Given a string, find the smallest window length with all distinct characters of the given string. For eg. str = “aabcbcdbca”, then the result would be 4 as of the smallest window will be “dbca” .

Examples:

Input : aabcbcdbca Output : dcba Explanation : dbca of length 4 is the smallest window with highest number of distinct characters. Input : aaab Output : ab Explanation : ab of length 2 is the smallest window with highest number of distinct characters.

Above problem states that we have to find the smallest window that contains all the distinct characters of the given string even if the smallest string contains repeating elements.

For example, in “aabcbcdb”, the smallest string that contains all the characters is “abcbcd”.

This problem reduces to Find the smallest window in a string containing all characters of another string.

In that problem we find the smallest window that contains all the characters of given pattern.

**1- ** Count all distinct characters in given string.

**2- ** Now follow the algorithm discussed in below post.

https://www.geeksforgeeks.org/find-the-smallest-window-in-a-string-containing-all-characters-of-another-string/

We basically maintain a window of characters. Whenever the window contains all characters of given string, we shrink the window from left side to remove extra characters and then compare its length with smallest window fount so far.

## C++

// C++ program to find the smallest window containing // all characters of a pattern. #include<bits/stdc++.h> using namespace std; const int MAX_CHARS = 256; // Function to find smallest window containing // all distinct characters string findSubString(string str) { int n = str.length(); // Count all distinct characters. int dist_count = 0; bool visited[MAX_CHARS] = {false}; for (int i=0; i<n; i++) { if (visited[str[i]] == false) { visited[str[i]] = true; dist_count++; } } // Now follow the algorithm discussed in below // post. We basically maintain a window of characters // that contains all characters of given string. // https://www.geeksforgeeks.org/find-the-smallest-window-in-a-string-containing-all-characters-of-another-string/ int start = 0, start_index = -1, min_len = INT_MAX; int count = 0; int curr_count[MAX_CHARS] = {0}; for (int j=0; j<n; j++) { // Count occurrence of characters of string curr_count[str[j]]++; // If any distinct character matched, // then increment count if (curr_count[str[j]] == 1 ) count++; // if all the characters are matched if (count == dist_count) { // Try to minimize the window i.e., check if // any character is occurring more no. of times // than its occurrence in pattern, if yes // then remove it from starting and also remove // the useless characters. while (curr_count[str[start]] > 1) { if (curr_count[str[start]] > 1) curr_count[str[start]]--; start++; } // Update window size int len_window = j - start + 1; if (min_len > len_window) { min_len = len_window; start_index = start; } } } // Return substring starting from start_index // and length min_len return str.substr(start_index, min_len); } // Driver code int main() { string str = "aabcbcdbca"; cout << "Smallest window containing all distinct" " characters is " << findSubString(str); return 0; }

## Java

// Java program to find the smallest window containing // all characters of a pattern. import java.util.Arrays; public class GFG { static final int MAX_CHARS = 256; // Function to find smallest window containing // all distinct characters static String findSubString(String str) { int n = str.length(); // Count all distinct characters. int dist_count = 0; boolean[] visited = new boolean[MAX_CHARS]; Arrays.fill(visited, false); for (int i=0; i<n; i++) { if (visited[str.charAt(i)] == false) { visited[str.charAt(i)] = true; dist_count++; } } // Now follow the algorithm discussed in below // post. We basically maintain a window of characters // that contains all characters of given string. // https://www.geeksforgeeks.org/find-the-smallest-window-in-a-string-containing-all-characters-of-another-string/ int start = 0, start_index = -1; int min_len = Integer.MAX_VALUE; int count = 0; int[] curr_count = new int[MAX_CHARS]; for (int j=0; j<n; j++) { // Count occurrence of characters of string curr_count[str.charAt(j)]++; // If any distinct character matched, // then increment count if (curr_count[str.charAt(j)] == 1 ) count++; // if all the characters are matched if (count == dist_count) { // Try to minimize the window i.e., check if // any character is occurring more no. of times // than its occurrence in pattern, if yes // then remove it from starting and also remove // the useless characters. while (curr_count[str.charAt(start)] > 1) { if (curr_count[str.charAt(start)] > 1) curr_count[str.charAt(start)]--; start++; } // Update window size int len_window = j - start + 1; if (min_len > len_window) { min_len = len_window; start_index = start; } } } // Return substring starting from start_index // and length min_len return str.substring(start_index, start_index+min_len); } // Driver code public static void main(String args[]) { String str = "aabcbcdbca"; System.out.println("Smallest window containing all distinct" + " characters is " + findSubString(str)); } } // This code is contributed by Sumit Ghosh

Output:

Smallest window containing all distinct characters is dbca

Related Article :

Length of the smallest sub-string consisting of maximum distinct characters

This article is contributed by **Sahil Chhabra**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.