Given an array of numbers, return array of products of all other numbers (no division)

Question

I was asked this question in a job interview, and I'd like to know how others would solve it. I'm most comfortable with Java, but solutions in other languages are welcome.

Given an array of numbers, nums, return an array of numbers products, where products[i] is the product of all nums[j], j != i.

Input : [1, 2, 3, 4, 5]
Output: [(2*3*4*5), (1*3*4*5), (1*2*4*5), (1*2*3*5), (1*2*3*4)]
      = [120, 60, 40, 30, 24]

You must do this in O(N) without using division.

Answer 1

php version
using array_product function without division.
If we set i value to 1 temporary, than array product will do exactly what we need

<?php
function product($key, $arr)
{
    $arr[$key] = 1;
    return array_product($arr);
};
$arr = [1, 2, 3, 4, 5];
$newarr = array();


foreach ($arr as $key => $value) {

    $newarr[$key] = product($key, $arr);
}
print_r($newarr);

Answer 2

Here is a small recursive function (in C++) to do the modification in-place. It requires O(n) extra space (on stack) though. Assuming the array is in a and N holds the array length, we have:

int multiply(int *a, int fwdProduct, int indx) {
    int revProduct = 1;
    if (indx < N) {
       revProduct = multiply(a, fwdProduct*a[indx], indx+1);
       int cur = a[indx];
       a[indx] = fwdProduct * revProduct;
       revProduct *= cur;
    }
    return revProduct;
}

Answer 3

An explanation of polygenelubricants method is:

The trick is to construct the arrays (in the case for 4 elements):

{              1,         a[0],    a[0]*a[1],    a[0]*a[1]*a[2],  }
{ a[1]*a[2]*a[3],    a[2]*a[3],         a[3],                 1,  }

Both of which can be done in O(n) by starting at the left and right edges respectively.

Then, multiplying the two arrays element-by-element gives the required result.

My code would look something like this:

int a[N] // This is the input
int products_below[N];
int p = 1;
for (int i = 0; i < N; ++i) {
    products_below[i] = p;
    p *= a[i];
}

int products_above[N];
p = 1;
for (int i = N - 1; i >= 0; --i) {
    products_above[i] = p;
    p *= a[i];
}

int products[N]; // This is the result
for (int i = 0; i < N; ++i) {
    products[i] = products_below[i] * products_above[i];
}

If you need the solution be O(1) in space as well, you can do this (which is less clear in my opinion):

int a[N] // This is the input
int products[N];

// Get the products below the current index
int p = 1;
for (int i = 0; i < N; ++i) {
    products[i] = p;
    p *= a[i];
}

// Get the products above the current index
p = 1;
for (int i = N - 1; i >= 0; --i) {
    products[i] *= p;
    p *= a[i];
}

Answer 4

def productify(arr, prod, i):
    if i < len(arr):
        prod.append(arr[i - 1] * prod[i - 1]) if i > 0 else prod.append(1)
        retval = productify(arr, prod, i + 1)
        prod[i] *= retval
        return retval * arr[i]
    return 1

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5]
    prod = []
    productify(arr, prod, 0)
    print(prod)

Answer 5

We're factoring elements of the array, first from before the index i.e. the prefix then after the index or postfix

class Solution:

   def productExceptSelf(nums):

      length = len(nums)


      result = [1] * length


      prefix_product = 1


      postfix_product = 1

# we initialize the result and products


      for i in range(length)

      result[i] *= prefix_product


       prefix_product *= nums[i]

#we multiply the result by each number before the index

      for i in range(length-1,-1,-1)

      result[i] *= postfix_product


      postfix_product *= nums[i]

#same for after index
   return result

sorry on mobile on my walk

Answer 6

def products(nums):
    prefix_products = []
    for num in nums:
        if prefix_products:
            prefix_products.append(prefix_products[-1] * num)
        else:
            prefix_products.append(num)

    suffix_products = []
    for num in reversed(nums):
        if suffix_products:
            suffix_products.append(suffix_products[-1] * num)
        else:
            suffix_products.append(num)
        suffix_products = list(reversed(suffix_products))

    result = []
    for i in range(len(nums)):
        if i == 0:
            result.append(suffix_products[i + 1])
        elif i == len(nums) - 1:
            result.append(prefix_products[i-1])
        else:
            result.append(
                prefix_products[i-1] * suffix_products[i+1]
            )
    return result

