Solution

Approach #1 (Brute Force) [Time Limit Exceeded]

Each time sumRange is called, we use a for loop to sum each individual element from index $$i$$ to $$j$$ .

private int[] data;

public NumArray(int[] nums) {
    data = nums;
}

public int sumRange(int i, int j) {
    int sum = 0;
    for (int k = i; k <= j; k++) {
        sum += data[k];
    }
    return sum;
}

Complexity analysis:

• Time complexity : $$O(n)$$ time per query. Each sumRange query takes $$O(n)$$ time.

• Space complexity : $$O(1)$$ . Note that data is a reference to nums and is not a copy of it.

Approach #2 (Caching) [Accepted]

Imagine that sumRange is called one thousand times with the exact same arguments. How could we speed that up?

We could trade in extra space for speed. By pre-computing all range sum possibilities and store its results in a hash table, we can speed up the query to constant time.

private Map<Pair<Integer, Integer>, Integer> map = new HashMap<>();

public NumArray(int[] nums) {
    for (int i = 0; i < nums.length; i++) {
        int sum = 0;
        for (int j = i; j < nums.length; j++) {
            sum += nums[j];
            map.put(Pair.create(i, j), sum);
        }
    }
}

public int sumRange(int i, int j) {
    return map.get(Pair.create(i, j));
}

Complexity analysis

• Time complexity : $$O(1)$$ time per query, $$O(n^2)$$ time pre-computation. The pre-computation done in the constructor takes $$O(n^2)$$ time. Each sumRange query's time complexity is $$O(1)$$ as the hash table's look up operation is constant time.

• Space complexity : $$O(n^2)$$ . The extra space required is $$O(n^2)$$ as there are $$n$$ candidates for both $$i$$ and $$j$$ .

Approach #3 (Caching) [Accepted]

The above approach takes a lot of space, could we optimize it?

Imagine that we pre-compute the cummulative sum from index $$0$$ to $$k$$ . Could we use this information to derive $$Sum(i, j)$$ ?

Let us define $$sum[k]$$ as the cumulative sum for $$nums[0 \cdots k-1]$$ (inclusive):

$$sum[k] = \left\{ \begin{array}{rl} \sum_{i=0}^{k-1}nums[i] & , k > 0 \\ 0 &, k = 0 \end{array} \right.$$

Now, we can calculate sumRange as following:

$$sumRange(i, j) = sum[j + 1] - sum[i]$$

private int[] sum;

public NumArray(int[] nums) {
    sum = new int[nums.length + 1];
    for (int i = 0; i < nums.length; i++) {
        sum[i + 1] = sum[i] + nums[i];
    }
}

public int sumRange(int i, int j) {
    return sum[j + 1] - sum[i];
}

Notice in the code above we inserted a dummy 0 as the first element in the sum array. This trick saves us from an extra conditional check in sumRange function.

Complexity analysis

• Time complexity : $$O(1)$$ time per query, $$O(n)$$ time pre-computation. Since the cumulative sum is cached, each sumRange query can be calculated in $$O(1)$$ time.

• Space complexity : $$O(n)$$ .