303. Range Sum Query - Immutable

LeetCode 303. Range Sum Query - Immutable

Description

Given an integer array nums, find the sum of the elements between indices left and right inclusive, where (left <= right).

Implement the NumArray class:

• NumArray(int[] nums) initializes the object with the integer array nums.

• int sumRange(int left, int right) returns the sum of the elements of the nums array in the range [left, right] inclusive (i.e., sum(nums[left], nums[left + 1], ... , nums[right])).

Example 1: Input ["NumArray", "sumRange", "sumRange", "sumRange"] [[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]] Output [null, 1, -1, -3] Explanation NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]); numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3) numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

Constraints:

• 1 <= nums.length <= 104

• -105 <= nums[i] <= 105

• 0 <= left <= right < nums.length

• At most 104 calls will be made to sumRange.

Tags

Dynamic Programming, Design

Solution

We can pre-compute a cumulative array sum with respect to the origin at index 0, with the formula: $sum[i]=sum[i-1]+nums[i]$. Then we can formulate the sum range with given i and j: $sumRange(i,j)=sums[j+1]−sums[i]$

Complexity

• Time complexity: $O(n)$for initializing sum; $O(1)$ for performing search;

• Space complexity: $O(n)$

Code

type NumArray struct {
	sum []int
}

func Constructor(nums []int) NumArray {
	if len(nums) == 0 {
		return NumArray{}
	}
	sum := make([]int, len(nums)+1)
	for i, num := range nums {
		sum[i+1] = sum[i] + num
	}
	return NumArray{sum: sum}
}

func (this *NumArray) SumRange(left int, right int) int {
	return this.sum[right+1] - this.sum[left]
}

Reference

1. 区域和检索 - 数组不可变