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 ")
