Check if an encoding represents a unique binary string
Last Updated :
18 Sep, 2023
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Given an encoding of a binary string of length k, the task is to find if the given encoding uniquely identifies a binary string or not. The encoding has counts of contiguous 1s (separated by 0s).
For example, encoding of 11111 is {5}, encoding of 01101010 is {2, 1, 1} and encoding of 111011 is {3, 2}.
Examples :
Input: encoding[] = {3, 3, 3}
Length, k = 12
Output: No
Explanation: There are more than one possible
binary strings. The strings are 111011101110
and 011101110111. Hence “No”
Input: encoding[] = {3, 3, 2}
Length, k = 10
Output: Yes
Explanation: There is only one possible encoding
that is 1110111011
A naive approach is to take an empty string and traverse through the n integers to no of 1s as given in encoding[0] and then add 1 zero to it, then encoding[1] 1s, and at the end check if the string length is equal to k then print “Yes” or print “No”
#include <iostream>
#include <vector>
#include <string>
using namespace std;
bool is_unique_encoding(vector<int> encoding, int k) {
string binary_str = "";
for (int count : encoding) {
binary_str += string(count, '1') + "0";
}
binary_str = binary_str.substr(0, binary_str.length() - 1);
if (binary_str.length() != k) {
return false;
}
return true;
}
int main() {
vector<int> encoding = {3, 3, 3};
int k = 12;
if (is_unique_encoding(encoding, k)) {
cout << "yes" << endl;
} else {
cout << "No" << endl;
}
return 0;
}
import java.util.ArrayList;
public class Main {
public static boolean isUniqueEncoding(ArrayList<Integer> encoding, int k) {
String binaryStr = "";
for (int count : encoding) {
binaryStr += "1".repeat(count) + "0";
}
binaryStr = binaryStr.substring(0, binaryStr.length() - 1);
if (binaryStr.length() != k) {
return false;
}
return true;
}
public static void main(String[] args) {
ArrayList<Integer> encoding = new ArrayList<Integer>();
encoding.add(3);
encoding.add(3);
encoding.add(3);
int k = 12;
if (isUniqueEncoding(encoding, k)) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
def isUniqueEncoding(encoding, k):
binaryStr = ""
for count in encoding:
binaryStr += "1" * count + "0"
binaryStr = binaryStr[:-1]
if len(binaryStr) != k:
return False
return True
if __name__ == '__main__':
encoding = [3, 3, 3]
k = 12
if isUniqueEncoding(encoding, k):
print("YES")
else:
print("NO")
using System;
using System.Collections.Generic;
class Program
{
static bool IsUniqueEncoding(List<int> encoding, int k)
{
string binaryStr = "";
foreach (int count in encoding)
{
binaryStr += new string('1', count) + "0";
}
binaryStr = binaryStr.Substring(0, binaryStr.Length - 1);
if (binaryStr.Length != k)
{
return false;
}
return true;
}
static void Main()
{
List<int> encoding = new List<int> { 3, 3, 3 };
int k = 12;
if (IsUniqueEncoding(encoding, k))
{
Console.WriteLine("yes");
}
else
{
Console.WriteLine("No");
}
}
}
function isUniqueEncoding(encoding, k) {
let binaryStr = "";
for (let count of encoding) {
binaryStr += "1".repeat(count) + "0";
}
binaryStr = binaryStr.substring(0, binaryStr.length - 1);
if (binaryStr.length !== k) {
return false;
}
return true;
}
let encoding = [3, 3, 3];
let k = 12;
if (isUniqueEncoding(encoding, k)) {
console.log("YES");
} else {
console.log("NO");
}
Output
no
An efficient approach will be to add all the n integers to sum, and then add (n-1) to sum and check if it is equal to K, as n-1 will be the number of zeros in between every 1’s. Check if sum is equal to k, to get exactly one string or else there are more or none.
Implementation:
// C++ program to check if given encoding
// represents a single string.
#include <bits/stdc++.h>
using namespace std;
bool isUnique(int a[], int n, int k)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += a[i];
sum += n - 1;
// Return true if sum becomes k
return (sum == k);
}
// Driver Code
int main()
{
int a[] = {3, 3, 3};
int n = sizeof(a) / sizeof(a[0]);
int k = 12;
if (isUnique(a, n, k))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java program to check if given encoding
// represents a single string.
import java.io.*;
class GFG
{
static boolean isUnique(int []a, int n, int k)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += a[i];
sum += n - 1;
// Return true if sum becomes k
return (sum == k);
}
// Driver Code
static public void main (String[] args)
{
int []a = {3, 3, 3};
int n = a.length;
int k = 12;
if (isUnique(a, n, k))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by vt_m
# Python 3 program to check if given
# encoding represents a single string.
def isUnique(a, n, k):
sum = 0
for i in range(0, n, 1):
sum += a[i]
sum += n - 1
# Return true if sum becomes k
return (sum == k)
# Driver Code
if __name__ == '__main__':
a = [3, 3, 3]
n = len(a)
k = 12
if (isUnique(a, n, k)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surndra_Gangwar
// C# program to check if given encoding
// represents a single string.
using System;
class GFG
{
static bool isUnique(int []a, int n, int k)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += a[i];
sum += n - 1;
// Return true if sum becomes k
return (sum == k);
}
// Driver Code
static public void Main ()
{
int []a = {3, 3, 3};
int n = a.Length;
int k = 12;
if (isUnique(a, n, k))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m
<script>
// javascript program to check if given encoding
// represents a single string.
function isUnique(a , n , k)
{
var sum = 0;
for (i = 0; i < n; i++)
sum += a[i];
sum += n - 1;
// Return true if sum becomes k
return (sum == k);
}
// Driver Code
var a = [ 3, 3, 3 ];
var n = a.length;
var k = 12;
if (isUnique(a, n, k))
document.write("Yes");
else
document.write("No");
// This code is contributed by Rajput-Ji
</script>
<?php
// PHP program to check
// if given encoding
// represents a single string
function isUnique( $a, $n, $k)
{
$sum = 0;
for ($i = 0; $i < $n; $i++)
$sum += $a[$i];
$sum += $n - 1;
// Return true if
// sum becomes k
return ($sum == $k);
}
// Driver Code
$a = array(3, 3, 3);
$n = count($a);
$k = 12;
if (isUnique($a, $n,$k))
echo"Yes";
else
echo "No";
// This code is contributed by anuj_67.
?>
Output
No
Time complexity : O(n)
Auxiliary Space : O(1)
