can't figure out missing corner cases

general
,
1

I have this question from a past contest that i participated in. I just can’t figure out what corner cases am I missing out on as two of the test cases ain’t giving correct answers. Plz help.
Problem link: https://www.hackerrank.com/contests/freshers-challenge-3/challenges/fcp
My Code: https://www.hackerrank.com/contests/freshers-challenge-3/challenges/fcp/submissions/code/4355231
Thanks

2

Link to test cases that I am getting wrong:
Input: https://hr-testcases-us-east-1.s3.amazonaws.com/15172/input04.txt?AWSAccessKeyId=AKIAJAMR4KJHHUS76CYQ&Expires=1449964738&Signature=WAV%2F%2FHvuN4vaS76Q6ZZ1p8O1VNk%3D&response-content-type=text%2Fplain
Output: https://hr-testcases-us-east-1.s3.amazonaws.com/15172/output04.txt?AWSAccessKeyId=AKIAJAMR4KJHHUS76CYQ&Expires=1449964963&Signature=Crn0N%2BPDlJzvMctQyyWs8B0pcao%3D&response-content-type=text%2Fplain

3

Plzz help someone!

4

Method 1: Using sorting

for _ in range(input()):
    n = input();i=1;temp=1;ans=0
    a=map(int,raw_input().split())
    a.sort()
    while(i<n):
        if(a[i]==a[i-1]):
            temp+=1
            i+=1
        else:
            ans+=temp*(temp-1)
            i+=1
            temp=1
    ans+=temp*(temp-1)
    print ans

Method 2: Counting frequency
for _ in range(input()):
a=[0]1000005
n=input()
ans=0
b=map(int,raw_input().split())
for i in range(n):
a[b[i]]+=1
for i in range(1000005):
ans+=a[i](a[i]-1)
print ans

I used above two methods and got accepted in all test cases for both. I could not access your code but hope above two python codes will help you.

1 Like
5

Thanks.

Can you review my code once?


int t,n,i,j,m,c,s;
vector<int>vec;
int main() {
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=0;i<n;++i){
            scanf("%d",&m);
            vec.push_back(m);
        }
        sort(vec.begin(),vec.end());
        s=0;
        for(i=0;i<n;++i){
            m=vec[i];
            c=1;
            
        for(j=i+1;(vec[j]<=m)&&(j<n);j++){
            c++;
        }
        s+=c*(c-1);
        i+=c-1;
            
        }
        printf("%d\n",s);
        vec.clear();
    }
    return 0;
}</pre>