Leetcode 303. Immutable range sum query

Calculate sum of array elements for any given indexes, keeping source array immutable at the same time Range Sum Query - Immutable.

Description

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note

  • You may assume that the array does not change.
  • There are many calls to sumRange function.

Caching solution

Cache sum of array elements for each combination of indexes into HashMap. Space complexity is O(n^2), time complexity is O(n^2).

class NumArray {

  Map<String, Integer> cache = 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];
        cache.put(i+","+j, sum);
      }
    }
  }

  public int sumRange(int i, int j) {
    return cache.get(i+","+j);
  }
}

We can do better

Sum of array elements between indexes i and j can be represented as following:
sum(i,j)=sum(0,j) - sum(0,i-1). We see that it is represented via zero index based sums. So we can calculate these sums once, and then use it for our calculations.

class NumArray {

  int[] zeroBasedSum;

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

  public int sumRange(int i, int j) {
    return zeroBasedSum[j] - (i-1 < 0 ? 0 : zeroBasedSum[i-1]);
  }
}

Obviously, space complexity is O(n), as well as time complexity.