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  1. Prepare
  2. Mathematics
  3. Probability
  4. Vertical Sticks
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Vertical Sticks

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31 Discussions

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  • aaron99ho
     + 1 comment
    def solve(y):
    V = 0
    n = len(y)
    result = []

    for num in y:
        count = sum(1 for other_num in y if other_num >= num)
        result.append(count)

    [V += (n+1)/ (result[i]+1) for i in range(n)]
    
    return ["%.2f" % V]

  • mark_ozerov
     + 0 comments
    def solve(y):
    y, n  = sorted(y,reverse=True), len(y) 
    num_replicas, total = 0, 1
    factorial_i_minus_1 = 1
    
    for i in range(2,n+1):
        factorial_i_minus_1 *= i-1
        is_replica = y[i-1]==y[i-2]
        num_replicas = is_replica * (is_replica + num_replicas)
        total += i * total + factorial_i_minus_1 * (i - num_replicas)
        
    return ["{0:.2f}".format(total/factorial_i_minus_1/n)]

  • submit_jane_doe
     + 0 comments

    The editorial has the most inefficient solution to this problem. Three for loops--are you sure? This was my linear python3 solution (I tried to make it as readable as possible and in line with the vars defined in the problem's description):

    def solve(n, y):
    V = 0
    y.sort()
    for i in range(n):
        if i == 0 or (i > 0 and y[i] > y[i-1]): 
            k = n-i
            v_i = (n+1)/(k+1) 
        V += v_i
    return ["%.2f" % V]

    This also exercises Linearity of Expectation, but all it's doing is sorting the input array y then walking through it to sum up v_i. If there's a repeating value, it'll use the first instance of v_i for that value (since v_i will continue to decrease as i increases).

  • thedaylan
     + 0 comments

    Error: Garbage collector could not allocate 16384 bytes of memory for major heap section.

    :/

  • wadezyw
     + 0 comments

    for javascript function processData(input) { //Enter your code here var loopNum=input.split("\n")[0];

    var inpuJump=1;
for(var i=0;i<loopNum;i++){
    var arrSize=parseInt(input.split("\n")[inpuJump]);
    var arr=input.split("\n")[inpuJump+1].split(" ");

    var sumArr=0;
    for(var j=0;j<arrSize;j++){
        sumArr+=(parseInt(arrSize)+1)/((parseInt(getLarger(arr,arr[j])))+1);
    }
 console.log(sumArr.toFixed(2));   

    inpuJump+=2;
}

function getLarger(arrt,curr){
    var count=0;
    for(var i=0;i<arrSize;i++){
        if(parseInt(arrt[i])>=curr) count++;
    }
    return count;
}

    }