Leetcode递增三元子序列：记忆化递归优化未全过用例求指导

我正在Leetcode上求解递增三元子序列问题。最初采用暴力递归解法，虽通过多数测试用例但出现超时问题，代码如下：

public static void main(String[] args) {
        IncreasingTriplet it = new IncreasingTriplet();
        int[] nums = { 1, 5, 0, 4, 1, 3 };
        System.out.println(it.increasingTriplet(nums));
    }

    public boolean increasingTriplet(int[] nums) {
        // if there are less than 3 numbers return false;
        if (nums.length < 3)
            return false;
        return helper(nums, nums.length, Integer.MIN_VALUE, 0, 0);
    }

    private boolean helper(int[] nums, int n, int prev, int index, int count) {
        if (index >= n) {
            if (count < 3) {
                return false;
            } else if (count == 3) {
                return true;
            }
            return false;
        } else if (count == 3) {
            return true;
        } else if (count > 3) {
            return false;
        }
        if (nums[index] > prev) {
            if (helper(nums, n, nums[index], index + 1, count + 1) || helper(nums, n, prev, index + 1, count)) {
                return true;
            }
            return false;
        }
        return helper(nums, n, prev, index + 1, count);
}

之后我尝试通过构建二维缓存cache[index][count]，记录每个索引作为递增三元序列第1、2、3位的状态来优化，但仍无法通过全部测试用例。请问该思路是否存在问题？恳请提供指导建议。优化后的代码如下：

public boolean increasingTriplet(int[] nums) {
        //if there are less than 3 numbers return false;
        if(nums.length < 3) return false;
        int[][]cache = new int[nums.length][4];
        for(int i = 0; i < cache.length; i++){
            for(int j = 0; j < cache[0].length; j++){
                cache[i][j] = -1;
            }
        }
        return helper(nums, nums.length, Integer.MIN_VALUE, 0, 0, cache);
    }
    //Let's optimize by keeping track of the index as either part of the sequence as 1st, 2nd or 3rd and then store that for future use.
    private boolean helper(int[] nums, int n, int prev, int index, int count, int[][]cache){
        if(index >= n){
            if(count < 3){
                return false;
            }
            if( count == 3){
                return true;
            }
            return false;
        }
        if(count == 3){
            return true;
        }
        if(count > 3){
            return false;
        }
        if(cache[index][count] != -1){
            return cache[index][count] == 1? true:false;
        }
        if(nums[index] > prev){
            if(helper(nums, n, nums[index], index + 1, count + 1, cache) || helper(nums, n, prev, index + 1, count, cache)){
                cache[index][count] = 1;
                return true;
            }
            cache[index][count] = 0;
            return false;
        }
        return helper(nums, n, prev, index + 1, count, cache);
    }

内容的提问来源于stack exchange，提问作者Spindoctor