Answer 7

I came up with 2 solutions in Javascript, one with division and one without

// without division
function methodOne(arr) {
  return arr.map(item => {
    return arr.reduce((result, num) => {
      if (num !== item) {
        result = result * num;
      }
      return result;
    },1)
  });
}

// with division
function methodTwo(arr) {
  var mul = arr.reduce((result, num) => {
    result = result * num;
    return result;
  },1)
 return arr.map(item => mul/item);
}

console.log(methodOne([1, 2, 3, 4, 5]));
console.log(methodTwo([1, 2, 3, 4, 5]));

Answer 8

We can exclude the nums[j] (where j != i) from list first, then get the product of the rest; The following is a python way to solve this puzzle:

from functools import reduce
def products(nums):
    return [ reduce(lambda x,y: x * y, nums[:i] + nums[i+1:]) for i in range(len(nums)) ]
print(products([1, 2, 3, 4, 5]))

[out]
[120, 60, 40, 30, 24]

Answer 9

Just 2 passes up and down. Job done in O(N)

private static int[] multiply(int[] numbers) {
        int[] multiplied = new int[numbers.length];
        int total = 1;

        multiplied[0] = 1;
        for (int i = 1; i < numbers.length; i++) {
            multiplied[i] = numbers[i - 1] * multiplied[i - 1];
        }

        for (int j = numbers.length - 2; j >= 0; j--) {
            total *= numbers[j + 1];
            multiplied[j] = total * multiplied[j];
        }

        return multiplied;
    }

Answer 10

A variation in JavaScript using reduce

const getProduct = arr => arr.reduce((acc, value) => acc * value);

const arrayWithExclusion = (arr, node) =>
  arr.reduce((acc, val, j) => (node !== j ? [...acc, val] : acc), []);

const getProductWithExclusion = arr => {
  let result = [];

  for (let i = 0; i < arr.length; i += 1) {
    result.push(getProduct(arrayWithExclusion(arr, i)));
  }

  return result;
};

Answer 11

int[] b = new int[] { 1, 2, 3, 4, 5 };            
int j;
for(int i=0;i<b.Length;i++)
{
  int prod = 1;
  int s = b[i];
  for(j=i;j<b.Length-1;j++)
  {
    prod = prod * b[j + 1];
  }
int pos = i;    
while(pos!=-1)
{
  pos--;
  if(pos!=-1)
     prod = prod * b[pos];                    
}
Console.WriteLine("\n Output is {0}",prod);
}

Answer 12

Here is a C implementation
O(n) time complexity.
INPUT

#include<stdio.h>
int main()
{
    int x;
    printf("Enter The Size of Array : ");
    scanf("%d",&x);
    int array[x-1],i ;
    printf("Enter The Value of Array : \n");
      for( i = 0 ; i <= x-1 ; i++)
      {
          printf("Array[%d] = ",i);
          scanf("%d",&array[i]);
      }
    int left[x-1] , right[x-1];
    left[0] = 1 ;
    right[x-1] = 1 ;
      for( i = 1 ; i <= x-1 ; i++)
      {
          left[i] = left[i-1] * array[i-1];
      }
    printf("\nThis is Multiplication of array[i-1] and left[i-1]\n");
      for( i = 0 ; i <= x-1 ; i++)
      {
        printf("Array[%d] = %d , Left[%d] = %d\n",i,array[i],i,left[i]);
      }
      for( i = x-2 ; i >= 0 ; i--)
      {
          right[i] = right[i+1] * array[i+1];
      }
   printf("\nThis is Multiplication of array[i+1] and right[i+1]\n");
      for( i = 0 ; i <= x-1 ; i++)
      {
        printf("Array[%d] = %d , Right[%d] = %d\n",i,array[i],i,right[i]);
      }
    printf("\nThis is Multiplication of Right[i] * Left[i]\n");
      for( i = 0 ; i <= x-1 ; i++)
      {
          printf("Right[%d] * left[%d] = %d * %d = %d\n",i,i,right[i],left[i],right[i]*left[i]);
      }
    return 0 ;
}

---

OUTPUT

    Enter The Size of Array : 5
    Enter The Value of Array :
    Array[0] = 1
    Array[1] = 2
    Array[2] = 3
    Array[3] = 4
    Array[4] = 5

    This is Multiplication of array[i-1] and left[i-1]
    Array[0] = 1 , Left[0] = 1
    Array[1] = 2 , Left[1] = 1
    Array[2] = 3 , Left[2] = 2
    Array[3] = 4 , Left[3] = 6
    Array[4] = 5 , Left[4] = 24

