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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Sherlock and The Beast
  5. Discussions

Sherlock and The Beast

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543 Discussions

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  • vutrantrung14821
     + 0 comments
    

    i = n // 3
while i >= 0:
    threes = n - 3 * i
    if threes % 5 == 0:
        print("5" * (3 * i) + "3" * threes)
        return
    i -= 1
print(-1)

    `

  • cass6896
     + 0 comments

    C++

    void decentNumber(int n) {
    for (auto i = 0; i <= n; ++i) {
        if (i % 5 == 0 && (n - i) % 3 == 0) {
            cout << string(n - i, '5') << string(i, '3') << endl;
            return;
        }
    }
    cout << -1 << endl;
}

  • nik_1492
     + 1 comment

    A few problems with this task 1) For n=15 both could work all 5 or all 3 but test just expect it to be all 5 2) 33333555 and 55533333 are valid solutions but again test expect it to be 55533333 3) for n = 100 there more then one solution it could be 10 times 5 and 90 times 3 but also could be 85 times 3 and 15 times 5

    All in all i've decided to not proceed because it doesn't make sense find all of the limitations that not descripbed in the Problem section.

    def decentNumber(n):
    for i in range(1,n//3+1):
        if (n - i*5) % 3 == 0:
            n5 = n - i*5
            n3 = n - n5
            return '5'*n5 + '3'*n3     
            
        if (n - i*3) % 5 == 0:
            n3 = n - i*3
            n5 = n - n3  
            # print('n3-n5', n3,n5)
            return '5'*n5 + '3'*n3
        
    return -1
``

    `

  • shiaramoon
     + 0 comments

    python 3

    def decentNumber(n):
    fives = n
    threes = 0
    
    while fives >= 0:
        if fives % 3 == 0 and threes % 5 == 0:
            break
        else:
            fives -= 1
            threes += 1
    print(-1 if fives < 0 else ('5' * fives) + ('3' * threes))

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can have the implementation link here : https://youtu.be/R01NQ4Zt2qA

    void decentNumber(int n) {
    int three = ((3 - (n % 3)) % 3 ) * 5;
    if(three > n) cout << -1;
    else cout << string(n - three, '5') << string(three, '3');
    cout << endl;
}