# Check if all the 1’s in a binary string are equidistant or not

Given a binary string str, the task is to check if all the 1’s in the string are equidistant or not. The term equidistant means that the distance between every two adjacent 1’s is same. Note that the string contains at least two 1’s.

Examples:

Input: str = “00111000”
Output: Yes
The distance between all the 1’s is same and is equal to 1.

Input: str = “0101001”
Output: No
The distance between the 1st and the 2nd 1’s is 2
and the distance between the 2nd and the 3rd 1’s is 3.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Store the position of all the 1’s in the string in a vector and then check if the difference between each two consecutive positions is same or not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if all the 1's ` `// in the binary string s are equidistant ` `bool` `check(string s, ``int` `l) ` `{ ` ` `  `    ``// Initialize vector to store ` `    ``// the position of 1's ` `    ``vector<``int``> pos; ` ` `  `    ``for` `(``int` `i = 0; i < l; i++) { ` ` `  `        ``// Store the positions of 1's ` `        ``if` `(s[i] == ``'1'``) ` `            ``pos.push_back(i); ` `    ``} ` ` `  `    ``// Size of the position vector ` `    ``int` `t = pos.size(); ` `    ``for` `(``int` `i = 1; i < t; i++) { ` ` `  `        ``// If condition isn't satisfied ` `        ``if` `((pos[i] - pos[i - 1]) != (pos - pos)) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Drivers code ` `int` `main() ` `{ ` `    ``string s = ``"100010001000"``; ` `    ``int` `l = s.length(); ` `    ``if` `(check(s, l)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function that returns true if all the 1's  ` `// in the binary string s are equidistant  ` `static` `boolean` `check(String s, ``int` `l)  ` `{  ` ` `  `    ``// Initialize vector to store  ` `    ``// the position of 1's  ` `    ``Vector pos = ``new` `Vector();  ` ` `  `    ``for` `(``int` `i = ``0``; i < l; i++)  ` `    ``{  ` ` `  `        ``// Store the positions of 1's  ` `        ``if` `(s.charAt(i)== ``'1'``)  ` `            ``pos.add(i);  ` `    ``}  ` ` `  `    ``// Size of the position vector  ` `    ``int` `t = pos.size();  ` `    ``for` `(``int` `i = ``1``; i < t; i++)  ` `    ``{  ` ` `  `        ``// If condition isn't satisfied  ` `        ``if` `((pos.get(i) - pos.get(i-``1``)) != (pos.get(``1``) - pos.get(``0``)))  ` `            ``return` `false``;  ` `    ``}  ` ` `  `    ``return` `true``;  ` `}  ` ` `  `// Drivers code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``String s = ``"100010001000"``;  ` `    ``int` `l = s.length();  ` `    ``if` `(check(s, l))  ` `        ``System.out.print(``"Yes"``);  ` `    ``else` `        ``System.out.print(``"No"``);  ` `} ` `}  ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function that returns true if all the 1's ` `# in the binary s are equidistant ` `def` `check(s, l): ` ` `  `    ``# Initialize vector to store ` `    ``# the position of 1's ` `    ``pos ``=` `[] ` ` `  `    ``for` `i ``in` `range``(l): ` ` `  `        ``# Store the positions of 1's ` `        ``if` `(s[i] ``=``=` `'1'``): ` `            ``pos.append(i) ` ` `  `    ``# Size of the position vector ` `    ``t ``=` `len``(pos) ` `    ``for` `i ``in` `range``(``1``, t): ` ` `  `        ``# If condition isn't satisfied ` `        ``if` `((pos[i] ``-`  `             ``pos[i ``-` `1``]) !``=` `(pos[``1``] ``-`  `                             ``pos[``0``])): ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Driver code ` `s ``=` `"100010001000"` `l ``=` `len``(s) ` `if` `(check(s, l)): ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) ` ` `  `# This code is contributed ` `# by mohit kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` ` `  `// Function that returns true if all the 1's  ` `// in the binary string s are equidistant  ` `static` `bool` `check(String s, ``int` `l)  ` `{  ` ` `  `    ``// Initialize vector to store  ` `    ``// the position of 1's  ` `    ``List<``int``> pos = ``new` `List<``int``>();  ` ` `  `    ``for` `(``int` `i = 0; i < l; i++)  ` `    ``{  ` ` `  `        ``// Store the positions of 1's  ` `        ``if` `(s[i]== ``'1'``) ` `        ``{ ` `             `  `            ``pos.Add(i);  ` `        ``} ` `    ``}  ` ` `  `    ``// Size of the position vector  ` `    ``int` `t = pos.Count;  ` `    ``for` `(``int` `i = 1; i < t; i++)  ` `    ``{  ` ` `  `        ``// If condition isn't satisfied  ` `        ``if` `((pos[i] - pos[i - 1]) != (pos - pos)) ` `            ``return` `false``; ` `    ``}  ` ` `  `    ``return` `true``;  ` `}  ` ` `  `// Drivers code  ` `public` `static` `void` `Main(String []args) ` `{  ` `    ``String s = ``"100010001000"``;  ` `    ``int` `l = s.Length;  ` `    ``if` `(check(s, l))  ` `        ``Console.WriteLine(``"Yes"``);  ` `    ``else` `        ``Console.WriteLine(``"No"``);  ` `} ` `}  ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

```Yes
```

Time Complexity: O(N) where N is the length of the string.

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