Find Consecutive Integers from a Data Stream

Problem

For a stream of integers, implement a data structure that checks if the last k integers parsed in the stream are equal to value.

Implement the DataStream class:

DataStream(int value, int k) Initializes the object with an empty integer stream and the two integers value and k.
boolean consec(int num) Adds num to the stream of integers. Returns true if the last k integers are equal to value, and false otherwise. If there are less than k integers, the condition does not hold true, so returns false.

Example 1:

Input
["DataStream", "consec", "consec", "consec", "consec"]
[[4, 3], [4], [4], [4], [3]]
Output
[null, false, false, true, false]

Explanation
DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3 
dataStream.consec(4); // Only 1 integer is parsed, so returns False. 
dataStream.consec(4); // Only 2 integers are parsed.
                      // Since 2 is less than k, returns False. 
dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True. 
dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
                      // Since 3 is not equal to value, it returns False.

Constraints:

1 <= value, num <= 109
1 <= k <= 105
At most 105 calls will be made to consec.

Solution (Java)

class DataStream {
    private int val, k, count;
    public DataStream(int value, int k) {
        this.val = value;
        this.k = k;
        this.count = 0;
    }

    public boolean consec(int num) {
        // if num == val, increment the streak else set it to 0
        count = (num == val)? count + 1 : 0;
        return count >= k;

    }
}

/**
 * Your DataStream object will be instantiated and called as such:
 * DataStream obj = new DataStream(value, k);
 * boolean param_1 = obj.consec(num);
 */

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).