Find all even length binary sequences with same sum of first and second half bits
Given a number n, find all binary sequences of length 2n such that sum of first n bits is same as sum of last n bits.
Examples:
Input: N = 2 Output: 0101 1111 1001 0110 0000 1010 Input: N = 3 Output: 011011 001001 011101 010001 101011 111111 110011 101101 100001 110101 001010 011110 010010 001100 000000 010100 101110 100010 110110 100100
The idea is to fix first and last bits and then recur for remaining 2*(n-1) bits. There are four possibilities when we fix first and last bits –
- First and last bits are 1, remaining n – 1 bits on both sides should also have the same sum.
- First and last bits are 0, remaining n – 1 bits on both sides should also have the same sum.
- First bit is 1 and last bit is 0, sum of remaining n – 1 bits on left side should be 1 less than the sum n-1 bits on right side.
- First bit is 0 and last bit is 1, sum of remaining n – 1 bits on left side should be 1 more than the sum n-1 bits on right side.
Below is implementation of above idea –
C++
// C++ program to print even length binary sequences // whose sum of first and second half bits is same #include <bits/stdc++.h> using namespace std; // Function to print even length binary sequences // whose sum of first and second half bits is same // diff --> difference between sums of first n bits // and last n bits // out --> output array // start --> starting index // end --> ending index void findAllSequences( int diff, char * out, int start, int end) { // We can't cover difference of more than n with 2n bits if ( abs (diff) > (end - start + 1) / 2) return ; // if all bits are filled if (start > end) { // if sum of first n bits and last n bits are same if (diff == 0) cout << out << " " ; return ; } // fill first bit as 0 and last bit as 1 out[start] = '0' , out[end] = '1' ; findAllSequences(diff + 1, out, start + 1, end - 1); // fill first and last bits as 1 out[start] = out[end] = '1' ; findAllSequences(diff, out, start + 1, end - 1); // fill first and last bits as 0 out[start] = out[end] = '0' ; findAllSequences(diff, out, start + 1, end - 1); // fill first bit as 1 and last bit as 0 out[start] = '1' , out[end] = '0' ; findAllSequences(diff - 1, out, start + 1, end - 1); } // Driver program int main() { // input number int n = 2; // allocate string containing 2*n characters char out[2 * n + 1]; // null terminate output array out[2 * n] = '\0' ; findAllSequences(0, out, 0, 2*n - 1); return 0; } |
Java
// Java program to print even length binary // sequences whose sum of first and second // half bits is same import java.io.*; import java.util.*; class GFG { // Function to print even length binary sequences // whose sum of first and second half bits is same // diff --> difference between sums of first n bits // and last n bits // out --> output array // start --> starting index // end --> ending index static void findAllSequences( int diff, char out[], int start, int end) { // We can't cover difference of more // than n with 2n bits if (Math.abs(diff) > (end - start + 1 ) / 2 ) return ; // if all bits are filled if (start > end) { // if sum of first n bits and // last n bits are same if (diff == 0 ) { System.out.print(out); System.out.print( " " ); } return ; } // fill first bit as 0 and last bit as 1 out[start] = '0' ; out[end] = '1' ; findAllSequences(diff + 1 , out, start + 1 , end - 1 ); // fill first and last bits as 1 out[start] = out[end] = '1' ; findAllSequences(diff, out, start + 1 , end - 1 ); // fill first and last bits as 0 out[start] = out[end] = '0' ; findAllSequences(diff, out, start + 1 , end - 1 ); // fill first bit as 1 and last bit as 0 out[start] = '1' ; out[end] = '0' ; findAllSequences(diff - 1 , out, start + 1 , end - 1 ); } // Driver program public static void main (String[] args) { // input number int n = 2 ; // allocate string containing 2*n characters char [] out = new char [ 2 * n + 1 ]; // null terminate output array out[ 2 * n] = '\0' ; findAllSequences( 0 , out, 0 , 2 *n - 1 ); } } // This code is contributed by Pramod Kumar |
