Hackers with Bits | Editorial by aksingh007

Just try to have a maximum window with only one allowed 0, keep another array which stores the count of no. of 1's to it's right, when updating the window size 
just check if the last 1 of the window has a count>0 corresponding to the auxiliary array.

**Code in C:**
```
#include <stdio.h>
 
int main()
{ 
    int n,arr[105],index[105 ],count=-1,max=0,blocks=0,pos,ans,res1,res2;
    
    scanf("%d",&n);
    
    for(int i=0;i<n;i++)
    scanf("%d",&arr[i]);
    
    for(int i=1;i<n-1;i++)
      if(arr[i]==0 && arr[i-1]==1 && arr[i+1]==1){
          index[++count]=i;
      }
     // printf("%d",count);
     // if(count>0){
          for(int i=0;i<=count;i++){
              pos=index[i]-1;
              int tot1=0,tot2=0;
              while(arr[pos]==1 && pos>=0){
                  tot1++;
                  pos--;
              }
              pos=index[i]+1;
              while(arr[pos]==1 && pos<n){
                  tot2++;
                  pos++;
              }
              //printf("%d %d\n",tot1,tot2);
              int tot=tot1+tot2;
              if(max<tot)
                max=tot;
          }
          if(count>=1)
            //printf("%d",max+1);
            res1=max+1;
             else
              //printf("__%d",max);
              res1=max;
           //return 0;
     // }
     // else
     // {  
     max=0;
          for(int i=0;i<n;i++){
              count=0;
              if(arr[i]==1 ){
                  while(arr[i]==1 && i<n){
                      i++;
                      count++;
                  }
                  if(count>max)
                    max=count;
                  blocks++;
              }
          }
          if(blocks>1)
            //printf("..%d",max+1);
            res2=max+1;
            else
            //printf("///%d",max);
          res2=max;
          //return 0;
      //}
      ans=(res1>res2)?res1:res2;
      printf("%d",ans);
    return 0;
}
```

Hackers with Bits | Editorial by aksingh007

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