Find a number with least sum of bit differences from given Array

Given an array arr[] and an integer L which represents the number of bits to be considered in bit representation of an array element. The task is to find a positive integer X, such that the sum of the bit difference of all the elements in arr[] with X is minimum.

Examples:

Input: N = 3, L = 5, arr[] = {18, 9, 21}
Output: 17
Explanation: arr[1] = (18)₁₀ = (10010)₂, arr[2] = (9)₁₀ = (01001)₂, arr[3] = (21)₁₀ = (10101)₂ 
Suppose X = (17)₁₀ = (10001)₂.
Difference between arr[1] and X : 2 (Different bits at index 3 and 4)
Difference between arr[2] and X : 2 (Different bits at index 0 and 1)
Difference between arr[3] and X : 1 (Different bits at index 2)
Therefore if X = 17, the sum of bit differences will be 2 + 2 +1 = 5, which is the minimum possible.

Input: N = 3, L = 5 and arr[] = {8, 8, 8}
Output: 8 

Approach: This problem can be solved using below idea:

• At bit position i, the bit in the resultant X should be the same as the majority bit at the ith position of all Array elements.
• This can be achieved using Hashing.

Illustration:

Suppose array is {5, 6, 4} 

Bit representation of array:
5 = 101
6 = 110
4 = 100

Now lets find the X using positionwise majority bit of Array:
At position 1: majority(1, 1, 1) = 1
At position 2: majority(0, 1, 0) = 0
At position 3: majority(1, 0, 0) = 0

Therefore X formed = 100 = 4

Now lets find the bit difference of X with Array elements:
BitDifference(X, 5) = BitDifference(100, 101) = 1
BitDifference(X, 6) = BitDifference(100, 110) = 1
BitDifference(X, 4) = BitDifference(100, 100) = 0

Sum of bit differences = 2, which will be the least possible. 
 

Follow the steps below to solve the given problem. 

• Initialize an array freq of length L which keeps the track of the number of bits set at every position in every given array element.
• Iterate freq and if the frequency of ith index is greater than N/2 then keep to bit set in the required number.
• If the frequency of the ith index is less than N/2 then keep the bit off in the required number.
• Convert the formed binary number into decimal number and return the value.
• Print the final result

Below is the implementation of the above approach.

// C++ program for above approach

#include <iostream>
using namespace std;

// Function to find number
// having smallest difference
// with N numbers
int smallestDifference(int N, int L, int arr[])
{
    // Initializing freq array
    // which keeps tracks of
    // number of set bits at every index
    int freq[L] = { 0 };

    // Making freq map of set bits
    for (int i = 0; i < L; i++) {

        // Traversing every element
        for (int j = 0; j < N; j++) {

            // If bit is on then
            // updating freq array
            if ((arr[j] & 1) > 0) {
                freq[i]++;
            }
            arr[j] >>= 1;
        }
    }

    // Converting binary form of needed
    // number into decimal form
    int number = 0;
    int p = 1;

    // Traversing freq array
    for (int i = 0; i < L; i++) {

        // If frequency of set bit
        // is greater than N/2
        // then we have to keep it set
        // in our answer
        if (freq[i] > N / 2) {
            number += p;
        }
        p *= 2;
    }

    // Returning numbers
    // having smallest difference
    // among N given numbers
    return number;
}
// Driver Code
int main()
{
    int N = 3;
    int L = 5;
    int arr[] = { 18, 9, 21 };

    // Function call
    int number = smallestDifference(N, L, arr);
    cout << number << endl;
    return 0;
}

// Java program for above approach

import java.util.*;

class GFG {

    // Function to find number
    // having smallest difference
    // with N numbers
    public static int smallestDifference(int N, int L,
                                         int[] arr)
    {
        // Initializing freq array
        // which keeps tracks of
        // number of set bits at every index
        int[] freq = new int[L];

        // Making freq map of set bits
        for (int i = 0; i < L; i++) {

            // Traversing every element
            for (int j = 0; j < N; j++) {

                // If bit is on then
                // updating freq array
                if ((arr[j] & 1) > 0) {
                    freq[i]++;
                }
                arr[j] >>= 1;
            }
        }

        // Converting binary form of needed
        // number into decimal form
        int number = 0;
        int p = 1;