Python3
# Python3 program to print even length binary sequences # whose sum of first and second half bits is same # Function to print even length binary sequences # whose sum of first and second half bits is same # diff --> difference between sums of first n bits # and last n bits # out --> output array # start --> starting index # end --> ending index def findAllSequences(diff, out, start, end): # We can't cover difference of more than n with 2n bits if ( abs (diff) > (end - start + 1 ) / / 2 ): return ; # if all bits are filled if (start > end): # if sum of first n bits and last n bits are same if (diff = = 0 ): print (''.join( list (out)),end = " " ); return ; # fill first bit as 0 and last bit as 1 out[start] = '0' ; out[end] = '1' ; findAllSequences(diff + 1 , out, start + 1 , end - 1 ); # fill first and last bits as 1 out[start] = out[end] = '1' ; findAllSequences(diff, out, start + 1 , end - 1 ); # fill first and last bits as 0 out[start] = out[end] = '0' ; findAllSequences(diff, out, start + 1 , end - 1 ); # fill first bit as 1 and last bit as 0 out[start] = '1' ; out[end] = '0' ; findAllSequences(diff - 1 , out, start + 1 , end - 1 ); # Driver program # input number n = 2 ; # allocate string containing 2*n characters out = [""] * ( 2 * n); findAllSequences( 0 , out, 0 , 2 * n - 1 ); # This code is contributed by mits |
C#
// C# program to print even length binary // sequences whose sum of first and second // half bits is same using System; class GFG { // Function to print even length binary // sequences whose sum of first and // second half bits is same // diff --> difference between sums of // first n bits // and last n bits // out --> output array // start --> starting index // end --> ending index static void findAllSequences( int diff, char []outt, int start, int end) { // We can't cover difference of // more than n with 2n bits if (Math.Abs(diff) > (end - start + 1) / 2) return ; // if all bits are filled if (start > end) { // if sum of first n bits and // last n bits are same if (diff == 0) { Console.Write(outt); Console.Write( " " ); } return ; } // fill first bit as 0 and last bit // as 1 outt[start] = '0' ; outt[end] = '1' ; findAllSequences(diff + 1, outt, start + 1, end - 1); // fill first and last bits as 1 outt[start] = outt[end] = '1' ; findAllSequences(diff, outt, start + 1, end - 1); // fill first and last bits as 0 outt[start] = outt[end] = '0' ; findAllSequences(diff, outt, start + 1, end - 1); // fill first bit as 1 and last // bit as 0 outt[start] = '1' ; outt[end] = '0' ; findAllSequences(diff - 1, outt, start + 1, end - 1); } // Driver program public static void Main () { // input number int n = 2; // allocate string containing 2*n // characters char []outt = new char [2 * n + 1]; // null terminate output array outt[2 * n] = '\0' ; findAllSequences(0, outt, 0, 2*n - 1); } } // This code is contributed by nitin mittal. |
PHP
<?php // PHP program to print even length binary sequences // whose sum of first and second half bits is same // Function to print even length binary sequences // whose sum of first and second half bits is same // diff --> difference between sums of first n bits // and last n bits // out --> output array // start --> starting index // end --> ending index function findAllSequences( $diff , $out , $start , $end ) { // We can't cover difference of more than n with 2n bits if ( abs ( $diff ) > (int)(( $end - $start + 1) / 2)) return ; // if all bits are filled if ( $start > $end ) { // if sum of first n bits and last n bits are same if ( $diff == 0) print (implode( "" , $out ). " " ); return ; } // fill first bit as 0 and last bit as 1 $out [ $start ] = '0' ; $out [ $end ] = '1' ; findAllSequences( $diff + 1, $out , $start + 1, $end - 1); // fill first and last bits as 1 $out [ $start ] = $out [ $end ] = '1' ; findAllSequences( $diff , $out , $start + 1, $end - 1); // fill first and last bits as 0 $out [ $start ] = $out [ $end ] = '0' ; findAllSequences( $diff , $out , $start + 1, $end - 1); // fill first bit as 1 and last bit as 0 $out [ $start ] = '1' ; $out [ $end ] = '0' ; findAllSequences( $diff - 1, $out , $start + 1, $end - 1); } // Driver program // input number $n = 2; // allocate string containing 2*n characters $out = array_fill (0,2* $n , "" ); findAllSequences(0, $out , 0, 2* $n - 1); // This code is contributed by chandan_jnu ?> |
Javascript
<script> // JavaScript program to print even length binary // sequences whose sum of first and second // half bits is same // Function to print even length binary // sequences whose sum of first and // second half bits is same // diff --> difference between sums of // first n bits // and last n bits // out --> output array // start --> starting index // end --> ending index function findAllSequences(diff, outt, start, end) { // We can't cover difference of // more than n with 2n bits if (Math.abs(diff) > parseInt((end - start + 1) / 2, 10)) return ; // if all bits are filled if (start > end) { // if sum of first n bits and // last n bits are same if (diff == 0) { document.write(outt.join( "" )); document.write( " " ); } return ; } // fill first bit as 0 and last bit // as 1 outt[start] = '0 '; outt[end] = ' 1 '; findAllSequences(diff + 1, outt, start + 1, end - 1); // fill first and last bits as 1 outt[start] = outt[end] = ' 1 '; findAllSequences(diff, outt, start + 1, end - 1); // fill first and last bits as 0 outt[start] = outt[end] = ' 0 '; findAllSequences(diff, outt, start + 1, end - 1); // fill first bit as 1 and last // bit as 0 outt[start] = ' 1 '; outt[end] = ' 0 '; findAllSequences(diff - 1, outt, start + 1, end - 1); } // input number let n = 2; // allocate string containing 2*n // characters let outt = new Array(2 * n + 1); // null terminate output array outt[2 * n] = ' \0'; findAllSequences(0, outt, 0, 2*n - 1); </script> |
Output
0101 1111 1001 0110 0000 1010
Time Complexity: O(()* N)
4^N because of 4 recursive calls, and N (simplified from 2N) for time spent printing strings of size 2N
Auxiliary Space: O(N)
There is another approach by which we generate all possible strings of length n and store them in a list at an index representing their sum. Then, we iterate through each list and generate the strings of size 2n by printing each string with all other strings in the list adding up to the same value.
C++
// C++ program to implement the approach #include <bits/stdc++.h> using namespace std; //function that generates the sequence void generateSequencesWithSum( int n, vector<vector<string> >& sumToString, vector<string> sequence, int sumSoFar) { // Base case, if there are no more binary digits to // include if (n == 0) { // add permutation to list of sequences with sum // corresponding to index string seq = "" ; for ( int i = 0; i < sequence.size(); i++) { seq = seq + sequence[i]; } vector<string> x = sumToString[sumSoFar]; x.push_back(seq); sumToString[sumSoFar] = x; return ; } // Generate sequence +0 sequence.push_back( "0" ); generateSequencesWithSum(n - 1, sumToString, sequence, sumSoFar); sequence.erase(sequence.begin()); // Generate sequence +1 sequence.push_back( "1" ); generateSequencesWithSum(n - 1, sumToString, sequence, sumSoFar + 1); sequence.erase(sequence.begin()); } // function to form permutations of the sequences void permuteSequences(vector<vector<string> > sumToString) { // There are 2^n substring in this list of lists for ( int sumIndexArr = 0; sumIndexArr < sumToString.size(); sumIndexArr++) { // Append for ( int sequence1 = 0; sequence1 < sumToString[sumIndexArr].size(); sequence1++) { for ( int sequence2 = 0; sequence2 < sumToString[sumIndexArr].size(); sequence2++) { if (sumIndexArr == sumToString.size() - 1 && sequence1 == sumToString[sumIndexArr] .size() - 1 && sequence2 == sumToString[sumIndexArr] .size() - 1) { cout << "1111 " ; } else { cout << sumToString[sumIndexArr] [sequence1] + sumToString[sumIndexArr] [sequence2] << " " ; } } } } } // function that finds all the subsequences void findAllSequences( int n) { vector<vector<string> > sumToString; for ( int i = 0; i < n + 1; i++) { sumToString.push_back( vector<string>()); // list of strings // where index // represents sum } generateSequencesWithSum(n, sumToString, vector<string>(), 0); permuteSequences(sumToString); } // Driver Code int main() { // Function Call findAllSequences(2); return 0; } // this code is contributed by phasing17 |
Java
// Java program to implement the approach import java.util.*; class GFG { // function that finds all the subsequences static void findAllSequences( int n) { ArrayList<ArrayList<String> > sumToString = new ArrayList<ArrayList<String> >(); for ( int i = 0 ; i < n + 1 ; i++) { sumToString.add( new ArrayList<String>()); // list of strings // where index // represents sum } generateSequencesWithSum( n, sumToString, new ArrayList<String>(), 0 ); permuteSequences(sumToString); } static void generateSequencesWithSum( int n, ArrayList<ArrayList<String> > sumToString, ArrayList<String> sequence, int sumSoFar) { // Base case, if there are no more binary digits to // include if (n == 0 ) { // add permutation to list of sequences with sum // corresponding to index String seq = "" ; for ( int i = 0 ; i < sequence.size(); i++) { seq = seq + sequence.get(i); } ArrayList<String> x = sumToString.get(sumSoFar); x.add(seq); sumToString.set(sumSoFar, x); return ; } // Generate sequence +0 sequence.add( "0" ); generateSequencesWithSum(n - 1 , sumToString, sequence, sumSoFar); sequence.remove( 0 ); // Generate sequence +1 sequence.add( "1" ); generateSequencesWithSum(n - 1 , sumToString, sequence, sumSoFar + 1 ); sequence.remove( 0 ); } // function to form permutations of the sequences static void permuteSequences( ArrayList<ArrayList<String> > sumToString) { // There are 2^n substring in this list of lists for ( int sumIndexArr = 0 ; sumIndexArr < sumToString.size(); sumIndexArr++) { // Append for ( int sequence1 = 0 ; sequence1 < sumToString.get(sumIndexArr).size(); sequence1++) { for ( int sequence2 = 0 ; sequence2 < sumToString.get(sumIndexArr).size(); sequence2++) { if (sumIndexArr == sumToString.size() - 1 && sequence1 == sumToString .get(sumIndexArr) .size() - 1 && sequence2 == sumToString .get(sumIndexArr) .size() - 1 ) { System.out.print( "1111" ); } else { System.out.println( sumToString.get(sumIndexArr) .get(sequence1) + sumToString.get(sumIndexArr) .get(sequence2)); } } } } } // Driver Code public static void main(String[] args) { // Function Call findAllSequences( 2 ); } // this code is contributed by phasing17 } |
Python3
def findAllSequences(n): sumToString = [[] for x in range (n + 1 )] # list of strings where index represents sum generateSequencesWithSum(n, sumToString, [], 0 ) permuteSequences(sumToString) def generateSequencesWithSum(n, sumToString, sequence, sumSoFar): #Base case, if there are no more binary digits to include if n = = 0 : sumToString[sumSoFar].append("".join(sequence)) #add permutation to list of sequences with sum corresponding to index return #Generate sequence +0 sequence.append( "0" ) generateSequencesWithSum(n - 1 , sumToString, sequence, sumSoFar) sequence.pop() #Generate sequence +1 sequence.append( "1" ) generateSequencesWithSum(n - 1 , sumToString, sequence, sumSoFar + 1 ) sequence.pop() def permuteSequences(sumToString): #There are 2^n substring in this list of lists for sumIndexArr in sumToString: # Append for sequence1 in sumIndexArr: for sequence2 in sumIndexArr: print (sequence1 + sequence2) findAllSequences( 2 ) #Contribution by Xavier Jean Baptiste |
