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‎SlidingWindow/Find_All_Anagarms_In_A_String.java

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//Leetcode 438. Find All Anagarms In a String
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// Question - https://leetcode.com/problems/find-all-anagrams-in-a-string/
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class Solution {
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public List<Integer> findAnagrams(String s, String p) {
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//to store the result
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List<Integer> res = new ArrayList<>();
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//to store frequency of each character in 'p'
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HashMap<Character,Integer> freqTable = new HashMap<>();
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for(int i=0;i<p.length();i++){
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int freq = freqTable.getOrDefault(p.charAt(i),0);
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freqTable.put(p.charAt(i),++freq);
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}
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int start = 0;
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int end = 0;
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int counter = freqTable.size();
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while(end<s.length()){
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//keep sliding the 'end' of the window to the right till all the characters of 'p' are present in the window
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char endChar = s.charAt(end);
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if(freqTable.containsKey(endChar)){
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int freq = freqTable.get(endChar);
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--freq;
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freqTable.put(endChar,freq);
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if(freq==0) counter--;
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}
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end++;
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//the current window has all the characters of 'p'. slide the 'start' of the window and check if the window contains anagram or not
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while(counter==0){
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//checking for anagram
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if(end-start == p.length()) res.add(start);
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char startChar = s.charAt(start);
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if(freqTable.containsKey(startChar)){
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int freq = freqTable.get(startChar);
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freq++;
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freqTable.put(startChar,freq);
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if(freq>0) counter++; //condition not met - window does not contain all characters of 'p' - stop sliding the 'start' of window to right and begin moving 'end' to the right
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}
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start++;
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}
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}
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return res;
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}
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}

‎SlidingWindow/Longest_Repeating_Character_Replacement.java

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//Leetcode 424. Longest Repeating Character Replacement
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//Question - https://leetcode.com/problems/longest-repeating-character-replacement/
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class Solution {
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public int characterReplacement(String s, int k) {
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//number of characters that can be replaced in a given window is
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//charsToBeReplaced = length Of Window - frequency Of Most Frequent Letter In Window
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//if charsToBeReplaced <= k then the window is valid - else window in invalid and we need to slide 'start' to right
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HashMap<Character,Integer> freqTable = new HashMap<>();
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int start = 0;
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int len = 0;
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int maxCount = 0;
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int charsToReplace = 0;
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for(int end = 0 ; end < s.length() ; end++){
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char endChar = s.charAt(end);
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int freq = freqTable.getOrDefault(endChar,0);
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freq++;
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freqTable.put(endChar,freq);
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maxCount = Math.max(maxCount, freq);
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charsToReplace = end - start - maxCount + 1;
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if(charsToReplace>k){
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char startChar = s.charAt(start);
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freq = freqTable.get(startChar);
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freq--;
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freqTable.put(startChar,freq);
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start++;
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}
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len = Math.max(len, end-start+1);
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}
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return len;
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}
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}

‎SlidingWindow/Longest_Substring_With_Atmost_Two_Distinct_Characters.java

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//Leetcode 159. Longest Substring With Atmost Two Distinct Characters
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//Question - https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/
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class Solution {
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public int lengthOfLongestSubstringTwoDistinct(String s){
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if(s.length()==0) return 0;
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//to store the frequency of characters
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HashMap<Character,Integer> freqTable = new HashMap<>();
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int start = 0;
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int end = 0;
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int distinctChars = 0;
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int len = 0;
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while(end<s.length()){
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char endChar = s.charAt(end);
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int freq = freqTable.getOrDefault(endChar,0);
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freq++;
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freqTable.put(endChar,freq); //increment count for repeated characters
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if(freq==1) distinctChars++; //count distinct characters in the window
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end++; //slide 'end' of window to right
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while(distinctChars>2){
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//slide 'start' of window to right as long as the number of distinct characters in window is greater than 2
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char startChar = s.charAt(start);
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if(freqTable.containsKey(startChar)){
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freq = freqTable.get(startChar);
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freq--;
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freqTable.put(startChar,freq);
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if(freq==0) distinctChars--; //startChar is no longer a part of our window and so the number of distinct characters must be decremented
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}
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start++;
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}
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len = Math.max(len,end-start); //calculate length after each slide
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}
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return len;
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}
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}

