Count subarrays having total distinct elements same as original array

// Javascript program Count total number of sub-arrays
// having total distinct elements same as that
// original array.

// Function to calculate distinct sub-array
function countDistinctSubarray(arr, n) {
  const unst1 = new Set(arr);
  const totalDist = unst1.size;
  let count = 0;

  for (let i = 0; i < n; i++) {
    const unst = new Set();
    for (let j = i; j < n; j++) {
      unst.add(arr[j]);
      if (unst.size === totalDist) {
        count += 1;
      }
    }
  }

  return count;
}

// Driver code
const arr = [2, 1, 3, 2, 3];
const n = arr.length;

console.log(countDistinctSubarray(arr, n));

<script>
// Javascript program Count total number of sub-arrays
// having total distinct elements same as that
// original array.
    
    // Method to calculate distinct sub-array
    function countDistictSubarray(arr,n)
    {
        // Count distinct elements in whole array
        let  vis = new Map();
          
        for (let i = 0; i < n; ++i)
            vis.set(arr[i], 1);
        let k = vis.size;
       
        // Reset the container by removing all elements
        let vid=new Map();
       
        // Use sliding window concept to find
        // count of subarrays having k distinct
        // elements.
        let ans = 0, right = 0, window = 0;
        for (let left = 0; left < n; left++)
        {
            while (right < n && window < k)
            {
                if(vid.has(arr[right]))
                    vid.set(arr[right], vid.get(arr[right]) + 1);
                else
                    vid.set(arr[right], 1);
       
                if (vid.get(arr[right])== 1)
                    window++;
       
                right++;
            }
               
            // If window size equals to array distinct 
            // element size, then update answer
            if (window == k)
                ans += (n - right + 1);
       
            // Decrease the frequency of previous element
            // for next sliding window
            if(vid.has(arr[left]))
                vid.set(arr[left], vid.get(arr[left])- 1);
       
            // If frequency is zero then decrease the
            // window size
            if (vid.get(arr[left]) == 0)
                    --window;
        }
        
        return ans;
        
    }
    
    // Driver method
    let arr=[2, 1, 3, 2, 3];
    document.write(countDistictSubarray(arr, arr.length));
    

// This code is contributed by patel2127
</script>