C#
using System; using System.Collections.Generic; class GFG { static void findAllSequences( int n) { List<List< string >> sumToString = new List<List< string >>(); for ( int i = 0; i < n + 1; i++) { sumToString.Add( new List< string >()); // list of strings where index represents sum } generateSequencesWithSum(n, sumToString, new List< string >(), 0); permuteSequences(sumToString); } static void generateSequencesWithSum( int n, List<List< string >> sumToString, List< string > sequence, int sumSoFar) { // Base case, if there are no more binary digits to include if (n == 0) { //add permutation to list of sequences with sum corresponding to index string seq = "" ; for ( int i = 0; i < sequence.Count; i++) { seq = seq + sequence[i]; } sumToString[sumSoFar].Add(seq); return ; } // Generate sequence +0 sequence.Add( "0" ); generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar); sequence.RemoveAt(0); // Generate sequence +1 sequence.Add( "1" ); generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar+1); sequence.RemoveAt(0); } static void permuteSequences(List<List< string >> sumToString) { // There are 2^n substring in this list of lists for ( int sumIndexArr = 0; sumIndexArr < sumToString.Count; sumIndexArr++) { // Append for ( int sequence1 = 0; sequence1 < sumToString[sumIndexArr].Count; sequence1++) { for ( int sequence2 = 0; sequence2 < sumToString[sumIndexArr].Count; sequence2++) { if (sumIndexArr == sumToString.Count-1 && sequence1 == sumToString[sumIndexArr].Count-1 && sequence2 == sumToString[sumIndexArr].Count-1) { Console.Write( "1111" ); } else { Console.WriteLine(sumToString[sumIndexArr][sequence1] + sumToString[sumIndexArr][sequence2]); } } } } } static void Main() { findAllSequences(2); } } // This code is contributed by divyesh072019. |
Javascript
<script> function findAllSequences(n) { let sumToString = []; for (let i = 0; i < n + 1; i++) { sumToString.push([]); // list of strings where index represents sum } generateSequencesWithSum(n, sumToString, [], 0); permuteSequences(sumToString); } function generateSequencesWithSum(n, sumToString, sequence, sumSoFar) { // Base case, if there are no more binary digits to include if (n == 0) { //add permutation to list of sequences with sum corresponding to index sumToString[sumSoFar].push(sequence.join( "" )); return ; } // Generate sequence +0 sequence.push( "0" ); generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar); sequence.shift(); // Generate sequence +1 sequence.push( "1" ); generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar+1); sequence.shift(); } function permuteSequences(sumToString) { // There are 2^n substring in this list of lists for (let sumIndexArr = 0; sumIndexArr < sumToString.length; sumIndexArr++) { // Append for (let sequence1 = 0; sequence1 < sumToString[sumIndexArr].length; sequence1++) { for (let sequence2 = 0; sequence2 < sumToString[sumIndexArr].length; sequence2++) { if (sumIndexArr == sumToString.length-1 && sequence1 == sumToString[sumIndexArr].length-1 && sequence2 == sumToString[sumIndexArr].length-1) { document.write( "1111" ); } else { document.write(sumToString[sumIndexArr][sequence1] + sumToString[sumIndexArr][sequence2] + "</br>" ); } } } } } findAllSequences(2); // This code is contributed by decode2207. </script> |
Output
0000 0101 0110 1001 1010 1111
Time complexity analysis:
generateSequencesWithSum = O((2N)*N)
- 2N: we generate all permutation of binary strings of size N
- N: convert the list of characters to a string and store into array. This is done in the base case.
