Given an array arr[] consisting of N integers, and an integer K, the task is to construct a binary string of length K satisfying the following conditions:
- The character at ithindex is ‘1′ if a subset with sum i can be formed from the array.
- Otherwise, the character at ithindex is ‘0’.
Examples:
Input: arr[] = {1, 4}, K = 5
Output: 10011
Explanation:
Character at 1st index can be made by ‘1’ considering the subset {1}.
Character at 4th index can be made by ‘1’ considering the subset {4}.
Character at 5th index can be made by ‘1’ considering the subset {1, 4}.Input: arr[] = {1, 6, 1}, K = 8
Output: 11000111
Approach: The idea is to use a greedy approach to solve this problem. Below are the steps:
- Initialize a bitset, say bit[], of size 105 + 5 and set bit[0] = 1.
- Traverse through the array and for each array element arr[i], update bit as bit |= bit << arr[i] to have bit p if p can be obtained as a subset sum.
- At ith iteration, bit[i] stores the initial sum and after performing bit << arr[i], all bits are shifted by arr[i]. Therefore, bit p becomes p + arr[i].
- Finally, bit | (bit << arr[i]) merges these two cases, whether to consider the ith position or not.
- Iterate from 1 to K and print every value bit[i] as the required binary string.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// To construct the // required binary string bitset<100003> bit; // Function to construct binary string // according to the given conditions void constructBinaryString( int arr[],
int N, int K)
{ // Initialize with 1
bit[0] = 1;
// Traverse the array
for ( int i = 0; i < N; i++) {
// To check if the i-th integer
// needs to be considered or not
bit |= bit << arr[i];
}
// Print the binary string
for ( int i = 1; i <= K; i++) {
cout << bit[i];
}
} // Driver Code int main()
{ // Given array
int arr[] = { 1, 6, 1 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
// Given K
int K = 8;
constructBinaryString(arr, N, K);
} |
// Java code to implement the approach import java.util.Arrays;
class BinaryString {
public static void main(String[] args) {
// Given array
int [] arr = { 1 , 6 , 1 };
// Size of the array
int N = arr.length;
// Given K
int K = 8 ;
constructBinaryString(arr, N, K);
}
// Function to construct binary string
// according to the given conditions
static void constructBinaryString( int [] arr, int N, int K) {
// Initialize with 1
int bit = 1 ;
// Traverse the array
for ( int i = 0 ; i < N; i++) {
// To check if the i-th integer
// needs to be considered or not
bit |= (bit << arr[i]);
}
// Print the binary string
String binaryString = Integer.toBinaryString(bit);
System.out.println(binaryString.substring( 1 , K));
}
} // This code is contributed by phasing17 |
# Python program for the above approach # To construct the # required binary string #bit = [0]*100003 # Function to construct binary string # according to the given conditions def constructBinaryString(arr,N, K):
# Initialize with 1
bit = 1
# Traverse the array
for i in range ( 0 , N):
# To check if the i-th eger
# needs to be considered or not
bit | = bit << arr[i]
# Print the binary string
#for i in range(1,K):
# print(bit[i])
bit = bin (bit).replace( "0b" , "")
print (bit[ 1 :K + 1 ])
# Driver Code # Given array arr = [ 1 , 6 , 1 ]
# Size of the array N = len (arr)
# Given K K = 8
constructBinaryString(arr, N, K) # This code is contributed by shubhamsingh10 |
using System;
using System.Linq;
namespace BinaryString {
class Program {
static void Main( string [] args)
{
// Given array
int [] arr = { 1, 6, 1 };
// Size of the array
int N = arr.Length;
// Given K
int K = 8;
ConstructBinaryString(arr, N, K);
}
// Function to construct binary string
// according to the given conditions
static void ConstructBinaryString( int [] arr, int N,
int K)
{
// Initialize with 1
int bit = 1;
// Traverse the array
for ( int i = 0; i < N; i++) {
// To check if the i-th integer
// needs to be considered or not
bit |= (bit << arr[i]);
}
// Print the binary string
string binaryString = Convert.ToString(bit, 2);
Console.WriteLine(binaryString.Substring(1, K));
}
}
} // This code is contributed by phasing17. |
// JavaScript program for the above approach // To construct the // required binary string //bit = [0]*100003 // Function to construct binary string // according to the given conditions function constructBinaryString(arr,N, K)
{ // Initialize with 1
let bit = 1
// Traverse the array
for ( var i = 0; i < N; i++)
// To check if the i-th eger
// needs to be considered or not
bit |= (bit << arr[i])
// Print the binary string
//for i in range(1,K):
// print(bit[i])
bit = bit.toString(2)
console.log(bit.substring(1, K + 1))
} // Driver Code // Given array let arr = [1, 6, 1] // Size of the array let N = arr.length // Given K let K = 8 constructBinaryString(arr, N, K) // This code is contributed by phasing17 |
11000111
Time Complexity: O(N)
Auxiliary Space: O(N)