# 327. Count of Range Sum

### Problem:

Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive. Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive.

Note: A naive algorithm of O(n2) is trivial. You MUST do better than that.

Example: Given nums = [-2, 5, -1], lower = -2, upper = 2, Return 3. The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

### Solutions:

public class Solution {
public int countRangeSum(int[] nums, int lower, int upper) {
long[] sums= new long[nums.length + 1];
for (int i = 0; i < nums.length; i ++) {
sums[i + 1] = sums[i] + nums[i];
}
return countMergeSort(sums, 0, sums.length, lower, upper);
}
int countMergeSort(long[] sums, int start, int end, int lower, int upper) {
if (end - start <= 1) {
return 0;
}
int mid = start + (end - start) / 2;
int count = countMergeSort(sums, start, mid, lower, upper) + countMergeSort(sums, mid, end, lower, upper);
int j = mid, k = mid, t = mid;
long[] cache = new long[end - start];
for (int i = start, r = 0; i < mid; i ++, r ++) {
while (k < end && sums[k] - sums[i] < lower) {
k ++;
}
while (j < end && sums[j] - sums[i] <= upper) {
j ++;
}
while (t < end && sums[t] < sums[i]) {
cache[r] = sums[t];
r ++;
t ++;
}
cache[r] = sums[i];
count += j - k;
}
for (int i = 0; i < t - start; i ++) {
sums[i + start] = cache[i];
}
return count;
}
}