Answer Posted / utkarsh

/*The maximum value of the product fr a set of 3 numbers positiv or negative can be
->either the product of 2 negative numbers( the ones with highest absolut values) and 1(biggest) positive number
->or product of three biggest positive numbers
->zero if all other numbers (except that 0) ar negativ
-> product of smallest 3 negativ number if all numbers are negative
>>>
<you can try it for ANY POSSIBLE VALUES..in th array and see..>
<<<<
*/

procedure
I)
so first IF ARRAY HAS LESS THAN 3 ELEMENTS THROW ERROR
else
sort the numbers in 2 arrays..
neg[]-> with negativ values in descending order
pos[]-> with positive values in descending order
->if there is a zero set a boolean value of a variabl ISZERO as TRUE

II) IF (neg[] has zero or only 1 element) THEN

/* product of the three highest values in pos[] (ie the first three values) is the answer*/

RETURN (pos[0]*pos[1]*pos[2]);

III) ELSE IF pos[] is empty
/*(meaning there are only negative values..multiply the smallest three negative numbers (ie say -1,-2,-3,-4,-5 means multiply -1,-2,-3 you get -6)..that is the answer */


if pos[] is NULL AND if there is a zero then
RETURN ZERO;
else //if pos[] is null and no element is zero then
RETURN neg[neg.length-1]*neg[neg.length-2]*neg[neg.length-3]

/*
//NOTE:

there are only negative values and zeros
highest value is 0
check that too from ISZERO variabl..
*/

IV)ONLY if neg[] array has more than 1 value THEN
multiply th biggest 2 negativ absolute numbers


/*
//(eg {-9,-7,-6,-5 } means...tak -9 and -7..so product is 63)

V) multiply th above result with th highest postiv value from positive array...(ie pos[])

VI) compare this value got in STEP FIVE with product of the three highest values in pos[](ie product of first three values) which ever is greate is the answer
ie
*/

IF (neg[0]*neg[1]*pos[0])>(pos[0]*pos[1]*pos[2]) THEN
RETURN (neg[0]*neg[1]*pos[0])
ELSE RETURN (pos[0]*pos[1]*pos[2])


IT IS VERY EASY TO CODE THIS BECAUSE THE STEPS ARE ALMOST IN PSUEDOCODE....hope it helps...

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