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I am solving this problem on leetcode. Can't figure out the time and space complexity for my solution.

Particularly, I can't understand how to apply Master Theorem here in case when we have FOR loop. What is a and b here? Since input divided multiple times and for different size of subproblems. Another complication is memoization.

class Solution {
    private Map<String, List<Integer>> cache = new HashMap<>();
    public List<Integer> diffWaysToCompute(String equation) {
        if (cache.containsKey(equation)) return cache.get(equation);
        if (!(equation.contains("+") || equation.contains("-") || equation.contains("*"))) return Collections.singletonList(Integer.valueOf(equation));
        List<Integer> result =  new ArrayList<>();

        for (int i = 0; i < equation.length();i++) {
            char ch = equation.charAt(i);

            if (ch == '+' || ch == '-' || ch == '*') {
                List<Integer> left = diffWaysToCompute(equation.substring(0, i));
                List<Integer> right = diffWaysToCompute(equation.substring(i+1, equation.length()));

                result.addAll(crossCalc(left, right, ch));
            }
        }

        cache.put(equation, result);

        return result;
    }

    private List<Integer> crossCalc(List<Integer> left, List<Integer> rigth, char sign) {
        List<Integer> result = new ArrayList<>();
        for (Integer l : left) {
            for (Integer r : rigth) {
                if (sign == '-') {
                    result.add(l - r);
                } else if (sign == '+') {
                    result.add(l + r);
                } else {
                    result.add(l*r);
                }
            }
        }
        return result;
    }
}

I am looking for explanation how to calculate time complexity, not only the answer. Preferably if you could explain complexity for both with and without memoization. Thanks!

John Smith
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  • I don't think you can apply master theorem here. master theorem is used if you divide your problem into two or more completely disjoint subproblems and merge their results at the end – mangusta Jan 08 '19 at 09:17

2 Answers2

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T(n) = Sum{T(i) + T(N-i)} for some index i <= 2(T(1) + T(2) + ... + T(n - 1))
=> T(n + 1) - T(n) = 2T(n) => T(n) <= O(3^n) worst case

where n is count of split numbers

chenyuandong
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The time complexity of your algorithm is equal to the number of expressions containing n pairs of parentheses which are correctly matched.

It's called a Catalan number, and it's equal to C(2 * n, n) / (n + 1) = (2 * n)! / ((n + 1)! * n!).

Also, there's a recursive formula for calculating a Catalan number:

f(n+1) = f(0)f(n) + f(1)f(n-1) + f(2)f(n-2) + ... + f(n-2)f(2) + f(n-1)f(1) + f(n)f(0)

And you know, it's the same as your algorithm time complexity equation!

T(n+1) = T(0)T(n) + T(1)T(n-1) + T(2)T(n-2) + ... + T(n-2)T(2) + T(n-1)T(1) + T(n)T(0)

The memory complexity of this algorithm can be as large as the time complexity of it because the number of elements of result ArrayList can be large. So in worst-case memory and time complexity would be the n-th Catalan number.

Source: https://en.wikipedia.org/wiki/Catalan_number

Erfan Alimohammadi
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