VK

VikashKale

Programmer and Softwere Engineer at None Updated Jan. 1, 2021, 6:25 p.m. ⋅ 115 views

TCS NQT/Digital Coding Program- 1 The product of Quadruplet


*A non-empty arrary A consisting of numeric values is given.

The product of quadruplet (P,Q,R,S) equates to A[P] * A[Q] * A[R] * A[S]

(0 <= P < Q < R < S < N).

For Example, Arrary A Such that.

A[0] = -3, A[1]  = 1,  A[2] = 2,  A[3] = -2, A[4] = 5, A[5] = 6, A[6] = 1

  • (0, 1, 2, 3) , product is  -3 * 1 * 2 * -2 = 12
  • (1, 2, 4, 5) , product is   1 * 2 * 5 * 6 = 60
  • (2, 4, 5, 6) , product is   2 * 5 * 6 * 1 = -60

60 is the product of quadruplet (2, 4, 5, 6), which is maximal.

Your goal is to find the maximal product of any quadruplet for input ArraryA[].

write an efficient algorithm for the following assumptions.

  • N is an integer/float within the range[4 ..... 100,000];

Solution Code:-

number = int (input ("Enter count of value you are going to insert:"))

print ("Enter", number, "Numbers :-")

elements = [int (input()) for i in range (0, number)]

print (elements)

val = [] 

if (number >= 4):

    for indexA in range (0, number):

        listA = []

        listB = []

        quad = 0

        if (number-indexA >= 4):

            print ("indexA =", indexA)

            listA.append (elements [indexA])

            listB.append (indexA)

            for indexB in range (indexA + 1, number):

                print ("index - quad", indexB, quad)

                if elements [indexA] <= elements [indexB]:

                    listA.append (elements [indexB])

                    listB.append (indexB)
                    
                    quad += 1

                    print ("inside quad", indexB, quad)

                    if quad == 3:

                        print ("break")

                        break

            if (len (listB) == 4):

                print (listB [0], listB [1], listB[2], listB [3],"......", listA[0],"*", listA[1], "*", listA[2], "*", listA[3], "-", listA[0] * listA[1] * listA[2] *listA[3])

                val.append (listA[0] * listA[1] * listA[2] *listA[3])

                listA.clear ()

                listB.clear ()

    print (" Maximum: ", max(val))

else:

    print (" Invalid Input ")

 



Go back to feed

HackersFriend Updates





View more


Events


Nov. 28, 2018, 5:30 p.m.

Python from zero to hero

place Delhi

View details


Aug. 13, 2018, 5:30 p.m.

Python from zero to hero

place Bangalore ( HackersFriend office BTM Layout)

View details