    This is Multiplication of array[i+1] and right[i+1]
    Array[0] = 1 , Right[0] = 120
    Array[1] = 2 , Right[1] = 60
    Array[2] = 3 , Right[2] = 20
    Array[3] = 4 , Right[3] = 5
    Array[4] = 5 , Right[4] = 1

    This is Multiplication of Right[i] * Left[i]
    Right[0] * left[0] = 120 * 1 = 120
    Right[1] * left[1] = 60 * 1 = 60
    Right[2] * left[2] = 20 * 2 = 40
    Right[3] * left[3] = 5 * 6 = 30
    Right[4] * left[4] = 1 * 24 = 24

    Process returned 0 (0x0)   execution time : 6.548 s
    Press any key to continue.

Answer 13

ruby solution

a = [1,2,3,4]
result = []
a.each {|x| result.push( (a-[x]).reject(&:zero?).reduce(:*)) }
puts result

Answer 14

Here is my concise solution using python.

from functools import reduce

def excludeProductList(nums_):
    after = [reduce(lambda x, y: x*y, nums_[i:]) for i in range(1, len(nums_))] + [1]
    before = [1] + [reduce(lambda x, y: x*y, nums_[:i]) for i in range(1, len(nums_))]
    zippedList =  list(zip(before, after))
    finalList = list(map(lambda x: x[0]*x[1], zippedList))
    return finalList

Answer 15

My first try, in Python. O(2n):

def product(l):
    product = 1
    num_zeroes = 0
    pos_zero = -1

    # Multiply all and set positions
    for i, x in enumerate(l):
        if x != 0:
            product *= x
            l[i] = 1.0/x
        else:
            num_zeroes += 1
            pos_zero = i

    # Warning! Zeroes ahead!
    if num_zeroes > 0:
        l = [0] * len(l)

        if num_zeroes == 1:
            l[pos_zero] = product

    else:
        # Now set the definitive elements
        for i in range(len(l)):
            l[i] = int(l[i] * product)

    return l


if __name__ == "__main__":
    print("[0, 0, 4] = " + str(product([0, 0, 4])))
    print("[3, 0, 4] = " + str(product([3, 0, 4])))
    print("[1, 2, 3] = " + str(product([1, 2, 3])))
    print("[2, 3, 4, 5, 6] = " + str(product([2, 3, 4, 5, 6])))
    print("[2, 1, 2, 2, 3] = " + str(product([2, 1, 2, 2, 3])))

Output:

[0, 0, 4] = [0, 0, 0]
[3, 0, 4] = [0, 12, 0]
[1, 2, 3] = [6, 3, 2]
[2, 3, 4, 5, 6] = [360, 240, 180, 144, 120]
[2, 1, 2, 2, 3] = [12, 24, 12, 12, 8]

Answer 16

import java.util.Arrays;

public class Pratik
{
    public static void main(String[] args)
    {
        int[] array = {2, 3, 4, 5, 6};      //  OUTPUT: 360  240  180  144  120
        int[] products = new int[array.length];
        arrayProduct(array, products);
        System.out.println(Arrays.toString(products));
    }

    public static void arrayProduct(int array[], int products[])
    {
        double sum = 0, EPSILON = 1e-9;

        for(int i = 0; i < array.length; i++)
            sum += Math.log(array[i]);

        for(int i = 0; i < array.length; i++)
            products[i] = (int) (EPSILON + Math.exp(sum - Math.log(array[i])));
    }
}

OUTPUT:

[360, 240, 180, 144, 120]

Time complexity : O(n)

Space complexity: O(1)

Answer 17

Here's a one-liner solution in Ruby.

nums.map { |n| (num - [n]).inject(:*) }

Answer 18

Here is simple Scala version in Linear O(n) time:

def getProductEff(in:Seq[Int]):Seq[Int] = {

   //create a list which has product of every element to the left of this element
   val fromLeft = in.foldLeft((1, Seq.empty[Int]))((ac, i) => (i * ac._1, ac._2 :+ ac._1))._2

   //create a list which has product of every element to the right of this element, which is the same as the previous step but in reverse
   val fromRight = in.reverse.foldLeft((1,Seq.empty[Int]))((ac,i) => (i * ac._1,ac._2 :+ ac._1))._2.reverse

   //merge the two list by product at index
   in.indices.map(i => fromLeft(i) * fromRight(i))

}

This works because essentially the answer is an array which has product of all elements to the left and to the right.

