Given an array arr[] consisting of N integers, the task is to find the total number of distinct subsequences having Binary Equivalence.
A subsequence has Binary Equivalence if the sum of the count of set and unset bits in the binary representations of all the decimal numbers across the subsequence are equal.
Examples:
Input: arr[] = {2, 7, 10}
Output: 0011
Explanation:
2 ? 0010?1’s = 1, 0’s = 3
7 ? 0111?1’s = 3, 0’s = 1
10 ? 1010?1’s = 2, 0’s = 2
The subsequence [2, 7, 10] has Binary Equivalence because the number of 0’s and 1’s across the subsequence is 6 each.
Similarly, [2, 7] also has Binary Equivalence of 4 each.
But [7, 10] does not have Binary Equivalence.
Likewise, [10] has Binary Equivalence of 2 each.
The total number of unique subsequences where Binary Equivalence is possible is 3.
Since 10 is the largest element in the given array and the number of bits required to represent 10 in binary is 4. Hence, the number of bits present in the output needs to be 4.Input: arr[] = {5, 7, 9, 12}
Output: 0111
Approach: The idea is to find the total number of bits required to represent the maximum element of the array.Follow these steps to solve this problem:
- Find the maximum element and the length of binary representation of the maximum element.
- Append 0 in the front other elements in binary representation, to make the number of bits in each element equal to the maximum number bits.
- Find all the subsequences of the given array.
- Find the total number of subsequences that have Binary Equivalence.
- Convert the total number into binary and append 0s if the length of the total number is less than the length of the maximum number to make both the lengths equal.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// C++ program for the above approach// Function to find the number of// subsequences having Binary Equivalencevoid numberOfSubsequence(vector<int> arr) { // Find the maximum array element int maxElement = INT_MIN; for(auto x: arr) maxElement = max(x, maxElement); // Convert the maximum element // to its binary equivalent int val = (int)(log2(maxElement)); string maxBinary = bitset<64>(maxElement).to_string().substr(64 - val - 1); // Dictionary to store the count of // set and unset bits of all array elements map<int,pair<int,int>> dic; for(auto i: arr){ int temp = (int)(log2(maxElement)); string str = bitset<64>(i).to_string().substr(64 - temp - 1); if (str.size() <= maxBinary.size()) { int diff = maxBinary.size() - str.size(); // Add the extra zeros before all // the elements which have length // smaller than the maximum element for(int indx = 0; indx < diff; indx++){ str = '0' + str; } } int zeros = 0; int ones = 0; for(int i = 0; i < str.size(); i++){ if(str[i] == '0') zeros++; else ones++; } // Fill the dictionary with number // of 0's and 1's dic[i] = {zeros, ones}; } vector<vector<int>> allCombinations; vector<int> curr_temp; allCombinations.push_back(curr_temp); // Find all the combinations for (int i : arr) { // cout << "hi" << endl; vector<vector<int>> newCombinations; for (auto j : allCombinations) { vector<int> combination = j; combination.push_back(i); newCombinations.push_back(combination); } for(auto curr_list: newCombinations){ allCombinations.push_back(curr_list); } } int count = 0; // Find all the combinations where // sum_of_zeros == sum_of_ones for (int i = 1; i < allCombinations.size(); i++) { int sum0 = 0; int sum1 = 0; for (auto j: allCombinations[i]) { sum0 += dic[j].first; sum1 += dic[j].second; } // Count the total combinations // where sum_of_zeros = sum_of_ones if (sum0 == sum1) { count += 1; } } // Convert the count number to its // binary equivalent int curr_val = (int)(log2(count)); string str = bitset<64>(count).to_string().substr(64 - curr_val - 1); if (str.size() <= maxBinary.size()) { int diff = maxBinary.size() - str.size(); // Append leading zeroes to // the answer if its length is // smaller than the maximum element for(int i = 0; i < diff; i++){ str = '0' + str; } } // Print the result cout << str << endl;}// Driver Codeint main(){ // Given arr[] arr. vector<int> arr = {5, 7, 9, 12}; // Function Call numberOfSubsequence(arr); return 0;}// The code is contributed by Nidhi goel. |