permuteSequences = O((2N) * N!/(N/2)!2 * N)
- 2N: we iterate through all the string generated of size n
- N!/(N/2)!2: This one is a bit challenging to explain
let’s take N = 2 as an example. Our array of possible sequence of size n would be:
array index | 0 | 1 | 2 |
list of strings | 00 | 01,10 | 11 |
In the list of strings which the index represents the sum, we get the count of strings of size 2n by using “n choose k” formula. I our case it would be nCk *nCk where k represents the number of 1s in each half of the string of size 2n:
k = 0, we have (2C0)^2 = 1 string (0000)
k = 1, we have (2C1)^2 string = 4 strings(0101 0110 1001 1010)
k = 2, we have (2c2)^2 = 1 string (1111)
We get our longest list of string when k = N/2, hence NCN/2 = N!/[(N/2)! * (N – N/2)!] which simplifies to NCN/2 = N!/(N/2)!2
Hence, for each element, we must iterate through, at most, NCN/2 for forming strings of length 2N
Without formal proof, if we graph 2^N and N!/(N/2)!2, we see that 2N has a faster growth rate than the latter. Therefore O(2N* N!/(N/2)2 ) < O(2N*2N) = O(22n) = O(4N)
- N: we must print each string of size 2N
Finally we can ignore the time complexity of generateSequencesWithSum because permuteSequence is the leading term
Time complexity: O(2N * N!/(N/2)!2 * N) (better than the first solution of O((4^N) * N, see explanation above for further details)
Auxiliary space: O(2N) because we store all binary string permutations of size N
C++
#include<bits/stdc++.h> using namespace std; class FirstHalf { public : string data; int sum; FirstHalf(string data, int sum) { this ->data = data; this ->sum = sum; } }; // MAP: Key -> sum of bits; Value -> All possible permutation with respective sum map< int , vector<string>> mp; // first N-half bits vector<FirstHalf> firstHalf; // function to find sum of the bits from a String int sumOfString(string s) { int sum = 0; // ex: converts "1" to 1 -> (ASCII('1') - ASCII('0') = 1) for ( auto c: s) { sum += (c - '0' ); } return sum; } void perm(string p, char * bin, int level, int n) { // p: processed string(processed permutation at current level) // bin: {'0', '1'} // l: current level of recursion tree (leaf/solution level = 0) // n: total levels if (level == 0) { // at solution level find sum of the current permutation int sum = sumOfString(p); // store current permutation to firstHalf list firstHalf.push_back(FirstHalf(p, sum)); // put current permutation to its respective sum value mp[sum].push_back(p); return ; } // generate calls for permutation // working: first solution with all 0s, // then replacing last 0 with 1 and so on... for ( int i = 0; i < n; i++) { char c = bin[i]; perm(p+c, bin, level-1, n); } } void result() { int i = 0; for ( auto first: firstHalf) { // for each firstHalf string // find sum of the bits of current string int sum = first.sum; // retrieve respective secondHalf from map based on sum key vector<string> secondHalf = mp[sum]; for ( auto second: secondHalf) { // append first and second half and print cout << first.data + second << " " ; // after every 6 solution line is changed in output // only for formatting below lines could be removed i++; if (i % 6 == 0) cout << endl; } } } int main(){ char up[2] = { '0' , '1' }; int n = 2; string x = "" ; perm(x, up, n, n); result(); return 0; } // This code is contributed by Nidhi goel. |
Java
import java.util.*; class GFG { static class FirstHalf { String data; int sum; FirstHalf(String data, int sum) { this .data = data; this .sum = sum; } } //MAP: Key -> sum of bits; Value -> All possible permutation with respective sum static Map<Integer, ArrayList<String>> map = new HashMap<>(); //first N-half bits static List<FirstHalf> firstHalf = new ArrayList<>(); //function to find sum of the bits from a String public static int sumOfString(String s) { int sum = 0 ; //ex: converts "1" to 1 -> (ASCII('1') - ASCII('0') = 1) for ( char c: s.toCharArray()) { sum += c - '0' ; } return sum; } public static void perm(String p, char [] bin, int level, int n) { //p: processed string(processed permutation at current level) //bin: {'0', '1'} //l: current level of recursion tree (leaf/solution level = 0) //n: total levels if (level == 0 ) { //at solution