Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max

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Question

Find the maximum product of three numbers in an array?
Eg. 9,5,1,2,3 
Max product= 9*5*3= 135
The array can hav negative numbers also..

Question Submitted By :: Bharghavi90

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Answer Posted By

Re: Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max product= 9*5*3= 135 The array can hav negative numbers also..

Answer
# 1

1)Sort the array in descending order.
2)Multiply for required number of indices.

CODE:

int mul=1;
for(i=0;i<3;i++)
{
mul=mul*a[i];
}
cout<<mul;

Is This Answer Correct ?

9 Yes

13 No

Sathish

Re: Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max product= 9*5*3= 135 The array can hav negative numbers also..

Answer
# 2

consider the array: 10,2,-7,-5,3
Descending order: 10,3,2,-5,-7
Then ur result will give: 10*3*2=60
But the max. product can be: 10*(-5)*(-7) =350

Is This Answer Correct ?

3 Yes

8 No

Bharghavi

Re: Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max product= 9*5*3= 135 The array can hav negative numbers also..

Answer
# 3

Just count the number of -ve symbols in the array before sorting.

CODE:
void main()
{
clrscr();
int i,j,count=0,a[5]={10,2,-7,-5,3};
for(i=0;i<5;i++)
if(a[i]<0)
count=count+1;
if((count%2)==0)
{
for(i=0;i<5;i++)
{

if(a[i]<0)
a[i]=-a[i];
}
}
for(i=0;i<5;i++)
{
for(j=i;j<5;j++)
{
if(a[i]<a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
int mul=1;
for(i=0;i<3;i++)
{
mul=mul*a[i];
}
cout<<mul;
getch();
}

Is This Answer Correct ?

8 Yes

7 No

Sathish

Re: Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max product= 9*5*3= 135 The array can hav negative numbers also..

Answer
# 4

/*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...

Is This Answer Correct ?

3 Yes

0 No

Utkarsh

Re: Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max product= 9*5*3= 135 The array can hav negative numbers also..

Answer
# 5

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 THEN
set a boolean value of a variabl ISZERO as TRUE
}

IF pos[] is empty
{
IF ISZERO==true THEN
// if there are only negativ values and a zero
RETURN ZERO;

ELSE
// all elements are -ve and no element is zero then
RETURN (neg[neg.length-1]*neg[neg.length-2]*neg[neg.length-3]);

}

IF (neg[] has zero or only 1 element) THEN
{
RETURN (pos[0]*pos[1]*pos[2]);
}
ELSE IF neg[] array has more than 1 value THEN
{
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]);

}

FOR DETAILS ABT LOGIC READ ABOVE POST

Is This Answer Correct ?

1 Yes

1 No

Utkarsh

Re: Find the maximum product of three numbers in an array? Eg. 9,5,1,2,3 Max product= 9*5*3= 135 The array can hav negative numbers also..

Answer
# 6

#include"stdafx.h"
#include<iostream>

using namespace std;
void sort(int[],int );
int maxproduct(int ar[],int n);
int main()
{
int arr[]={2,7,9,5,3,6,-12};

cout<<maxproduct(arr,7);
system("calc");

system("pause");
return 0;
}
void sort(int array1[],int n)
{
for(int i=1;i<n;i++)
for(int r=0;r<n-1;r++)

if(array1[r]>array1[r+1])
{
int temp=array1[r];
array1[r]=array1[r+1];
array1[r+1]=temp;
}

}
int maxproduct(int ar[],int n)
{
sort(ar,n);
int max=1;
for(int i=n-1;i>=n-3;i--)
max*=ar[i];
return max;
}

Is This Answer Correct ?

1 Yes

1 No

Jack

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