Runtime: 12 ms (Top 87.36%) | Memory: 75.4 MB (Top 41.15%)

Author: AnasImloul

scripts/algorithms/C/Count Number of Nice Subarrays/Count Number of Nice Subarrays.java

@@ -1,30 +1,30 @@
+// Runtime: 12 ms (Top 87.36%) | Memory: 75.4 MB (Top 41.15%)
 class Solution {
     public int numberOfSubarrays(int[] nums, int k) {
         int i = 0;
         int j = 0;
         int odd = 0;
         int result = 0;
         int temp = 0;
-        
-        
-        /* 
+
+        /*
             Approach : two pointer + sliding window technique
-            
+
             step 1 : we have fix i and moving j until our count of odd numbers == k
             step 2 : when(odd == count) we are counting every possible subarray by reducing the size of subarray from i
-            
+
         why temp?
         from reducing the size of subarray we will count all the possible subarray from between i and j
         but when i encounter a odd element the odd count will reduce and that while will stop executing
-        
+
         now there are two possible cases
         1.The leftover elements have a odd number
         2.The leftover elements do not have any odd number
-        
-        1. if our leftover elements have a odd number 
+
+        1. if our leftover elements have a odd number
                 then we cannot include our old possible subarrays into new possible subarrays because now new window for having odd == k is formed
                 that's why temp = 0;
-                
+
         2. if out leftover elements do not have any odd element left
             then our leftover elements must also take in consideration becuase they will also contribute in forming subarrays
         */
@@ -45,8 +45,8 @@ public int numberOfSubarrays(int[] nums, int k) {
         //so include them
             result += temp;
             j++;
-            
+
         }
         return result;
     }
-}
+}