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  1. Practice
  2. Algorithms
  3. Implementation
  4. Birthday Chocolate
  5. Discussions

Birthday Chocolate

  • Sentinal_prime + 1 comment

    please explain the logic

    • khirulislam_cse + 7 comments

      Here is the Java implemetation. We just need the starting index that we are working on. and every time there is an ArrayIndexOutofBoundsException that means we don't need the loop anymore and we break. We count the total of consecutive squares and check if it's equal to the d. 

      static int solve(int n, int[] s, int d, int m){
        int startingIndex =0;
        int result =0;
        int total = 0;
        
        for(int i=0;i<n;i++){
            startingIndex = i;
            
            try{
            for(int j=0;j<m;j++){
                total += s[startingIndex + j];
            }   
            }catch(Exception e){
                break;
            }
            
            if(total == d){
                result++;
            }
            total =0;
        }
        return result;
    }
      • codertoaster + 5 comments

        I have the same idea as you, but why are most of my test cases failing ?

        static int solve(int n, int[] s, int d, int m)
    {
        int start=0,end=m-1;
        int sum=0,num=0;
        for(int i=start ; i<=end && end<s.length ; i++)
        {
            sum+=s[i];
            if(i==end)
            {
                if(sum == d)
                {
                    num++;
                    sum=0;
                    i=start+1;
                    start++;
                    end++;
                }
                else
                {
                    sum=0;
                    i=start+1;
                    start++;
                    end++;
                }
            }
        }
        return num;
    }
        • pankajbabu + 3 comments

          i use yur code,but test case is fails.....so improve your codess.. thanks

          • pooniaarvind29 + 0 comments

            thug life

          • rearbreed + 0 comments

            lol

          • btiwari004 + 0 comments

            lol

        • aishwarya217 + 1 comment

          Hey, it's probably not working because in this code i is getting changed to start + 1 and then getting incremented again as part of the for loop. 

          if(sum == d)
                {
                    num++;
                    sum=0;
                    i=start+1;
                    start++;
                    end++;
                }
                else
                {
                    sum=0;
                    i=start+1;
                    start++;
                    end++;
                }

          Hope this helps.

          • moihawk + 0 comments

            thats what it was for me, thank you

        • rvrishav7 + 1 comment
          int main()
{ int n;
 cin>>n;
 int ar[n];
 int i,c=0;
 
 for(i=0;i<n;i++)
     cin>>ar[i];
 int d,m;
 cin>>d>>m;
 int s=0,l=1; i=0;
 
       while(m<=n)
       {l=0;s=0;
           while(l<m)
           {
               s=s+ar[l+i];
               l=l+1;
           }
           if(s==d)
           {
               c=c+1;
           } 
        i=i+1;
        n=n-1;
        
       } cout<<c;
        return 0;
}
          • abhi_satya1357 + 0 comments

            include

            using namespace std;

            int main() { int i,n,sum=0,date,month,num[1000],count=0,j; cin>>n; for(i=0;i>num[i]; } cin>>date>>month; for(i=0;i

        • romulo8000 + 0 comments

          A good constructive tip: extract common code like:

          if(sum == d) {
    num++;
} else {
    start++;
}

i=start+1;
sum=0;
start++;
end++;

          It's More readable

        • kevinhs472 + 0 comments

          Nice logic man, i learn some new tips there.thank you for posting this.

      • krmatalia + 1 comment
        int c=0;
        for(int i=0;i<n;i++)
        {
            int sum=0,j=i,lp=0;
             if((j+m)>=n && (j+m)!=1)
                    break;
            else
            {
               while(lp<m)
               {
                    sum+=s[j++];
                    lp++;
                    if(sum==d)
                        ++c;     
                }
            }
        }
        return c;

        even this is failing i don't know why when i purchased the testcases i was getting the right output.

        • codertoaster + 0 comments

          yeah that happens a lot. my outputs are same as the test cases still shows failure.

      • harshitha2922 + 0 comments

        thanks for an easy understood algorithm !

      • dparthib + 0 comments

        Nice code:

        If you change your code as below suggestion then no need to handle the exception. 

        for(int j=0;(j<n)&&(j<m);j++)
      • ramji_vishnu_26 + 0 comments

        @Khirulislam_cse You can avoid using the entire try catch block by replacing with simple while loop statement.

      • Shubham_203 + 0 comments

        Hie,see this java implementation

        static int solve(int n, int[] s, int d, int m){
        // Complete this function
        int sum,count=0,j,k;
        for(int i=0;i<=n-m;i++)
        {
            k=i;
            sum=0;
            j=1;
         while(j<=m)
         {
             sum=sum+s[k];
             k++;
             j++;
         }
            if(sum==d)
                count++;
        }
        return count;
    }
      • phatngo1606 + 0 comments

        Your code is so wonderfullll