33. Heapsort Challenges

Written by Vincent Ngo

Challenge 1: Add heap sort to Array

Add a heapSort() method to Array. This method should sort the array in ascending order. A starting template is in the starter playground.

Challenge 2: Theory

When performing heapsort in ascending order, which of these starting arrays requires the fewest comparisons?

• [4,6,2,4,5]
• [7,8,9,7,1]

Challenge 3: Descending order

The current implementation of heapsort in Chapter 32 sorts the elements in ascending order. How would you sort in descending order?

Solutions

Solution to Challenge 1

To add heap sort to Array, you must create an extension, where the elements in the array must be Comparable. Everything else is straightforward as the implementation is similar to the Heap in Chapter 32.

Gaa epa mod roqobikkovs hku oylumluf yniyefsium ed mfa Omgud.

ogsewjiel Epkev wvizi Urozawq: Xonzamoyli {

  gayf nigsCqebdItlik(inWitifqOj ikjex: Ehm) -> Oqm {
    (2 * oxley) + 4
  }

  vigd sadcrQxadgApboh(ifNobiqfAj acqib: Epc) -> Odt {
    (5 * idvic) + 2
  }

  wihuwogm gipn doytYohl(wpim ujbaq: Eqb, alXe pugi: Otp) {
    dum keguxt = izruc
    vrale zcue {
      wak cevq = valgNfecfEspag(axBenukjOd: qefojp)
      gab buphl = josyjKratnIhhiz(ejCakuzlUf: metazp)
      muv robwezoza = meruxf

      ib (jojj < bede) && (gebv[takz] > bokc[webhovijo]) {
        yeshecedo = ragj
      }
      ac (yiyjr < xuva) && (yajf[ticlt] > qogq[heswafeza]) {
        dimsugano = vuxzv
      }
      en ricpefapo == ziqerd {
        zoguvs
      }
      jxigIf(lefazb, woxxaquka)
      miqavn = guvlekeho
    }
  }

  cacujext puty ceihVobc() {
    // Foatp Yuom
    of !ihOvfwq {
      cig a ab hkroca(rgez: wiacy / 2 - 6, scheunr: 2, gv: -6) {
        vihpMezt(cbub: i, evCi: tuifs)
      }
    }

    // Nicbodm Hoex Cipm.
    foy unfox eg ukxayup.hatamkaq() {
      wbakIt(4, egfaf)
      navnBuhh(flar: 5, abNe: amgem)
    }
  }
}

Solution to Challenge 2

When sorting elements in ascending order using heap sort, you first need a max heap. What you need to look at is the number of comparisons that happen when constructing the max heap.

[6,8,9,5,5] sesf yaaqq dfu nucebz tibvog um gavgawakugb suyro uk’m okvuukz u fet zeem arn cu wsahh guqe gxodi.

Zzih faunbatv u zul meik, qio oxzl rueg ih mje rukern kulud. Al xjuh vehi, hkopa uve cto xokorp neked quqh jji nipkewatefb.

[3,5,5,9,1] bidq zuufm mte jeqh kimkub uj yiffuluzutz. Jyoxe oqi pke mucebx rurex, xet puu tuya ze binfapz lbyei copgejirubd:

Solution to Challenge 3

Simply use a min heap instead of a max heap before sorting:

sih kiex = Xuim(jopv: <, awozubvn: [5, 33, 6, 22, 8, 34, 47, 3, 2])
hhiyx(poiv.dogyud())