‎SlidingWindow/Longest_Substring_Without_Repeating_Characters.java

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//Leetcode 3. Longest Substring Without Repeating Characters
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// Question - https://leetcode.com/problems/longest-substring-without-repeating-characters/
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class Solution {
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public int lengthOfLongestSubstring(String s) {
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if(s.length()==0) return 0;
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//a Map to store the last-seen index of all characters scanned - we will move the 'start' of our window to right according to the data in the map
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HashMap<Character,Integer> indexTable = new HashMap<>();
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int start = 0;
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int end = 0;
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int len = s.length();
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int result = 0;
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for(start = 0, end = 0; end<len ; end++){
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char currChar = s.charAt(end);
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if(indexTable.containsKey(currChar)){
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start = Math.max(start,indexTable.get(currChar)+1);
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//the max condition helps us to check if 'currChar' is in the current window or not.
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}
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//update indexTable to store the current index of currChar
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indexTable.put(currChar,end);
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result = Math.max(result,end-start+1);
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}
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return result;
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}
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}

‎SlidingWindow/Minimum_Window_Substring.java

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//Leetcode 76. Minimum Window Substring
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//Question - https://leetcode.com/problems/minimum-window-substring/
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class Solution {
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public String minWindow(String s, String t) {
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//creating a storage to check if the threshold frequency of each character in 't' is met.
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HashMap<Character, Integer> freqTable = new HashMap<Character,Integer>();
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//fill the frequency table
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for(int i=0;i<t.length();i++){
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char curr = t.charAt(i);
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int freq = 1;
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if(freqTable.containsKey(curr)){
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freq = freqTable.get(curr);
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freq++;
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}
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freqTable.put(curr,freq);
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}
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//initialize the window
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int start = 0;
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int end = 0;
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String ans = "";
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int windowLen = Integer.MAX_VALUE;
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//keeps the count of distinct characters in 't' which are not yet in the window
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int unmatchedChars = freqTable.size();
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//start sliding the window
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while(end<s.length()){
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char endChar = s.charAt(end);
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if(freqTable.containsKey(endChar)){
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int freq = freqTable.get(endChar);
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freq--;
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freqTable.put(endChar, freq);
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//minimum threshold reached for 'endChar'
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if(freq==0) unmatchedChars--;
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}
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end++;
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//as long as the threshold of all characters in 't' is maintained. Keep sliding the start of the window to the right. Trimming unnecessarcy characters to minimize window size.
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//The sliding of start of the window to the right is triggered only when the threshold of all characters in 't' is met.
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while(start<s.length() && unmatchedChars==0){
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char startChar = s.charAt(start);
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//update window length and answer after each slide.
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if(end-start<windowLen){
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windowLen = end-start;
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ans = s.substring(start,end);
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}
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if(freqTable.containsKey(startChar)){
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int freq = freqTable.get(startChar);
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freq++;
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freqTable.put(startChar,freq);
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//threshold for startChar is not met. sliding the start of window has to be stopped.
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if(freq>0) unmatchedChars++;
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}
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start++;
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}
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}
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return ans;
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}
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}

‎SlidingWindow/Permutation_in_String.java

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//Leetcode 567. Permutation in String
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//Question - https://leetcode.com/problems/permutation-in-string/
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class Solution {
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public boolean checkInclusion(String s1, String s2) {
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if(s1.length()==0 && s2.length()==0) return true;
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else if(s1.length()==0) return true;
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else if(s2.length()==0) return false;
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//We are supposed to check if 's2' contains the anagram of 's1'
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HashMap<Character,Integer> freqTable = new HashMap<>();
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int freq = -1;
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for(int i=0;i<s1.length();i++){
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char curr = s1.charAt(i);
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freq = freqTable.getOrDefault(curr,0);
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freq++;
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freqTable.put(curr,freq);
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}
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//System.out.println(freqTable);
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int end = 0;
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int start = 0;
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int counter = freqTable.size();
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while(end < s2.length()){
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char endChar = s2.charAt(end);
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if(freqTable.containsKey(endChar)){
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freq = freqTable.get(endChar);
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freq--;
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freqTable.put(endChar,freq);
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if(freq==0) counter--;
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}
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end++;
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while(counter==0){
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char startChar = s2.charAt(start);
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if(end-start==s1.length()){
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return true;
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}
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if(freqTable.containsKey(startChar)){
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freq = freqTable.get(startChar);
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freq++;
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freqTable.put(startChar,freq);
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if(freq>0) counter++;
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}
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start++;
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}
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}
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return false;
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}
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}

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