CODEWITHSUNDEEP
javaalgorithmoptimizationiterationtime-complexity
Viktor Plane
Viktor Plane

Reputation: 127

Time complexity on iterative and recursive solution

I'm trying to solve the following problem: Programming problem

I feel like I've given it a lot of thoughts and tried a lot of stuff. I manage to solve it, and produce correct values but the problem is that it isn't time efficient enough. It completes 2 out of the Kattis tests and fails on the 3 because of the time limit 1 second was exceeded. There is noway for me to see what the input was that they tested with I'm afraid.

I started out with a recursive solution and finished that. But then I realised that it wasn't time efficient enough so I instead tried to switch to an iterative solution.

I start with reading input and add those to an ArrayList. And then I call the following method with target as 1000.

public static int getCorrectWeight(List<Integer> platesArr, int target) {
    /* Creates two lists, one for storing completed values after each iteration,
    one for storing new values during iteration. */
    List<Integer> vals = new ArrayList<>();
    List<Integer> newVals = new ArrayList<>();

    // Inserts 0 as a first value so that we can start the first iteration.
    int best = 0;
    vals.add(best);

    for(int i=0; i < platesArr.size(); i++) {
        for(int j=0; j < vals.size(); j++) {
            int newVal = vals.get(j) + platesArr.get(i);
            if (newVal <= target) {
                newVals.add(newVal);
                if (newVal > best) {
                    best = newVal;
                }
            } else if ((Math.abs(target-newVal) < Math.abs(target-best)) || (Math.abs(target-newVal) == Math.abs(target-best) && newVal > best)) {
                best = newVal;
            }
        }
        vals.addAll(newVals);
    }
    return best;
}

My question is, is there some way that I can reduce the time complexity on this one for large number of data?

Upvotes: 0

Views: 353

Answers (3)

ExMathGuy
ExMathGuy

Reputation: 208

If you think too generally about this type of problem, you may think you have to check all possible combinations of input (each weight can be included or excluded), giving you 2n combinations to test if you have n inputs. This is, however, rather beside the point. Rather, the key here is that all weights are integers, and that the goal is 1000.

Let's examine corner cases first, because that limits the search space.

If all weights are >= 1000, pick the smallest.

If there is at least one weight < 1000, that is always better than any weight >= 2000, so you can ignore any weight >= 1000 for combination purposes.

Then, apply dynamic programming. Keep a set (you got HashSet as suggestion from other poster, but BitSet is even better since the maximum value in it is so small) of all combinations of the first k inputs, and increase k by combining all previous solutions with the k+1'th input.

When you have considered all possibilities, just search the bit vector for the best response.

static int count() {
    int[] weights = new int[]{900, 500, 498, 4};

    // Check for corner case to limit search later
    int min = Integer.MAX_VALUE;
    for (int weight : weights) min = Math.min(min, weight);
    if (min >= 1000) {
        return min;
    }        
    // Get all interesting combinations
    BitSet combos = new BitSet();
    for (int weight : weights) {  
        if (weight < 1000) {
            for (int t = combos.previousSetBit(2000 - weight) ; t >= 0; t = combos.previousSetBit(t-1)) {
                combos.set(weight + t);
            }
            combos.set(weight);
        }
    }
    // Pick best combo
    for (int distance = 0; distance <= 1000; distance++) {
        if (combos.get(1000 + distance)) {
            return 1000 + distance;
        }
        if (combos.get(1000 - distance)) {
            return 1000 - distance;
        }
    }
    return 0;
}

Upvotes: 0

Benson Lin
Benson Lin

Reputation: 1404

You only need to store a DP table of size 2001 (0 to 2000) Let dp[i] represent if it is possible to form ikg of weights. If the weight goes over the array bounds, ignore it. For example:

dp[0] = 1;      
for (int i = 0; i < values.size();  i++){
    for (int j = 2000; j >= values[i]; j--){  
        dp[j] = max(dp[j],dp[j-values[i]);
    }
}

Here, values is where all the original weights are stored. All values of dp are to be set to 0 except for dp[0].

Then, check 1000 if it is possible to make it. If not, check 999 and 1001 and so on. This should run in O(1000n + 2000) time, since n is at most 1000 this should run in time.

By the way, this is a modified knapsack algorithm, you might want to look up some other variants.

Upvotes: 0

fgb
fgb

Reputation: 18559

The main problem is that the size of vals and newVals can grow very quickly, as each iteration can double their size. You only need to store 1000 or so values which should be manageable. You're limiting the values but because they're stored in an ArrayList, it ends up with a lot of duplicate values.

If instead, you used a HashSet, then it should help the efficiency a lot.

Upvotes: 1

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