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Last Updated: 6 Mar, 2021

Difficulty: Moderate

```
1. There will be no leading zeros in any string in the list ‘BINARYNUMS’.
```

```
Consider N = 5 and the list ‘binaryNums’= [“0”, “01”, “010”, “100”, “101”]. This list consists of the binary representation of numbers [0, 1, 2, 4, 5]. Clearly, the missing number is 3 and its binary representation will be “11”. So you should return string “11”.
```

```
The first line of input contains an integer ‘T’ denoting the number of test cases. then ‘T’ test cases follow.
The first line contains single integers ‘N’ represent the size of the list ‘BINARYNUMS’.
The second line contains ‘N’ space-separated string representing the list ‘BINARYNUMS’.
```

```
For each test case, print a single line containing a single string that represents this ‘Missing Integer’ in binary without leading zeros.
The output of each test case will be printed in a separate line.
```

```
You do not need to print anything, it has already been taken care of. Just implement the given function.
```

```
1 <= T <= 50
1 <= N <= 10 ^ 4
Where ‘T’ is the total number of test cases and ‘N’ is the size of list ‘BINARYNUMS’
Time limit: 1 sec.
```

In this approach, we will create an array of integers ‘**nums**’, by converting each string present in list **‘binaryNums’** in the integers they represent.

A binary string can be converted into a respective integer by simply traversing from rightmost character (least significant bit) to leftmost character (most significant bit) and for each position of set bit i.e (index where char is ‘1’) we add the power of 2 according to its position in the result.

Let’s name the method that does this conversion be **convertToInt(binStr),** where ‘**binStr’ **is the binary string whose integer equivalent is needed to be determined, then it can be implemented as follows:

- Initialize an integer variable
**num: = 0**. - Run a loop where ‘i’ ranges from len - 1 to 0 (where len is the length of ‘binStr’) and for reach ‘i’ do the following -:
- If
**binStr[i] =’ 1’,**do**num += 2 ^ (len - i - 1).**

- If
**Return ‘num’.**

**We can convert any integer to binary by repeatedly dividing it by 2. **Let’s name the method that converts an integer to binary string be **convertToBin(num),** where ‘**num’ **is the integer which we convert into a binary string, then it can be implemented as follows:

- Create an empty string
**‘binStr’**. - If ‘
**num’ = 0**then**return “0”.** - Run a loop till ‘
**num’ > 0**and in each iteration do the following -:- If
**‘num’**is odd then append ‘1’ in**‘binStr’**otherwise append ‘0’ in**‘binStr’**. - Do
**num: = num / 2.**

- If
**Reverse ‘binStr’ and then return it.**

**Algorithm**

- Create an integer array ‘
**nums**’. - Run a loop where ‘
**i’**ranges from**0 to N - 1**and for each**‘i’**do the following -:- Convert
**binaryNums[i]**in integer using method**convertToInt()**as described above and add it in array**nums.**

- Convert
- Sort
**nums**in ascending order. - Initialize an integer variable
**‘missingNum’ := N.** - Run a loop where ‘i’ ranges from ‘0’ to N - 1, and if for any ‘i’
**nums[i] != ‘i’**, then assign**missingNum’: = i and break the loop.** **Convert missingNum to the binary string using method convertToBin() as described above and return this binary string.**

In this approach, we will create a boolean array ‘**present**’ of size **‘N + 1**’, where **present[i] **will be **true, **only if the binary representation of integer** ‘i’ i**s present in the list** binaryNums. **

We can fill the boolean array ‘**present**’ by simply traversing over the list **binaryNums **and convert each string present in it using method **convertToInt() **defined in the previous approach, and mark the corresponding index **true**. In the end, only one index remains false, which will be the missing integer.

**Algorithm**

- Create a boolean array ‘
**present**’ of size**‘N’ + 1**. - Run a loop where ‘
**i’**ranges from**0 to N - 1**and for each**‘i’**do the following -:- Convert
**binaryNums[i]**in integer using method**convertToInt()**as described in Approach - 1 - Let ‘
**num**’ be the integer represented by**binaryNums[i],**then do**present[num]:= true**

- Convert
- Initialize an integer variable
**‘missingNum’: = -1.** - Run a loop where ‘i’ ranges from ‘0’ to ‘N’, and if for any ‘i’
**present[i] = false**, then assign**missingNum’:= i and break the loop.** **Convert missingNum to the binary string using method convertToBin() as described in the Approach - 1 and return this binary string**

Let ‘**xorAll**’ be the bitwise xor of all the integers from** 0 to ‘N’** and ‘**xorNum**’ be the bitwise xor of all the integers represented by the strings in the list **binaryNums. **Then missing integer will be** xorAll ^ xorNum, **where** ^ **is the bitwise xor operator.

**Algorithm**

- Initialize two integer variable ‘
**xorAll**’**: = 0**and ‘**xorNum’: = 0.** **Run a loop where ‘i’ ranges from 0 to N and for each ‘i’ do xorAll:= xorAll ^ i**- Run a loop where ‘
**i’**ranges from**0 to N - 1**and for each**‘i’**do the following -:- Convert
**binaryNums[i]**in integer using method**convertToInt()**as described in Approach - 1 - Let ‘
**num**’ be the integer represented by**binaryNums[i],**then do**xorNum:= xorNum ^ num.**

- Convert
- Initialize an integer variable
**‘missingNum’:= xorAll ^ xorNum .** **Convert missingNum to the binary string using method convertToBin() as described above and return this binary string**

The number of bits required to represent an integer ‘**N’** will be **floor(logN) + 1** (log on base 2).

You can observe that number of integers between 0 to N (inclusive) that have the 0th-bit set will be **floor((N + 1)/2)**.

The number of integers between 0 to N (inclusive) that have the 1st-bit set will be **2 * floor((N + 1)/4) + max(0, (N + 1) % 4 - 2) **

The number of integers between 0 to N (inclusive) that have the 2nd-bit set will be **4 * floor((N + 1) / 8) + max(0, (N + 1) % 8 - 4)**

In general, we can say that number of integers between 0 to N (inclusive that have ith bit-set will be ** 2 ^ i * floor((N + 1) / 2 ^ (i + 1)) + max(0, (N + 1) % 2 ^ (i + 1) - 2 ^ i). **

**Algorithm**

- Initialize an integer variable
**totalBits:= floor(logN) + 1.** **Create an empty string ‘missingNum’**- Run a loop where
**‘i’**ranges from**0**to**totalBits - 1,**and for each**‘i’**do the following -:- Initialize an integer variable
**countReq: = 2 ^ i * floor((N + 1) / 2 ^ (i + 1)) + max(0, (N + 1) % 2 ^ (i + 1) - 2 ^ i).** - Initialize another integer variable
**count:= 0.** - Run a loop where
**‘j’**ranges from**0 to N - 1**and for each**‘j’**do the following -:- Let
**len =****binaryNums[j].length** - If
**len > i**and**binaryNums[j][len - i - 1] = ‘1’,**then increment**count**by 1.

- Let
- If
**count != countReq,**then append ‘1’ in**missingNum’**otherwise append ‘0’ in**‘missingNum’**.

- Initialize an integer variable
- Reverse
**‘missingNum’**and remove all leading ‘0’s. **Return ‘missingNum’.**

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