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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:

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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
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class NumArray {

    public int[] nums;

    public NumArray(int[] nums) {
        this.nums = nums; // notice the use of this refers to nums of the NumArray class
        for(int i=1; i < nums.length; i++) {
            nums[i] = nums[i-1] + nums[i];
        }
        
    }
    
    public int sumRange(int left, int right) {
        if(left == 0) return nums[right] ;
        return nums[right] - nums[left - 1]; //!!!!!! left - 1 not left
    }
}

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