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Java-based LeetCode algorithm problem solutions, regularly updated
package g0301_0400.s0303_range_sum_query_immutable;
// #Easy #Array #Design #Prefix_Sum #Programming_Skills_I_Day_12_Class_and_Object
// #2022_07_07_Time_7_ms_(100.00%)_Space_44.9_MB_(96.34%)
/**
* 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`.
**/
public class NumArray {
private int[] sums;
public NumArray(int[] nums) {
sums = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
if (i == 0) {
sums[i] = nums[i];
} else {
sums[i] = sums[i - 1] + nums[i];
}
}
}
public int sumRange(int i, int j) {
if (i == 0) {
return sums[j];
}
return sums[j] - sums[i - 1];
}
}
/*
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(left,right);
*/
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