Count the Number of Good Subarrays - Sliding Window [JS]

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Description 

Solution: Sliding Window

If a subarray has at least k good pairs, then we know that expanding the subarray will always also result in at least k good pairs.
Keep track of the count of occurances of each number.
We have two pointers i and j for our sliding window.

  • As we move j up, we will gain another count[nums[j]] good pairs.
  • Move i up while the good pairs >= k. When we move i up, we subtract count[nums[i]] - 1 good pairs.
  • Then, we will know that all subarrays with start index between 0 and i - 1, and end index at j will have at least k good pairs.

Time Complexity: O(n)
Space Complexity: O(n)

var countGood = function(nums, k) {
  let n = nums.length, count = {}, goodPairs = 0, ans = 0;
  for (let j = 0, i = 0; j < n; j++) {
    goodPairs += count[nums[j]] || 0;
    count[nums[j]] = (count[nums[j]] || 0) + 1;
    while (goodPairs >= k) {
      goodPairs -= --count[nums[i]];
      i++;
    }
    ans += i;
  }
  return ans;
};
Location: Auckland, New Zealand

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