level find sum of the current permutation int sum = sumOfString(p); //store current permutation to firstHalf list firstHalf.add( new FirstHalf(p, sum)); //put current permutation to its respective sum value map.putIfAbsent(sum, new ArrayList<String>()); map.get(sum).add(p); return ; } //generate calls for permutation //working: first solution with all 0s, then replacing last 0 with 1 and so on... for ( char c: bin) { perm(p+c, bin, level- 1 , n); } } public static void result() { int i = 0 ; for (FirstHalf first: firstHalf) { //for each firstHalf string //find sum of the bits of current string int sum = first.sum; //retrieve respective secondHalf from map based on sum key ArrayList<String> secondHalf = map.get(sum); for (String second: secondHalf) { //append first and second half and print System.out.print(first.data+second+ " " ); //after every 6 solution line is changed in output //only for formatting below lines could be removed i++; if (i % 6 == 0 ) System.out.println(); } } } public static void main(String[] args) { char [] up = { '0' , '1' }; int n = 2 ; perm( "" , up, n, n); result(); } } //Code contributed by Animesh Singh |
Python3
# Python code implementation class FirstHalf: def __init__( self , data, sum ): self .data = data self . sum = sum # MAP: Key -> sum of bits; Value -> All possible permutation with respective sum map = {} # first N-half bits firstHalf = [] # function to find sum of the bits from a String def sumOfString(s): sum = 0 # ex: converts "1" to 1 -> (ASCII('1') - ASCII('0') = 1) for i in range ( len (s)): sum + = ord (s[i]) - ord ( '0' ) return sum def perm(p, bin , level, n): # p: processed string(processed permutation at current level) # bin: ['0', '1'] # l: current level of recursion tree (leaf/solution level = 0) # n: total levels if level = = 0 : # at solution level find sum of the current permutation sum = sumOfString(p) # store current permutation to firstHalf list firstHalf.append(FirstHalf(p, sum )) # put current permutation to its respective sum value if sum not in map : map [ sum ] = [] map [ sum ].append(p) return # generate calls for permutation # working: first solution with all 0s, then replacing last 0 with 1 and so on... for i in range ( len ( bin )): perm(p + bin [i], bin , level - 1 , n) def result(): i = 0 for j in range ( len (firstHalf)): # for each firstHalf string # find sum of the bits of current string sum = firstHalf[j]. sum # retrieve respective secondHalf from map based on sum key secondHalf = map [ sum ] for k in range ( len (secondHalf)): # append first and second half and print print (firstHalf[j].data + secondHalf[k] + " " , end = "") # after every 6 solution line is changed in output # only for formatting below lines could be removed i = i + 1 if (i % 6 = = 0 ): print ( '\n' ) up = [ '0' , '1' ] n = 2 perm("", up, n, n) result() # The code is contributed by Nidhi goel. |
C#
using System; using System.Collections.Generic; class FirstHalf { public string data; public int sum; public FirstHalf( string data, int sum) { this .data = data; this .sum = sum; } } class Gfg { // MAP: Key -> sum of bits; Value -> All possible permutation with respective sum static Dictionary< int , List< string >> mp = new Dictionary< int , List< string >>(); // first N-half bits static List<FirstHalf> firstHalf = new List<FirstHalf>(); // function to find sum of the bits from a String static int sumOfString( string s) { int sum = 0; // ex: converts "1" to 1 -> (ASCII('1') - ASCII('0') = 1) foreach ( char c in s) { sum += (c - '0' ); } return sum; } static void perm( string p, char [] bin, int level, int n) { // p: processed string(processed permutation at current level) // bin: {'0', '1'} // l: current level of recursion tree (leaf/solution level = 0) // n: total levels if (level == 0) { // at solution level find sum of the current permutation int sum = sumOfString(p); // store current permutation to firstHalf list firstHalf.Add( new FirstHalf(p, sum)); // put current permutation to its respective sum value if (mp.ContainsKey(sum)) { mp[sum].Add(p); } else { mp.Add(sum, new List< string > { p }); } return ; } // generate calls for permutation // working: first solution with all 