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# 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.