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;\n\npublic NumArray(int[] nums) {\n    data = nums;\n}\n\npublic int sumRange(int i, int j) {\n    int sum = 0;\n    for (int k = i; k <= j; k++) {\n        sum += data[k];\n    }\n    return sum;\n}\n

Complexity analysis:

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  Time complexity : $$O(n)$$ time per query.\nEach sumRange query takes $$O(n)$$ time.

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

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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<>();\n\npublic NumArray(int[] nums) {\n    for (int i = 0; i < nums.length; i++) {\n        int sum = 0;\n        for (int j = i; j < nums.length; j++) {\n            sum += nums[j];\n            map.put(Pair.create(i, j), sum);\n        }\n    }\n}\n\npublic int sumRange(int i, int j) {\n    return map.get(Pair.create(i, j));\n}\n

Complexity analysis

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  Time complexity : $$O(1)$$ time per query, $$O(n^2)$$ time pre-computation.\nThe 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.

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

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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):

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

Now, we can calculate sumRange as following:

\n $$\nsumRange(i, j) = sum[j + 1] - sum[i]\n$$ \n

private int[] sum;\n\npublic NumArray(int[] nums) {\n    sum = new int[nums.length + 1];\n    for (int i = 0; i < nums.length; i++) {\n        sum[i + 1] = sum[i] + nums[i];\n    }\n}\n\npublic int sumRange(int i, int j) {\n    return sum[j + 1] - sum[i];\n}\n

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

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

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  Space complexity : $$O(n)$$ .

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