Smallest distinct window

# include # include # include using namespace std ; string smallestWindow(string str)

QUESTION: Smallest distinct window Given two strings S and P. Find the smallest window in the string S consisting of all the characters(including duplicates) of the string P. Return "-1" in case there is no such window present. In case there are multiple such windows of same length, return the one with the least starting index. Example 1: Input: S = "timetopractice" P = "toc" Output: toprac Explanation: "toprac" is the smallest substring in which "toc" can be found. Example 2: Input: S = "zoomlazapzo" P = "oza" Output: apzo Explanation: "apzo" is the smallest substring in which "oza" can be found. Your Task: You don't need to read input or print anything. Your task is to complete the function smallestWindow() which takes two string S and P as input paramters and returns the smallest window in string S having all the characters of the string P. In case there are multiple such windows of same length, return the one with the least starting index. Expected Time Complexity: O(|S|) Expected Auxiliary Space: O(1) Constraints: 1 ≤ |S|, |P| ≤ 105 SOLUTION: // Driver Code Ends • Copy lines • Copy permalink • • Go Footer

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: dbca Explanation: Possible substrings= Of the set of possible substrings 'ab' is the shortest substring having all the distinct characters of given string. Solution: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. Method 1 :This is the Brute Force method of solving the problem using HashMap. • Approach :For solving the problem we first have to find out all the distinct characters present in the string. This can be done using a • • HashMap is a part of Java's collection since Java 1.2. It provides the basic implementation of the Map interface of Java. It stores the data in (Key, Value) pairs. To access a value one must know its key. HashMap is known as HashMap because it uses a technique called Hashing. Hashing is a technique of converting a large String to small String that represents the same String. A shorter value helps in indexing and faster searches. HashSet also uses HashMap internally. It internally uses a link list to store key-value pairs already explained in HashSet in detail and further articles. • Algorithm : • Store ...

A tag already exists with the provided branch name. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Are you sure you want to create this branch? Cancel Create dsa-levelUp-solutions-cpp / Section 04 - sliding window / smallestDistinctWindow.cpp # include # include using namespace std ; const int MAX_CHARS = 256; //Algorithm : https://www.geeksforgeeks.org/smallest-window-contains-characters-string/#:~:text=Solution%3A%20Above%20problem%20states%20that,the%20characters%20is%20%E2%80%9Cabcbcd%E2%80%9D. // Function to find smallest window containing // all distinct characters string smallestWindow(string str) • Copy lines • Copy permalink • • Go Footer

We are given an array consisting of n integers and an integer k. We need to find the minimum range in array [l, r] (both l and r are inclusive) such that there are exactly k different numbers. If such subarray doesn’t exist print “Invalid k”. Examples: Input : arr[] = k = 3 Output : Invalid k • Pick each of the elements from the given array as the starting element [ i-th element ] of our required subarray. • In each iteration initialize an empty set to store the distinct elements of the subarray • Pick each remaining element [ i, i+1,..n – 1] from the array as the last element [ j-th element ]. • Add the current element to the set. • If the set size equals k then update the results and break from the inner loop (already found k distinct elements increasing the size of the subarray has 2 possibilities either will get more distinct elements, or increase the subarray size with repeated elements which are not to be considered in the required results). • If (j == n) or j = size of the array, i.e. we have not found any desired subarray starting from i-th index and going forward we will be having fewer elements to consider. ( For example : consider given array is 4 5 5 4 5and k = 3, when start from 0th index we will not find any subarray of k size and j will reach end so that means we won’t get any element that can make a k = 3 size required subarray). So, Break from the outer loop. • Print the output if found, otherwise, print “Invalid k”. Implementation: Output 0 2 Time Co...

A tag already exists with the provided branch name. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Are you sure you want to create this branch? Cancel Create Smallest-window-that-contains-all-characters-of-string-itself / Smallest-window-that-contains-all-characters-of-string-itself.cpp /* This is the Brute Force method of solving the problem using HashMap. Approach : For solving the problem we first have to find out all the distinct characters present in the string. This can be done using a HashMap. The next thing is to generate all the possible substrings. This follows by checking whether a substring generated has all the required characters(stored in the hash_map) or not. If yes, then compare its length with the minimum substring length which follows the above constraints, found till now. HashMap: HashMap is a part of Java’s collection since Java 1.2. It provides the basic implementation of the Map interface of Java. It stores the data in (Key, Value) pairs. To access a value one must know its key. HashMap is known as HashMap because it uses a technique called Hashing. Hashing is a technique of converting a large String to small String that represents the same String. A shorter value helps in indexing and faster searches. HashSet also uses HashMap internally. It internally uses a link list to store key-value pairs already explained in HashSet in detail and further articles. Time Complexity: O(N^2). This time is...