commit

resources/sliding-window.md

@@ -8,6 +8,8 @@
 - [Longest Substring With K Unique Characters / Variable Size Sliding Window](/src/com/problems/slidingwindow/LargestSubstringWithKUniqueCharacters.java)
 - [Longest Substring With Without Repeating Characters / Variable Size Sliding Window](/src/com/problems/slidingwindow/LongestSubstringWithoutRepeatingCharacters.java)
 - [Fruit Into Baskets / Pick Toys / An Interesting Sliding Window Problem](/src/com/problems/slidingwindow/FruitIntoBaskets.java)
+- [Minimum Number of Flips to Make the Binary String Alternating](/src/com/problems/slidingwindow/MinimumNumberOfFlipsToMakeTheBinaryStringAlternating.java)
+- [Maximum Number of Vowels in a Substring of Given Length](/src/com/problems/slidingwindow/MaximumNumberOfVowelsInASubstringOfGivenLength.java)
 - [Minimum Window Substring / Variable Size Sliding Window](/src/com/problems/slidingwindow/MinimumWindowSubstring.java)
 - [Minimum Window Subsequence](/src/com/problems/slidingwindow/MinimumWindowSubsequence.java)
 - [Maximum Sum of Distinct Subarrays With Length K](/src/com/problems/slidingwindow/MaximumSumOfDistinctSubarraysWithLengthK.java)
@@ -16,6 +18,7 @@
 - [Number of Substrings Containing All Three Characters](/src/com/problems/slidingwindow/NumberOfSubstringsContainingAllThreeCharacters.java)
 - [Max Consecutive Ones III](/src/com/problems/slidingwindow/MaxConsecutiveOnes3.java)
 - [Longest Repeating Character Replacement](/src/com/problems/slidingwindow/LongestRepeatingCharacterReplacement.java)
+- [Permutation in String](/src/com/problems/slidingwindow/PermutationInString.java)
 - [Binary Subarrays With Sum](/src/com/problems/slidingwindow/BinarySubarraysWithSum.java)
 - [Count Number of Nice Subarrays](/src/com/problems/slidingwindow/CountNumberOfNiceSubarrays.java)
 - [Maximum Points You Can Obtain from Cards](/src/com/problems/slidingwindow/MaximumPointsYouCanObtainFromCards.java)

src/com/problems/slidingwindow/FrequencyOfTheMostFrequentElement.java

@@ -7,19 +7,23 @@
  * https://leetcode.com/problems/frequency-of-the-most-frequent-element/description/
  *
  * Solution link :
- * Neetcode : https://www.youtube.com/watch?v=vgBrQ0NM5vE
- * Aryan Mittal : https://www.youtube.com/watch?v=e0AFPpKcjw0
+ * Neetcode
+ * https://www.youtube.com/watch?v=vgBrQ0NM5vE
+ * Aryan Mittal :
+ * https://www.youtube.com/watch?v=e0AFPpKcjw0
  *
  * https://takeuforward.org/arrays/find-the-highest-lowest-frequency-element/
+ * https://neetcode.io/solutions/frequency-of-the-most-frequent-element
  */
+
 public class FrequencyOfTheMostFrequentElement {
     public static void main(String[] args) {
         type1();
     }
 
-    // using the sliding window approach
-    // time complexity O(nlog(n) + 2n)
-    // the intuition is kind of greedy
+    // todo using the sliding window approach
+    //  time complexity O(nlog(n) + 2n)
+    //  the intuition is kind of greedy
     // lets say if the nums array is [1,1,1,3,3,5]
     // and we are operating on 5 currently, and we need to choose other
     // numbers to convert them into 5, if we change 1 into 5 it will take 4 unit
@@ -30,7 +34,7 @@ public static void main(String[] args) {
     // let's say we are on j th element, we will assume that the left side numbers
     // will be converted into this, if the left point is on i
     // so the total range is (j-i+1)
-    // and our criteria will be nums[j]*(j-i+1) - range-sum(j-i+1) <= k
+    // and our criteria will be (nums[j] * (j - i + 1)) - range-sum(j - i + 1) <= k
     // if it is less than equal to k then it is possible to convert all the numbers in that range
     // else we have to increment left till it does not fit into the criteria
     // we will also need to carry the range sum
@@ -44,20 +48,23 @@ private static void type1() {
     public static int maxFrequency1(int[] nums, int k) {
         // we need to sort at first
         Arrays.sort(nums);
+        int n = nums.length;
+
         long sum = 0;
         int max = 0;
-        int n = nums.length, left = 0, right = 0;
-        while (right < n) {
+        // we will go from 0 to n and choose current number to be target number for all the left side numbers
+        for (int right = 0, left = 0; right < n; right++) {
             // adding the current element in the range sum
             long num = nums[right];
-            sum += num;
+            sum += num; // updating the sum
+            // if num * range - range sum > k then we can not transform all the numbers to the current num
+            // so we will decrease from left and shrink the window
             while (num * (right - left + 1) - sum > k) {
                 // decrementing the left most element from that range
                 sum -= nums[left++];
             }
             // checking the max frequency
             max = Math.max(max, right - left + 1);
-            right++;
         }
         return max;
     }