Rearrange given Array such that all set bit positions have higher value than others

Given an array B1[] and a binary array B2[] each of size N, the task is to rearrange array B1[] in a way such that for all setbit position i of B2[] the value of B1[i] will be greater than the values where bit is not set in B2[] i.e for all (i, j)B1[i] > B1[j] when B2[i] = 1, B2[j] = 0 (0 ≤ i, j < N).

Note: If multiple arrangements are possible, print any one of them.

Examples:

Input: N = 7, B1[] = {1, 2, 3, 4, 5, 6, 7}, B2[] = {0, 1, 0, 1, 0, 1, 0}
Output: 1 5 2 6 3 7 4
Explanation: Values having 1 in B2[] array are = {2, 4, 6} and values with 0s are = {1, 3, 5, 7}.
So, in correspondent to B2[] array, B1[] can be rearranged = {1, 5, 2, 6, 3, 7, 4}
B1[] = {1, 5, 2, 6, 3, 7, 4}
B2[] = {0, 1, 0, 1, 0, 1, 0}, now all places in B1[i] > B1[j] where B2[i] = 1, and B2[j] = 0.
It also can be rearranged as {1, 7, 2, 6, 3, 5, 4}

Input: N = 3, B1[] = {4, 3, 5}, B2[] = {1, 1, 1}
Output: 5 4 3

Approach: The problem can be solved based on the following idea:

To arrange B1[] as per set bit positions of B2[], sort the values of B1[] and arrange the higher values in positions with bits set and the lower values in positions in other positions.

Follow the steps mentioned below to implement the above idea:

• Make a list of pairs (say v) with the bit value of B2[] and the corresponding value at B1[],
• Sort the list of pairs in ascending order of bits value of B2[].
• Rearrange B1[] using two pointer approach.
  • Declare left pointer with 0 and right pointer with N-1.
  • When B2[i] is 0 (0 ≤ i < N), set B1[] as v[left] and increment left.
  • Otherwise, set B1[i] as v[right] and decrement right.
• Return B1[] as the final answer.

Below is the implementation of the above approach :

5 4 3 

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

Customized Sorting and Placement Strategy Based on Flag Array

The approach described here is utilized to reorder an array B1 in a manner dictated by a corresponding flag array B2. Elements in B1 are rearranged such that those associated with a '1' in B2 are placed first, sorted in descending order, and those associated with a '0' are sorted in ascending order and placed afterwards. This method is particularly effective when you need to prioritize certain elements based on binary conditions while maintaining an ordered sequence within those priorities. It is suitable for applications where dynamic reordering is necessary based on external or updated criteria, such as in task scheduling, prioritizing resources, or data processing tasks where order affects output quality or processing efficiency.

Follow the steps to solve the problem:

• Generate lists of indices where bits are set and unset in B2.
• Separate B1 values into two groups based on the bit values in B2.
• Sort the values where bits are set in descending order and where bits are unset in ascending order.
• Reinsert the sorted values back into their respective positions in B1 according to the layout specified by B2.

Below is the implementation of above approach:

5 4 3 

Time Complexity: O(NlogN)

Auxilary Space: O(N)