Range Sum Query - Immutable

Problem

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.

Solution

DP Design: O(1) time, O(n) space

class NumArray {

    int[] sum;

    public NumArray(int[] nums) {
        this.sum = new int[nums.length + 1];
        for (int i = 0; i < nums.length; i++) sum[i + 1] = sum[i] + nums[i]; //sum[i] is sum until i, exclusive i
    }

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

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

Analysis

Since sumRange() will be called multiple times, we store the sum in DP way
sum[i] is the sum from 0 (inclusive) to i (exclusive)
In the construction method, we init int[] sum and then in sumRange(i, j) we just return:
sum[j + 1] - sum[i] since j+1 is exclusive, it will return the range from i to j