Java
import java.util.*;class Main { public static void main(String[] args) { int[] arr = {5, 7, 9, 12}; numberOfSubsequence(arr); } // Function to find the number of subsequences // having Binary Equivalence static void numberOfSubsequence(int[] arr) { // Find the maximum array element int maxElement = Arrays.stream(arr).max().getAsInt(); // Convert the maximum element to its binary equivalent String maxBinary = Integer.toBinaryString(maxElement); // Dictionary to store the count of set and unset bits // of all array elements Map<Integer, int[]> dic = new HashMap<>(); for (int i : arr) { String str = Integer.toBinaryString(i); if (str.length() < maxBinary.length()) { int diff = maxBinary.length() - str.length(); // Add extra zeros before all the elements which have length // smaller than the maximum element str = "0".repeat(diff) + str; } // Fill the dictionary with number of 0's and 1's int zeros = str.length() - str.replaceAll("0", "").length(); int ones = str.length() - str.replaceAll("1", "").length(); dic.put(i, new int[]{zeros, ones}); } List<List<Integer>> allCombinations = new ArrayList<>(); allCombinations.add(new ArrayList<>()); // Find all the combinations for (int i : arr) { List<List<Integer>> newCombinations = new ArrayList<>(); for (List<Integer> j : allCombinations) { List<Integer> combination = new ArrayList<>(j); combination.add(i); newCombinations.add(combination); } allCombinations.addAll(newCombinations); } int count = 0; // Find all the combinations where sum_of_zeros == sum_of_ones for (int i = 1; i < allCombinations.size(); i++) { int sum0 = 0; int sum1 = 0; for (int j : allCombinations.get(i)) { sum0 += dic.get(j)[0]; sum1 += dic.get(j)[1]; } // Count the total combinations where sum_of_zeros = sum_of_ones if (sum0 == sum1) { count += 1; } } // Convert the count number to its binary equivalent String str = Integer.toBinaryString(count); if (str.length() < maxBinary.length()) { int diff = maxBinary.length() - str.length(); // Append leading zeroes to the answer if its length is // smaller than the maximum element str = "0".repeat(diff) + str; } // Print the result System.out.println(str); }} |
Python3
# Python program for the above approachimport itertools# Function to find the number of# subsequences having Binary Equivalencedef numberOfSubsequence(arr): # Find the maximum array element Max_element = max(arr) # Convert the maximum element # to its binary equivalent Max_Binary = "{0:b}".format(int( Max_element)) # Dictionary to store the count of # set and unset bits of all array elements Dic = {} for i in arr: Str = "{0:b}".format(int(i)) if len(Str) <= len(Max_Binary): diff = len(Max_Binary)-len(Str) # Add the extra zeros before all # the elements which have length # smaller than the maximum element Str = ('0'*diff)+Str zeros = Str.count('0') ones = Str.count('1') # Fill the dictionary with number # of 0's and 1's Dic[int(i)] = [zeros, ones] all_combinations = [] # Find all the combination for r in range(len(arr)+1): comb = itertools.combinations(arr, r) comlist = list(comb) all_combinations += comlist count = 0 # Find all the combinations where # sum_of_zeros == sum_of_ones for i in all_combinations[1:]: sum0 = 0 sum1 = 0 for j in i: sum0 += Dic[j][0] sum1 += Dic[j][1] # Count the total combinations # where sum_of_zeros = sum_of_ones if sum0 == sum1: count += 1 # Convert the count number to its # binary equivalent Str = "{0:b}".format(int(count)) if len(Str) <= len(Max_Binary): diff = len(Max_Binary)-len(Str) # Append leading zeroes to # the answer if its length is # smaller than the maximum element Str = ('0'*diff) + Str # Print the result print(Str)# Driver Code# Give array arr[]arr = [5, 7, 9, 12]# Function CallnumberOfSubsequence(arr) |
C#