0s, // then replacing last 0 with 1 and so on... for ( int i = 0; i < n; i++) { char c = bin[i]; perm(p + c, bin, level - 1, n); } } static void result() { int i = 0; foreach (FirstHalf first in firstHalf) { // for each firstHalf string // find sum of the bits of current string int sum = first.sum; // retrieve respective secondHalf from map based on sum key List< string > secondHalf = mp[sum]; foreach ( string second in secondHalf) { // append first and second half and print Console.Write(first.data + second + " " ); // after every 6 solution line is changed in output // only for formatting below lines could be removed i++; if (i % 6 == 0) Console.WriteLine(); } } } static void Main( string [] args) { char [] up = { '0' , '1' }; int n = 2; string x = "" ; perm(x, up, n, n); result(); } } |
Javascript
class FirstHalf { constructor(data, sum) { this .data = data; this .sum = sum; } } // MAP: Key -> sum of bits; Value -> All possible permutation with respective sum const map = new Map(); // first N-half bits const firstHalf = []; // function to find sum of the bits from a String function sumOfString(s) { let sum = 0; //ex: converts "1" to 1 -> (ASCII('1') - ASCII('0') = 1) for (let i = 0; i < s.length; i++) { sum += s.charCodeAt(i) - '0' .charCodeAt(0); } return sum; } function perm(p, bin, level, n) { // p: processed string(processed permutation at current level) // bin: ['0', '1'] // l: current level of recursion tree (leaf/solution level = 0) // n: total levels if (level == 0) { // at solution level find sum of the current permutation let sum = sumOfString(p); // store current permutation to firstHalf list firstHalf.push( new FirstHalf(p, sum)); // put current permutation to its respective sum value if (!map.has(sum)) map.set(sum, []); map.get(sum).push(p); return ; } // generate calls for permutation // working: first solution with all 0s, then replacing last 0 with 1 and so on... for (let i = 0; i < bin.length; i++) { perm(p+bin[i], bin, level-1, n); } } function result() { let i = 0; for (let j = 0; j < firstHalf.length; j++) { // for each firstHalf string // find sum of the bits of current string let sum = firstHalf[j].sum; // retrieve respective secondHalf from map based on sum key let secondHalf = map.get(sum); for (let k = 0; k < secondHalf.length; k++) { // append first and second half and print process.stdout.write(firstHalf[j].data + secondHalf[k] + " " ); // after every 6 solution line is changed in output // only for formatting below lines could be removed i++; if (i % 6 == 0) process.stdout.write( '\n' ); } } } const up = [ '0' , '1' ]; const n = 2; perm( "" , up, n, n); result(); |
Output
0000 0101 0110 1001 1010 1111
Algorithm:
1. Generate all binary permutations of size n
2. Calculate sum of the bits of each permutation and remember it for second half
[for ex: for n=2, remember there are two strings with sum = 1 i.e. “01”, “10” ]
3. Iterate all the generated permutations and for each of them append the second half according to the sum of the bits
Time complexity analysis:
sumOfString() = O(N) : traverse each bit and add it to sum
perm() = O(2N * N)
2N * N : we generate all permutations of binary bits of size N and find sum of the bits for each permutation
result() = O((2N) * (N!/(N/2)!)2)
2N: we iterate through all possible permutations of size N (first-Half)
NCN/2 = N!/(N/2)!2 : (second-Half maximum size) : explanation below:
let’s take N = 4 as an example.:
//Hash-Map looks like
0 -> [0000] …………………………..(list-size: 4C0 = 1)
1 -> [0001, 0010, 0100, 1000] …………………………..(list-size: 4C1 = 4)
2 -> [0011, 0101, 0110, 1001, 1010, 1100] …………………………..(list-size: 4C2 = 6)
3 -> [0111, 1011, 1101, 1110] …………………………..(list-size: 4C3 = 4)
4 -> [1111] …………………………..(list-size: 4C4 = 1)
We observe here that each list has a size of N choose Key which will be maximum at N choose N/2
Since we are iterating all the 2N permutations and appending second half from the map. The map has the maximum sized list at N/2 position.
Worst case occurs in N/2 position where we’ve to traverse NCN/2 = N!/(N/2)!2 permutations.
Time complexity: O(2N * N!/(N/2)!2 )
Auxiliary space: O(2N) because we store all binary string permutations of size N