Count Number of Nice Subarrays LeetCode solution in Java

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class Solution {
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  public int numberOfSubarrays(int[] nums, int k) {
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    int i = 0;
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    int j = 0;
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    int odd = 0;
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    int result = 0;
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    int temp = 0;
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    /*
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        Approach : two pointer + sliding window technique
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        step 1 : we have fix i and moving j until our count of odd numbers == k
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        step 2 : when(odd == count) we are counting every possible subarray by reducing the size of subarray from i
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    why temp?
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    from reducing the size of subarray we will count all the possible subarray from between i and j
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    but when i encounter a odd element the odd count will reduce and that while will stop executing
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    now there are two possible cases
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    1.The leftover elements have a odd number
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    2.The leftover elements do not have any odd number
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    1. if our leftover elements have a odd number
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            then we cannot include our old possible subarrays into new possible subarrays because now new window for having odd == k is formed
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            that's why temp = 0;
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    2. if out leftover elements do not have any odd element left
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        then our leftover elements must also take in consideration becuase they will also contribute in forming subarrays
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    */
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    while (j < nums.length) {
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      if (nums[j] % 2 != 0) {
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        odd++;
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        // if leftover elements have odd element then new window is formed so we set temp = 0;
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        temp = 0;
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      }
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      while (odd == k) {
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        temp++;
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        if (nums[i] % 2 != 0) odd--;
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        i++;
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      }
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      // if no leftover elements is odd, each element will part in forming subarray
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      // so include them
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      result += temp;
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      j++;
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    }
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    return result;
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  }
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}