Party to friends Or Not-CodePudding

Ashutosh has N friends and he wants to give a party to his C friends on his birthday. Ashutosh knows the amount he will have to spend on each of his N friends in party. Given the expense amount of each of the N friends, comment if it is possible for Ashutosh to give a party to his C friends if he has just R rupees.

Input Method

First line contains three space separated integers: N, C and R respectively

Second line contains N space separated integers which represents the amount he needs to spend on each of his N friends.

Constraints

N <= 1000

C is lesser than N

R < 10000

Output Output Method

If it is possible to give party to C friends, print "Party"

Else print "Sad".

Sample Input 1

5 3 24 6 4 21 20 13

Sample Output 1

Party

function solve(N,C,R,arr){
    //console.log(N,C,R,arr)
    var flag=false;
    function partyOrNot(N,C,R,arr,position,newarr){
        var sum=0;
        for(var j=0;j<newarr.length;j  ){
            sum =newarr[j];
        }
        
        if(newarr.length==C && sum<=R){
            flag=true;
            return;
        }
        if(sum>=R||newarr.length>=C){
            return;
        }
        else{
            for(var i=position;i<N;i  ){
                newarr.push(arr[i]);
                partyOrNot(N,C,R,arr,i 1,newarr);
                newarr.pop();
            }
            return;
        }
    }
    
    
    partyOrNot(N,C,R,arr,0,[]);
    //console.log(flag);
    if(flag){
        console.log("Party")
    }
    else{
        console.log("Sad")
    }
}

I tried this but not get correct answer some test cases passed but still partial accepted.

CodePudding user response：

You Just need to remove = from this condition.

 if(sum>R||newarr.length>C){
            return;
 }

After this it will work fine.

Use can solve this problem using sorting. first sort the complete array
And take Sum variable and Start the Loop 0 to n

for(let i=0;i<n;i  ){
   sum =arr[i];
   if(sum<=R){
     print "Party"
   }else{
     print "Sad"
   }
}

This will work as well Thanks!