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Smallest non-zero substring which has any permutation divisible by 2^K
• Last Updated : 13 May, 2021

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:
Smallest substring that can be jumbled is “10”.
Input: S = “1111”, k = 2
Output: -1

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`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)

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