# Smallest non-zero substring which has any permutation divisible by 2^K

Given a binary string S of length N and an integer K, the task is to find the smallest non-zero sub-string of S that can be jumbled to produce a binary string divisible by 2K. If no such sub-string exists then print -1. Note that K is always greater than 0.

Examples:

Input: S = “100”, k = 1
Output: 2
Smallest substring that can be jumbled is “10”.

Input: S = “1111”, k = 2
Output: -1

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

Approach: Let’s look at the condition of the permutation of a string being divisible by 2K.

1. The string must have at least K number of 0s.
2. The string must have at least one 1.

This can be implemented using two-pointer technique. For every index i, try to find the smallest index j such that the substring S[i…j-1] satisfies the above two conditions.
Let’s say the left pointer is pointing at index i and the right pointer is pointing at j and ans stores the length of the smallest required substring.
If the condition is not satisfied then increment j, else increment i.
While iterating, find the minimum (j – i) satisfying the above two conditions and update the answer as ans = min(ans, j – i).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length of the ` `// smallest substring divisible by 2^k ` `int` `findLength(string s, ``int` `k) ` `{ ` `    ``// To store the final answer ` `    ``int` `ans = INT_MAX; ` ` `  `    ``// Left pointer ` `    ``int` `l = 0; ` ` `  `    ``// Right pointer ` `    ``int` `r = 0; ` ` `  `    ``// Count of the number of zeros and ` `    ``// ones in the current substring ` `    ``int` `cnt_zero = 0, cnt_one = 0; ` ` `  `    ``// Loop for two pointers ` `    ``while` `(l < s.size() and r <= s.size()) { ` ` `  `        ``// Condition satisfied ` `        ``if` `(cnt_zero >= k and cnt_one >= 1) { ` ` `  `            ``// Updated the answer ` `            ``ans = min(ans, r - l); ` ` `  `            ``// Update the pointer and count ` `            ``l++; ` `            ``if` `(s[l - 1] == ``'0'``) ` `                ``cnt_zero--; ` `            ``else` `                ``cnt_one--; ` `        ``} ` ` `  `        ``else` `{ ` ` `  `            ``// Update the pointer and count ` `            ``if` `(r == s.size()) ` `                ``break``; ` `            ``if` `(s[r] == ``'0'``) ` `                ``cnt_zero++; ` `            ``else` `                ``cnt_one++; ` `            ``r++; ` `        ``} ` `    ``} ` ` `  `    ``if` `(ans == INT_MAX) ` `        ``return` `-1; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"100"``; ` `    ``int` `k = 2; ` ` `  `    ``cout << findLength(s, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `    ``static` `final` `int` `INT_MAX = Integer.MAX_VALUE; ` `     `  `    ``// Function to return the length of the  ` `    ``// smallest substring divisible by 2^k  ` `    ``static` `int` `findLength(String s, ``int` `k)  ` `    ``{  ` `        ``// To store the final answer  ` `        ``int` `ans = INT_MAX;  ` `     `  `        ``// Left pointer  ` `        ``int` `l = ``0``;  ` `     `  `        ``// Right pointer  ` `        ``int` `r = ``0``;  ` `     `  `        ``// Count of the number of zeros and  ` `        ``// ones in the current substring  ` `        ``int` `cnt_zero = ``0``, cnt_one = ``0``;  ` `     `  `        ``// Loop for two pointers  ` `        ``while` `(l < s.length() && r <= s.length()) ` `        ``{  ` `     `  `            ``// Condition satisfied  ` `            ``if` `(cnt_zero >= k && cnt_one >= ``1``) ` `            ``{  ` `     `  `                ``// Updated the answer  ` `                ``ans = Math.min(ans, r - l);  ` `     `  `                ``// Update the pointer and count  ` `                ``l++;  ` `                ``if` `(s.charAt(l - ``1``) == ``'0'``)  ` `                    ``cnt_zero--;  ` `                ``else` `                    ``cnt_one--;  ` `            ``}  ` `            ``else` `            ``{  ` `     `  `                ``// Update the pointer and count  ` `                ``if` `(r == s.length())  ` `                    ``break``;  ` `                ``if` `(s.charAt(r) == ``'0'``)  ` `                    ``cnt_zero++;  ` `                ``else` `                    ``cnt_one++;  ` `                ``r++;  ` `            ``}  ` `        ``}  ` `        ``if` `(ans == INT_MAX)  ` `            ``return` `-``1``;  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``String s = ``"100"``;  ` `        ``int` `k = ``2``;  ` `     `  `        ``System.out.println(findLength(s, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the length of the ` `# smallest subdivisible by 2^k ` `def` `findLength(s, k): ` `     `  `    ``# To store the final answer ` `    ``ans ``=` `10``*``*``9` ` `  `    ``# Left pointer ` `    ``l ``=` `0` ` `  `    ``# Right pointer ` `    ``r ``=` `0` ` `  `    ``# Count of the number of zeros and ` `    ``# ones in the current substring ` `    ``cnt_zero ``=` `0` `    ``cnt_one ``=` `0` ` `  `    ``# Loop for two pointers ` `    ``while` `(l < ``len``(s) ``and` `r <``=` `len``(s)): ` ` `  `        ``# Condition satisfied ` `        ``if` `(cnt_zero >``=` `k ``and` `cnt_one >``=` `1``): ` ` `  `            ``# Updated the answer ` `            ``ans ``=` `min``(ans, r ``-` `l) ` ` `  `            ``# Update the pointer and count ` `            ``l ``+``=` `1` `            ``if` `(s[l ``-` `1``] ``=``=` `'0'``): ` `                ``cnt_zero ``-``=` `1` `            ``else``: ` `                ``cnt_one ``-``=` `1` ` `  `        ``else``: ` ` `  `            ``# Update the pointer and count ` `            ``if` `(r ``=``=` `len``(s)): ` `                ``break` `            ``if` `(s[r] ``=``=` `'0'``): ` `                ``cnt_zero ``+``=` `1` `            ``else``: ` `                ``cnt_one ``+``=` `1` `            ``r ``+``=` `1` ` `  `    ``if` `(ans ``=``=` `10``*``*``9``): ` `        ``return` `-``1` `    ``return` `ans ` ` `  `# Driver code ` `s ``=` `"100"` `k ``=` `2` ` `  `print``(findLength(s, k)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``static` `int` `INT_MAX = ``int``.MaxValue; ` `     `  `    ``// Function to return the length of the  ` `    ``// smallest substring divisible by 2^k  ` `    ``static` `int` `findLength(``string` `s, ``int` `k)  ` `    ``{  ` `        ``// To store the final answer  ` `        ``int` `ans = INT_MAX;  ` `     `  `        ``// Left pointer  ` `        ``int` `l = 0;  ` `     `  `        ``// Right pointer  ` `        ``int` `r = 0;  ` `     `  `        ``// Count of the number of zeros and  ` `        ``// ones in the current substring  ` `        ``int` `cnt_zero = 0, cnt_one = 0;  ` `     `  `        ``// Loop for two pointers  ` `        ``while` `(l < s.Length && r <= s.Length) ` `        ``{  ` `     `  `            ``// Condition satisfied  ` `            ``if` `(cnt_zero >= k && cnt_one >= 1) ` `            ``{  ` `     `  `                ``// Updated the answer  ` `                ``ans = Math.Min(ans, r - l);  ` `     `  `                ``// Update the pointer and count  ` `                ``l++;  ` `                ``if` `(s[l - 1] == ``'0'``)  ` `                    ``cnt_zero--;  ` `                ``else` `                    ``cnt_one--;  ` `            ``}  ` `            ``else` `            ``{  ` `     `  `                ``// Update the pointer and count  ` `                ``if` `(r == s.Length)  ` `                    ``break``;  ` `                     `  `                ``if` `(s[r] == ``'0'``)  ` `                    ``cnt_zero++;  ` `                ``else` `                    ``cnt_one++;  ` `                ``r++;  ` `            ``}  ` `        ``}  ` `        ``if` `(ans == INT_MAX)  ` `            ``return` `-1;  ` `             `  `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main ()  ` `    ``{  ` `        ``string` `s = ``"100"``;  ` `        ``int` `k = 2;  ` `     `  `        ``Console.WriteLine(findLength(s, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

Time Complexity: O(N)

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