Answer 19

I'm use to C#:

    public int[] ProductExceptSelf(int[] nums)
    {
        int[] returnArray = new int[nums.Length];
        List<int> auxList = new List<int>();
        int multTotal = 0;

        // If no zeros are contained in the array you only have to calculate it once
        if(!nums.Contains(0))
        {
            multTotal = nums.ToList().Aggregate((a, b) => a * b);

            for (int i = 0; i < nums.Length; i++)
            {
                returnArray[i] = multTotal / nums[i];
            }
        }
        else
        {
            for (int i = 0; i < nums.Length; i++)
            {
                auxList = nums.ToList();
                auxList.RemoveAt(i);
                if (!auxList.Contains(0))
                {
                    returnArray[i] = auxList.Aggregate((a, b) => a * b);
                }
                else
                {
                    returnArray[i] = 0;
                }
            }
        }            

        return returnArray;
    }

Answer 20

Adding my javascript solution here as I didn't find anyone suggesting this. What is to divide, except to count the number of times you can extract a number from another number? I went through calculating the product of the whole array, and then iterate over each element, and substracting the current element until zero:

//No division operation allowed
// keep substracting divisor from dividend, until dividend is zero or less than divisor
function calculateProducsExceptCurrent_NoDivision(input){
  var res = [];
  var totalProduct = 1;
  //calculate the total product
  for(var i = 0; i < input.length; i++){
    totalProduct = totalProduct * input[i];
  }
  //populate the result array by "dividing" each value
  for(var i = 0; i < input.length; i++){
    var timesSubstracted = 0;
    var divisor = input[i];
    var dividend = totalProduct;
    while(divisor <= dividend){
      dividend = dividend - divisor;
      timesSubstracted++;
    }
    res.push(timesSubstracted);
  }
  return res;
}

Answer 21

I got asked this question recently, and whilst I couldn't get O(N) during it, I had a different approach (unfortunately O(N^2)) but thought I'd share anyway.

Convert to List<Integer> first.

Loop through original array array.length() times.

Use a while loop to multiple the next set of required numbers:

while (temp < list.size() - 1) {
    res *= list.get(temp);
    temp++;
}

Then add res to a new array (which of course you've declared earlier), then add the value at array[i] to the List, and continue so forth.

I know this won't be of great use, but it's what I came up with under the pressures of an interview :)

    int[] array = new int[]{1, 2, 3, 4, 5};
    List<Integer> list = Arrays.stream(array).boxed().collect(Collectors.toList());
    int[] newarray = new int[array.length];
    int res = 1;
    for (int i = 0; i < array.length; i++) {
        int temp = i;
        while (temp < list.size() - 1) {
            res *= list.get(temp);
            temp++;
        }
        newarray[i] = res;
        list.add(array[i]);
        res = 1;
    }

Output: [24, 120, 60, 40, 30]

Answer 22

Here is the ptyhon version

  # This solution use O(n) time and O(n) space
  def productExceptSelf(self, nums):
    """
    :type nums: List[int]
    :rtype: List[int]
    """
    N = len(nums)
    if N == 0: return

    # Initialzie list of 1, size N
    l_prods, r_prods = [1]*N, [1]*N

    for i in range(1, N):
      l_prods[i] = l_prods[i-1] * nums[i-1]

    for i in reversed(range(N-1)):
      r_prods[i] = r_prods[i+1] * nums[i+1]

    result = [x*y for x,y in zip(l_prods,r_prods)]
    return result

  # This solution use O(n) time and O(1) space
  def productExceptSelfSpaceOptimized(self, nums):
    """
    :type nums: List[int]
    :rtype: List[int]
    """
    N = len(nums)
    if N == 0: return

    # Initialzie list of 1, size N
    result = [1]*N

    for i in range(1, N):
      result[i] = result[i-1] * nums[i-1]

    r_prod = 1
    for i in reversed(range(N)):
      result[i] *= r_prod
      r_prod *= nums[i]

    return result

Answer 23

I have a solution with O(n) space and O(n^2) time complexity provided below,

public static int[] findEachElementAsProduct1(final int[] arr) {

        int len = arr.length;

//        int[] product = new int[len];
//        Arrays.fill(product, 1);

        int[] product = IntStream.generate(() -> 1).limit(len).toArray();


        for (int i = 0; i < len; i++) {

            for (int j = 0; j < len; j++) {

                if (i == j) {
                    continue;
                }

                product[i] *= arr[j];
            }
        }

        return product;
    }

Answer 24

Based on Billz answer--sorry I can't comment, but here is a scala version that correctly handles duplicate items in the list, and is probably O(n):

val list1 = List(1, 7, 3, 3, 4, 4)
val view = list1.view.zipWithIndex map { x => list1.view.patch(x._2, Nil, 1).reduceLeft(_*_)}
view.force

returns:

List(1008, 144, 336, 336, 252, 252)

Answer 25

Here is another simple concept which solves the problem in O(N).

        int[] arr = new int[] {1, 2, 3, 4, 5};
        int[] outArray = new int[arr.length]; 
        for(int i=0;i<arr.length;i++){
            int res=Arrays.stream(arr).reduce(1, (a, b) -> a * b);
            outArray[i] = res/arr[i];
        }
        System.out.println(Arrays.toString(outArray));

Answer 26

Coded up using EcmaScript 2015

'use strict'

/*
Write a function that, given an array of n integers, returns an array of all possible products using exactly (n - 1) of those integers.
*/
/*
Correct behavior:
- the output array will have the same length as the input array, ie. one result array for each skipped element
- to compare result arrays properly, the arrays need to be sorted
- if array lemgth is zero, result is empty array
- if array length is 1, result is a single-element array of 1

input array: [1, 2, 3]
1*2 = 2
1*3 = 3
2*3 = 6
result: [2, 3, 6]
*/
class Test {
  setInput(i) {
    this.input = i
    return this
  }
  setExpected(e) {
    this.expected = e.sort()
    return this
  }
}

class FunctionTester {
  constructor() {
    this.tests = [
      new Test().setInput([1, 2, 3]).setExpected([6, 3, 2]),
      new Test().setInput([2, 3, 4, 5, 6]).setExpected([3 * 4 * 5 * 6, 2 * 4 * 5 * 6, 2 * 3 * 5 * 6, 2 * 3 * 4 * 6, 2 * 3 * 4 * 5]),
    ]
  }

  test(f) {
    console.log('function:', f.name)
    this.tests.forEach((test, index) => {
      var heading = 'Test #' + index + ':'
      var actual = f(test.input)
      var failure = this._check(actual, test)

      if (!failure) console.log(heading, 'input:', test.input, 'output:', actual)
      else console.error(heading, failure)

      return !failure
    })
  }

  testChain(f) {
    this.test(f)
    return this
  }

  _check(actual, test) {
      if (!Array.isArray(actual)) return 'BAD: actual not array'
      if (actual.length !== test.expected.length) return 'BAD: actual length is ' + actual.length + ' expected: ' + test.expected.length
      if (!actual.every(this._isNumber)) return 'BAD: some actual values are not of type number'
      if (!actual.sort().every(isSame)) return 'BAD: arrays not the same: [' + actual.join(', ') + '] and [' + test.expected.join(', ') + ']'

      function isSame(value, index) {
        return value === test.expected[index]
      }
  }

  _isNumber(v) {
    return typeof v === 'number'
  }
}

/*
Efficient: use two iterations of an aggregate product
We need two iterations, because one aggregate goes from last-to-first
The first iteration populates the array with products of indices higher than the skipped index
The second iteration calculates products of indices lower than the skipped index and multiplies the two aggregates

input array:
1 2 3
   2*3
1*    3
1*2

input array:
2 3 4 5 6
    (3 * 4 * 5 * 6)
(2) *     4 * 5 * 6
(2 * 3) *     5 * 6
(2 * 3 * 4) *     (6)
(2 * 3 * 4 * 5)

big O: (n - 2) + (n - 2)+ (n - 2) = 3n - 6 => o(3n)
*/
function multiplier2(ns) {
  var result = []

  if (ns.length > 1) {
    var lastIndex = ns.length - 1
    var aggregate

    // for the first iteration, there is nothing to do for the last element
    var index = lastIndex
    for (var i = 0; i < lastIndex; i++) {
      if (!i) aggregate = ns[index]
      else aggregate *= ns[index]
      result[--index] = aggregate
    }