// Following is the C# equivalent of the above Java codeusing System;using System.Collections.Generic;using System.Linq;namespace NumberOfSubsequence{ class Program { static void Main(string[] args) { int[] arr = { 5, 7, 9, 12 }; NumberOfSubsequence(arr); } // Function to find the number of subsequences // having Binary Equivalence static void NumberOfSubsequence(int[] arr) { // Find the maximum array element int maxElement = arr.Max(); // Convert the maximum element to its binary equivalent string maxBinary = Convert.ToString(maxElement, 2); // Dictionary to store the count of set and unset bits // of all array elements Dictionary<int, int[]> dic = new Dictionary<int, int[]>(); foreach(int i in arr) { string strr = Convert.ToString(i, 2); if(strr.Length < maxBinary.Length) { int diff = maxBinary.Length - strr.Length; // Add extra zeros before all the elements which have length // smaller than the maximum element strr = new string('0', diff) + strr; } // Fill the dictionary with number of 0's and 1's int zeros = strr.Length - strr.Replace("0", "").Length; int ones = strr.Length - strr.Replace("1", "").Length; dic.Add(i, new int[] { zeros, ones }); } List<List<int>> allCombinations = new List<List<int>>(); allCombinations.Add(new List<int>()); // Find all the combinations foreach(int i in arr) { List<List<int>> newCombinations = new List<List<int>>(); foreach(List<int> j in allCombinations) { List<int> combination = new List<int>(j); combination.Add(i); newCombinations.Add(combination); } allCombinations.AddRange(newCombinations); } int count = 0; // Find all the combinations where sum_of_zeros == sum_of_ones for(int i = 1; i < allCombinations.Count; i++) { int sum0 = 0; int sum1 = 0; foreach(int j in allCombinations[i]) { sum0 += dic[j][0]; sum1 += dic[j][1]; } // Count the total combinations where sum_of_zeros = sum_of_ones if(sum0 == sum1) { count += 1; } } // Convert the count number to its binary equivalent string str = Convert.ToString(count, 2); if(str.Length < maxBinary.Length) { int diff = maxBinary.Length - str.Length; // Append leading zeroes to the answer if its length is // smaller than the maximum element str = new string('0', diff) + str; } // Print the result Console.WriteLine(str); } }} |
Javascript
// JavaScript program for the above approach// Function to find the number of// subsequences having Binary Equivalencefunction numberOfSubsequence(arr) { // Find the maximum array element let maxElement = Math.max(...arr); // Convert the maximum element // to its binary equivalent let maxBinary = maxElement.toString(2); // Dictionary to store the count of // set and unset bits of all array elements let dic = {}; for (let i of arr) { let str = i.toString(2); if (str.length <= maxBinary.length) { let diff = maxBinary.length - str.length; // Add the extra zeros before all // the elements which have length // smaller than the maximum element str = '0'.repeat(diff) + str; } let zeros = str.split('0').length - 1; let ones = str.split('1').length - 1; // Fill the dictionary with number // of 0's and 1's dic[i] = [zeros, ones]; } let allCombinations = [[]]; // Find all the combination for (let i of arr) { let newCombinations = []; for (let j of allCombinations) { newCombinations.push(j.concat(i)); } allCombinations = allCombinations.concat(newCombinations); } let count = 0; // Find all the combinations where // sum_of_zeros == sum_of_ones for (let i = 1; i < allCombinations.length; i++) { let sum0 = 0; let sum1 = 0; for (let j of allCombinations[i]) { sum0 += dic[j][0]; sum1 += dic[j][1]; } // Count the total combinations // where sum_of_zeros = sum_of_ones if (sum0 == sum1) { count += 1; } } // Convert the count number to its // binary equivalent let str = count.toString(2); if (str.length <= maxBinary.length) { let diff = maxBinary.length - str.length; // Append leading zeroes to // the answer if its length is // smaller than the maximum element str = '0'.repeat(diff) + str; } // Print the result console.log(str);}// Driver Code// Give array arr[]let arr = [5, 7, 9, 12];// Function CallnumberOfSubsequence(arr); |
Output:
0111
Time Complexity: O(2N)
Auxiliary Space: O(N2)
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