Range Sum Query - Immutable Problem
Description
LeetCode Problem 303.
Given an integer array nums, handle multiple queries of the following type:
- 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
Constraints:
- 1 <= nums.length <= 10^4
- -10^5 <= nums[i] <= 10^5
- 0 <= left <= right < nums.length
- At most 10^4 calls will be made to sumRange.
Sample C++ Code
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class NumArray {
public:
vector<int> dp;
NumArray(vector<int>& nums) {
if (nums.size() == 0)
return;
dp.push_back(nums[0]);
for (int i = 1; i < nums.size(); i ++) {
dp.push_back(dp[i-1] + nums[i]);
}
}
int sumRange(int i, int j) {
if (i == 0)
return dp[j];
else
return (dp[j]-dp[i-1]);
}
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray* obj = new NumArray(nums);
* int param_1 = obj->sumRange(i,j);
*/