smallest window

Author: sachuverma

DSA Crack Sheet/README.md

@@ -84,6 +84,8 @@
 - [Second most repeated string in a sequence](https://practice.geeksforgeeks.org/problems/second-most-repeated-string-in-a-sequence0534/1# "view question") - [Cpp Solution](./solutions/Second%20most%20repeated%20string%20in%20a%20sequence.cpp)
 - [Minimum Swaps for Bracket Balancing](https://practice.geeksforgeeks.org/problems/minimum-swaps-for-bracket-balancing/0# "view question") - [Cpp Solution](./solutions/Minimum%20Swaps%20for%20Bracket%20Balancing.cpp)
 - [Longest Common Subsequence](https://practice.geeksforgeeks.org/problems/longest-common-subsequence-1587115620/1 "view question") - [Cpp Solution](./solutions/Longest%20Common%20Subsequence.cpp)
+- [Program to generate all possible valid IP addresses from given string](https://www.geeksforgeeks.org/program-generate-possible-valid-ip-addresses-given-string/ "view topic")
+- [Smallest distinct window](https://practice.geeksforgeeks.org/problems/smallest-distant-window3132/1 "view question") - [Cpp Solution](./solutions/Smallest%20distinct%20window.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Searching & Sorting

DSA Crack Sheet/solutions/Smallest distinct window.cpp

@@ -0,0 +1,78 @@
+/*
+Smallest distinct window
+========================
+
+Given a string 's'. The task is to find the smallest window length that contains all the characters of the given string at least one time.
+For eg. A = “aabcbcdbca”, then the result would be 4 as of the smallest window will be “dbca”.
+
+Example 1:
+Input : "AABBBCBBAC"
+Output : 3
+Explanation : Sub-string -> "BAC"
+
+Example 2:
+Input : "aaab"
+Output : 2
+Explanation : Sub-string -> "ab"
+
+Example 3:
+Input : "GEEKSGEEKSFOR"
+Output : 8
+Explanation : Sub-string -> "GEEKSFOR"
+
+Your Task:  
+You don't need to read input or print anything. Your task is to complete the function findSubString() which takes the string  S as inputs and returns the length of the smallest such string.
+
+Expected Time Complexity: O(256.N)
+Expected Auxiliary Space: O(256)
+
+Constraints:
+1 ≤ |S| ≤ 105
+String may contain both type of English Alphabets.
+*/
+
+string findSubString(string str)
+{
+  int n = str.size();
+  int dist_count = 0;
+  bool visited[256] = {false};
+
+  for (int i = 0; i < n; i++)
+  {
+    if (visited[str[i]] == false)
+    {
+      visited[str[i]] = true;
+      dist_count++;
+    }
+  }
+
+  int start = 0, start_index = -1, min_len = INT_MAX;
+
+  int count = 0;
+  int curr_count[256] = {0};
+
+  for (int i = 0; i < n; ++i)
+  {
+    curr_count[str[i]]++;
+    if (curr_count[str[i]] == 1)
+      count++;
+    if (count == dist_count)
+    {
+      while (curr_count[str[start]] > 1)
+      {
+        if (curr_count[str[start]] > 1)
+          curr_count[str[start]]--;
+        start++;
+      }
+
+      int len_window = i - start + 1;
+      if (min_len > len_window)
+      {
+        min_len = len_window;
+        start_index = start;
+      }
+    }
+  }
+
+  return str.substr(start_index, min_len);
+}