Replace the array element with the multiples of all other elements throughout the array with O(n)

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I have an array arr = [3,4,2,5,6,8,2]. Every element in the array has to be replaced with the multiples of all other elements in the array. e.g index=0, value 3 must be replaced with 4*2*5*6*8*2 and index=3, value 5 must be replaced with 3*4*2*6*8*2. And this must be done in O(n). I have solutions with dequeue and stack but i end up doing n*n-1 iterations resulting in O(n^2). Any help?

First, calculate the total Product of this array. Then, for each element in the array divide by that value and assign the result to its position. So the total product is done in O(n), the assignment is also done in O(n).

=> ALG: O(2*n) = O(n)

Replace every array element by multiplication of previous and next , Given an array of integers, update every element with multiplication of previous and next elements with following exceptions. An efficient solution can solve the problem in O(n) time and O(1) space. Replace every element of the array with BitWise XOR of all other � Replace each element of Array with it's� Naive approach would be to calculate product of all elements in the left sub-array and right sub-array for each element of the array. The time complexity of above solution is O(n 2 ) . We can solve this problem in linear time by using two auxiliary arrays left[] and right[] where left[i] stores the product of all elements in the sub-array A[0

  1. Form two arrays: one cumulative product from the left, and one cumulative product from the right. They can each be constructed in O(2*n).
  2. Take the left-cumulative product for the element to the left and multiply by the right-cumulative product to the right. That takes O(n).

Something like:

leftCumulative[1]=1;
for(j=2;j<=n;++j) leftCumulative[j]=leftCumulative[j-1]*arr[j];
rightCumulative[n]=1;
for(j=n-1;j>=1;--j) rightCumulative[j]=arr[j]*rightCumulative[j+1];

arr[1]=rightCumulative[2];
arr[n]=leftCumulative[n-1];
for(j=2;j<=n-1;++j) arr[j]=leftCumulative[j-1]*rightCumulative[j+1];

It is possible to use less extra memory, and make the loop cleaner, but this solves the stated problem. It even works if multiplication is non-commutative!

Replace every element of the array by product of all other elements , Replace every element by the product of all other elements of the array First, take the product of all the element of the array. void ReplaceElements( int arr[], int n) array such that frequency sum of all repeating elements in product is anything incorrect by clicking on the "Improve Article" button below. Algorithm to replace each element of an array by multiplication of previous and next. Declare and initialize an array. For the first element multiply arr[0] and arr[1] and save to arr[0]. For the last element multiply arr[n – 1] and arr[n – 2] and save to arr[n – 1] declare a variable for the previous element. Consider previous.

Using java,

  int[] arr = {3, 4, 2, 5, 6, 8, 2};

  int wholeProduct = arr[0];   //  wholeProduct = arr[0] * arr[1] * ... * arr[n-1]
  for(int i=1; i<arr.length; i++)
     wholeProduct *= arr[i];

  for(int i=0; i<arr.length; i++) 
     arr[i] = wholeProduct/arr[i];

Same technique described by @Lecagy.

Please Note, when the contains many elements, the wholeProduct may not contains the result, you need to change the type to long or BigInteger.

Replace each element of an array with product of every other , of the array. The time complexity of above solution is O(n2). of every other element without using division operator right[i] stores the product of all elements in sub-array[i+1..n-1] Now, multiple array1 and array2 at each index to get the final array (array2 can be avoided by taking care of the multiplication inline). Naive approach would be to calculate product of all elements in the left sub-array and right sub-array for each element of the array. The time complexity of above solution is O(n2). We can solve this problem in linear time by using two auxiliary arrays left[] and right[] where left[i] stores the product of all elements in the sub-array A[0..i-1

Chapter 7: Arrays, When creating an array, you specify the number of elements in the array as follows: where the index can be any expression that results in an int. If we changed the numbers array to have a different number of elements, this code would no A method for replacing a value in an array can be based on the same code. All� The idea is to store each element’s index in an ordered map (Since the array contains all distinct integers, we can use array elements and their index as key-value pair in the map). Since elements are stored in sorted order in ordered map, if we iterate through the map, we get elements in increasing order.

5. Working with Arrays and Loops - JavaScript Cookbook , On the positive side, though, array literals can replace the need for temporary variables, An array, whether literal or object, can hold values of different data types: Arrays in JavaScript are zero-based, which means the first element index is zero, and the last Not all array elements have to be defined when created. Time complexity O(n*n). Method 2: (Optimized tricky solution) Replace all elements using one traversal of the array. Start from the rightmost element, move to the left side one by one, and keep track of the maximum element. Replace every element with the maximum element.

Work with object and array data, represent all changes to a given path and its subproperties, including array mutations. Array item paths (like foo.11 ) represent an item in an array. Likewise, calling set on an array property won't cause Polymer to pick up These methods perform the mutation action on the array, and then notify other elements that may� Let's say we have an array of ints like this: const int size = 100000; int array[size]; //set some items to 0 and other items to 1 I'd like to replace all items that have value of 1 with another

Comments
  • And what if the array contains one zero?
  • well, thats a good point. one could loop once through the array and determine if there is non, one or multiples zeros: if non, the above solution hold. if multiple, all entries are zero. if there is only one zero, just multiply all entries except zero und assign the value in there. all others are zero nontheless
  • Had the same doubt of a zero occurance.
  • "It even works if multiplication is non-commutative!" I know that you mean it even works for other operations which aren't necessarily commutative, but this statement is quite amusing if taken literally. This solution is nice, anyway, because it doesn't require handling special cases for how many zeroes are in the array. +1
  • i suppose you mean ++j in line 2 and --j in line 4?