        // Traversing freq array
        for (int i = 0; i < L; i++) {

            // If frequency of set bit
            // is greater than N/2
            // then we have to keep it set
            // in our answer
            if (freq[i] > N / 2) {
                number += p;
            }
            p *= 2;
        }

        // Returning numbers
        // having smallest difference
        // among N given numbers
        return number;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 3;
        int L = 5;
        int[] arr = { 18, 9, 21 };

        // Function call
        int number = smallestDifference(N, L, arr);
        System.out.println(number);
    }
}

# Python program for above approach

# Function to find number
# having smallest difference
# with N numbers
def smallestDifference(N, L, arr):

    # Initializing freq array
    # which keeps tracks of
    # number of set bits at every index
    freq = []
    for i in range(0, L):
        freq.append(0)

    # Making freq map of set bits
    for i in range(0, L):

        # Traversing every element
        for j in range(0, N):

            # If bit is on then
            # updating freq array
            if ((arr[j] & 1) > 0):
                freq[i] += 1

            arr[j] >>= 1

    # Converting binary form of needed
    # number into decimal form
    number = 0
    p = 1

    # Traversing freq array
    for i in range(0, L):

        # If frequency of set bit
        # is greater than N/2
        # then we have to keep it set
        # in our answer
        if (freq[i] > N // 2):
            number += p

        p *= 2

    # Returning numbers
    # having smallest difference
    # among N given numbers
    return number

# Driver Code
N = 3
L = 5
arr = [18, 9, 21]

# Function call
number = smallestDifference(N, L, arr)
print(number)

# This code is contributed by Samim Hossain Mondal.

using System;

public class GFG{

  // Function to find number
  // having smallest difference
  // with N numbers
  public static int smallestDifference(int N, int L,
                                       int[] arr)
  {
    // Initializing freq array
    // which keeps tracks of
    // number of set bits at every index
    int[] freq = new int[L];

    // Making freq map of set bits
    for (int i = 0; i < L; i++) {

      // Traversing every element
      for (int j = 0; j < N; j++) {

        // If bit is on then
        // updating freq array
        if ((arr[j] & 1) > 0) {
          freq[i]++;
        }
        arr[j] >>= 1;
      }
    }

    // Converting binary form of needed
    // number into decimal form
    int number = 0;
    int p = 1;

    // Traversing freq array
    for (int i = 0; i < L; i++) {

      // If frequency of set bit
      // is greater than N/2
      // then we have to keep it set
      // in our answer
      if (freq[i] > N / 2) {
        number += p;
      }
      p *= 2;
    }

    // Returning numbers
    // having smallest difference
    // among N given numbers
    return number;
  }

  // Driver Code
  static public void Main ()
  {
    int N = 3;
    int L = 5;
    int[] arr = { 18, 9, 21 };

    // Function call
    int number = smallestDifference(N, L, arr);
    Console.Write(number);
  }
}

// This code is contributed by hrithikgarg03188.

<script>
// Javascript program for above approach

// Function to find number
// having smallest difference
// with N numbers
function smallestDifference(N, L, arr)
{

    // Initializing freq array
    // which keeps tracks of
    // number of set bits at every index
    let freq = [];
    for(let i = 0; i < L; i++) {
        freq[i] = 0;
    }

    // Making freq map of set bits
    for (let i = 0; i < L; i++) {

        // Traversing every element
        for (let j = 0; j < N; j++) {

            // If bit is on then
            // updating freq array
            if ((arr[j] & 1) > 0) {
                freq[i]++;
            }
            arr[j] >>= 1;
        }
    }

    // Converting binary form of needed
    // number into decimal form
    let number = 0;
    let p = 1;

    // Traversing freq array
    for (let i = 0; i < L; i++) {

        // If frequency of set bit
        // is greater than N/2
        // then we have to keep it set
        // in our answer
        if (freq[i] > N / 2) {
            number += p;
        }
        p *= 2;
    }

    // Returning numbers
    // having smallest difference
    // among N given numbers
    return number;
}

// Driver Code
let N = 3;
let L = 5;
let arr = [ 18, 9, 21 ];

// Function call
let number = smallestDifference(N, L, arr);
document.write(number);

// This code is contributed by Samim Hossain Mondal.
</script>

17

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

kdheeraj

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Article Tags :

• Bit Magic
• Greedy
• Mathematical
• DSA
• Arrays
• Maths

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Practice Tags :

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