    // for second iteration, there is nothing to do for element 0
    // aggregate does not require multiplication for element 1
    // no multiplication is required for the last element
    for (var i = 1; i <= lastIndex; i++) {
      if (i === 1) aggregate = ns[0]
      else aggregate *= ns[i - 1]
      if (i !== lastIndex) result[i] *= aggregate
      else result[i] = aggregate
    }
  } else if (ns.length === 1) result[0] = 1

  return result
}

/*
Create the list of products by iterating over the input array

the for loop is iterated once for each input element: that is n
for every n, we make (n - 1) multiplications, that becomes n (n-1)
O(n^2)
*/
function multiplier(ns) {
  var result = []

  for (var i = 0; i < ns.length; i++) {
    result.push(ns.reduce((reduce, value, index) =>
      !i && index === 1 ? value // edge case: we should skip element 0 and it's the first invocation: ignore reduce
      : index !== i ? reduce * value // multiply if it is not the element that should be skipped
      : reduce))
  }

  return result
}

/*
Multiply by clone the array and remove one of the integers

O(n^2) and expensive array manipulation
*/
function multiplier0(ns) {
  var result = []

  for (var i = 0; i < ns.length; i++) {
    var ns1 = ns.slice() // clone ns array
    ns1.splice(i, 1) // remove element i
    result.push(ns1.reduce((reduce, value) => reduce * value))
  }

  return result
}

new FunctionTester().testChain(multiplier0).testChain(multiplier).testChain(multiplier2)

run with Node.js v4.4.5 like:

node --harmony integerarrays.js

function: multiplier0
Test #0: input: [ 1, 2, 3 ] output: [ 2, 3, 6 ]
Test #1: input: [ 2, 3, 4, 5, 6 ] output: [ 120, 144, 180, 240, 360 ]
function: multiplier
Test #0: input: [ 1, 2, 3 ] output: [ 2, 3, 6 ]
Test #1: input: [ 2, 3, 4, 5, 6 ] output: [ 120, 144, 180, 240, 360 ]
function: multiplier2
Test #0: input: [ 1, 2, 3 ] output: [ 2, 3, 6 ]
Test #1: input: [ 2, 3, 4, 5, 6 ] output: [ 120, 144, 180, 240, 360 ]

Answer 27

Here is my solution in modern C++. It makes use of std::transform and is pretty easy to remember.

Online code (wandbox).

#include<algorithm>
#include<iostream>
#include<vector>

using namespace std;

vector<int>& multiply_up(vector<int>& v){
    v.insert(v.begin(),1);
    transform(v.begin()+1, v.end()
             ,v.begin()
             ,v.begin()+1
             ,[](auto const& a, auto const& b) { return b*a; }
             );
    v.pop_back();
    return v;
}

int main() {
    vector<int> v = {1,2,3,4,5};
    auto vr = v;

    reverse(vr.begin(),vr.end());
    multiply_up(v);
    multiply_up(vr);
    reverse(vr.begin(),vr.end());

    transform(v.begin(),v.end()
             ,vr.begin()
             ,v.begin()
             ,[](auto const& a, auto const& b) { return b*a; }
             );

    for(auto& i: v) cout << i << " "; 
}

Answer 28

Try this!

import java.util.*;
class arrProduct
{
 public static void main(String args[])
     {
         //getting the size of the array
         Scanner s = new Scanner(System.in);
            int noe = s.nextInt();

        int out[]=new int[noe];
         int arr[] = new int[noe];

         // getting the input array
         for(int k=0;k<noe;k++)
         {
             arr[k]=s.nextInt();
         }

         int val1 = 1,val2=1;
         for(int i=0;i<noe;i++)
         {
             int res=1;

                 for(int j=1;j<noe;j++)
                 {
                if((i+j)>(noe-1))
                {

                    int diff = (i+j)-(noe);

                    if(arr[diff]!=0)
                    {
                    res = res * arr[diff];
                    }
                }

                else
                {
                    if(arr[i+j]!=0)
                    {
                    res= res*arr[i+j];
                    }
                }


             out[i]=res;

         }
         }

         //printing result
         System.out.print("Array of Product: [");
         for(int l=0;l<out.length;l++)
         {
             if(l!=out.length-1)
             {
            System.out.print(out[l]+",");
             }
             else
             {
                 System.out.print(out[l]);
             }
         }
         System.out.print("]");
     }

}

Answer 29

1. Travel Left->Right and keep saving product. Call it Past. -> O(n)
2. Travel Right -> left keep the product. Call it Future. -> O(n)
3. Result[i] = Past[i-1] * future[i+1] -> O(n)
4. Past[-1] = 1; and Future[n+1]=1;

O(n)

Answer 30

A neat solution with O(n) runtime:

1. For each element calculate the product of all the elements that occur before that and it store in an array "pre".
2. For each element calculate the product of all the elements that occur after that element and store it in an array "post"

3. Create a final array "result", for an element i,

   result[i] = pre[i-1]*post[i+1];