Example 2：

Input ：answerKey = “TFFT”, k = 1
Output ：3
explain ： We can put the front ‘T’ Switch to ‘F’ , obtain answerKey = “FFFT” .
perhaps , We can put the second ‘T’ Switch to ‘F’ , obtain answerKey = “TFFF” .
In both cases , There are three consecutive ‘F’ .

Example 3：

Input ：answerKey = “TTFTTFTT”, k = 1
Output ：5
explain ： We can put the first ‘F’ Switch to ‘T’ , obtain answerKey = “TTTTTFTT” .
Or we can put the second ‘F’ Switch to ‘T’ , obtain answerKey = “TTFTTTTT” .
In both cases , There are five consecutive ‘T’ .

Although I did this problem with sliding windows , But it's too complicated , Write your own code , Felt it was very difficult , The solution code is as follows ：

int maxConsecutiveAnswers(char * answerKey, int k){
    
    int n=0;
    int sum=0;
    int max=0;

    int i=0;
    k=k+1;
    int r[k];
    for(i=0;i<k;i++){
    
        r[i]=0;
    }
    int num=0;
    int knum=0;
    i=0;
   int j;
  
  while(answerKey[i]!='\0'){
    
     
      if(answerKey[i]=='F'){
    
          if(knum<k-1){
    
              r[knum]=num+1;
                 max=num+1+max;
                 sum=num+1+sum;
          }
          else{
    
              if(knum==k-1){
    
                    max=num+max;
                     r[knum]=num+1;
                     sum=max;
                  
              }
              if(knum>k-1){
    
                 
           
                        sum=sum+num+1-r[knum%k];
                        if(sum>max){
    
                            max=sum;
                        }
                    
                       r[knum%k]=num+1;

              }
          }
       


          num=0;
          knum++;

         
      }
      else{
    
          num++;
      }
      i++;

            
  }
 
if(answerKey[i-1]=='T'){
    
    if(knum<=k-1){
    
        max=max+num;
       
    }
    else{
    

   
      sum=sum+num+1-r[knum%k];
                if(sum>max){
    
                            max=sum;
                        }
    }
  }

num=0;
knum=0;
for(i=0;i<k;i++){
    
        r[i]=0;
    }
i=0;
int max1=0;
sum=0;
  while(answerKey[i]!='\0'){
    
      if(answerKey[i]=='T'){
    
          if(knum<k-1){
    
              r[knum]=num+1;
                 max1=num+1+max1;
                  sum=num+1+sum;
          }
          else{
    
              if(knum==k-1){
    
                    max1=num+max1;
                     r[knum]=num+1;
                     sum=max1;
                     
              }
              if(knum>k-1){
    
                  
                   
                        sum=sum+num+1-r[knum%k];
                   
                    if(sum>max1){
    
                        max1=sum;
                    }
                       r[knum%k]=num+1;

              }
          }
          num=0;
          knum++;  
      }
      else{
    
          num++;
      }
      i++;

            
  }
  if(answerKey[i-1]=='F'){
    
    if(knum<=k-1){
    
        max1=max1+num;
       
    }
    else{
    
      sum=sum+num+1-r[knum%k];
                if(sum>max1){
    
                            max1=sum;
                        }
    }
  }

if(max1>max){
    
    return max1;
}

else{
    
  return max;
}

  



}