LeetCode in Kotlin

303. Range Sum Query - Immutable

Easy

Given an integer array nums, handle multiple queries of the following type:

1. Calculate the sum of the elements of nums 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 nums between indices left and right inclusive (i.e. 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 (-2) + 0 + 3 = 1 numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1 numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints:

• 1 <= nums.length <= 104
• -105 <= nums[i] <= 105
• 0 <= left <= right < nums.length
• At most 104 calls will be made to sumRange.

Solution

class NumArray(nums: IntArray) {
    private val sums: IntArray

    init {
        sums = IntArray(nums.size)
        for (i in nums.indices) {
            if (i == 0) {
                sums[i] = nums[i]
            } else {
                sums[i] = sums[i - 1] + nums[i]
            }
        }
    }

    fun sumRange(i: Int, j: Int): Int {
        return if (i == 0) {
            sums[j]
        } else sums[j] - sums[i - 1]
    }
}

/*
 * Your NumArray object will be instantiated and called as such:
 * var obj = NumArray(nums)
 * var param_1 = obj.sumRange(left,right)
 */

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