659. Split Array into Consecutive Subsequences

Greedily append the element to previous sequences
Always try to append
Only create a single sequence if we are not able to append

class Solution {
    public boolean isPossible(int[] nums) {
        // For every distinct number, we keep track of then consective sequence end with num
        // whose length are 1 2 3
        int prev = Integer.MIN_VALUE, p1 = 0, p2 = 0, p3 = 0;
        for (int i = 0; i < nums.length; i++) {
            int curr = nums[i];
            int c1 = 0, c2 = 0, c3 = 0;
            int count = 1;
            // accumulate numbers with same value as curr
            while (i < nums.length - 1 && nums[i + 1] == curr) {
                count++;
                i++;
            }
           
            // cannot append to previous number
            if (curr != prev + 1) {
                if (p1 != 0 || p2 != 0) return false;
                c1 = count;
            } else { // append to p1 first
                if (count < p1 + p2) return false;
                c2 = p1;
                c3 = p2 + Math.min(p3, count - (p1 + p2));
                c1 = Math.max(0, count - (p1 + p2 + p3));
            }
            prev = curr;
            p1 = c1;
            p2 = c2;
            p3 = c3;
        }
        return p1 == 0 && p2 == 0;
    }
}