LeetCode Solutions
327. Count of Range Sum
Time: $O(n\log n)$ Space: $O(n)$
class Solution {
public:
int countRangeSum(vector<int>& nums, int lower, int upper) {
const int n = nums.size();
int ans = 0;
vector<long> prefix(n + 1);
for (int i = 0; i < n; ++i)
prefix[i + 1] = prefix[i] + nums[i];
mergeSort(prefix, 0, n, lower, upper, ans);
return ans;
}
private:
void mergeSort(vector<long>& prefix, int l, int r, int lower, int upper,
int& ans) {
if (l >= r)
return;
const int m = (l + r) / 2;
mergeSort(prefix, l, m, lower, upper, ans);
mergeSort(prefix, m + 1, r, lower, upper, ans);
merge(prefix, l, m, r, lower, upper, ans);
}
void merge(vector<long>& prefix, int l, int m, int r, int lower, int upper,
int& ans) {
int lo = m + 1; // 1st index s.t. prefix[lo] - prefix[i] >= lower
int hi = m + 1; // 1st index s.t. prefix[hi] - prefix[i] > upper
// For each index i in range [l, m], add hi - lo to ans
for (int i = l; i <= m; ++i) {
while (lo <= r && prefix[lo] - prefix[i] < lower)
++lo;
while (hi <= r && prefix[hi] - prefix[i] <= upper)
++hi;
ans += hi - lo;
}
vector<long> sorted(r - l + 1);
int k = 0; // sorted's index
int i = l; // left's index
int j = m + 1; // right's index
while (i <= m && j <= r)
if (prefix[i] < prefix[j])
sorted[k++] = prefix[i++];
else
sorted[k++] = prefix[j++];
// Put possible remaining left part to the sorted array
while (i <= m)
sorted[k++] = prefix[i++];
// Put possible remaining right part to the sorted array
while (j <= r)
sorted[k++] = prefix[j++];
copy(begin(sorted), end(sorted), begin(prefix) + l);
}
};
class Solution {
public int countRangeSum(int[] nums, int lower, int upper) {
final int n = nums.length;
long[] prefix = new long[n + 1];
for (int i = 0; i < n; ++i)
prefix[i + 1] = (long) nums[i] + prefix[i];
mergeSort(prefix, 0, n, lower, upper);
return ans;
}
private int ans = 0;
private void mergeSort(long[] prefix, int l, int r, int lower, int upper) {
if (l >= r)
return;
final int m = (l + r) / 2;
mergeSort(prefix, l, m, lower, upper);
mergeSort(prefix, m + 1, r, lower, upper);
merge(prefix, l, m, r, lower, upper);
}
private void merge(long[] prefix, int l, int m, int r, int lower, int upper) {
int lo = m + 1; // 1st index s.t. prefix[lo] - prefix[i] >= lower
int hi = m + 1; // 1st index s.t. prefix[hi] - prefix[i] > upper
// For each index i in range [l, m], add hi - lo to ans
for (int i = l; i <= m; ++i) {
while (lo <= r && prefix[lo] - prefix[i] < lower)
++lo;
while (hi <= r && prefix[hi] - prefix[i] <= upper)
++hi;
ans += hi - lo;
}
long[] sorted = new long[r - l + 1];
int k = 0; // sorted's index
int i = l; // left's index
int j = m + 1; // right's index
while (i <= m && j <= r)
if (prefix[i] < prefix[j])
sorted[k++] = prefix[i++];
else
sorted[k++] = prefix[j++];
// Put possible remaining left part to the sorted array
while (i <= m)
sorted[k++] = prefix[i++];
// Put possible remaining right part to the sorted array
while (j <= r)
sorted[k++] = prefix[j++];
System.arraycopy(sorted, 0, prefix, l, sorted.length);
}
}
class Solution:
def countRangeSum(self, nums: List[int], lower: int, upper: int) -> int:
n = len(nums)
self.ans = 0
prefix = [0] + list(itertools.accumulate(nums))
self._mergeSort(prefix, 0, n, lower, upper)
return self.ans
def _mergeSort(self, prefix: List[int], l: int, r: int, lower: int, upper: int) -> None:
if l >= r:
return
m = (l + r) // 2
self._mergeSort(prefix, l, m, lower, upper)
self._mergeSort(prefix, m + 1, r, lower, upper)
self._merge(prefix, l, m, r, lower, upper)
def _merge(self, prefix: List[int], l: int, m: int, r: int, lower: int, upper: int) -> None:
lo = m + 1 # 1st index s.t. prefix[lo] - prefix[i] >= lower
hi = m + 1 # 1st index s.t. prefix[hi] - prefix[i] > upper
# For each index i in range [l, m], add hi - lo to ans
for i in range(l, m + 1):
while lo <= r and prefix[lo] - prefix[i] < lower:
lo += 1
while hi <= r and prefix[hi] - prefix[i] <= upper:
hi += 1
self.ans += hi - lo
sorted = [0] * (r - l + 1)
k = 0 # sorted's index
i = l # left's index
j = m + 1 # right's index
while i <= m and j <= r:
if prefix[i] < prefix[j]:
sorted[k] = prefix[i]
k += 1
i += 1
else:
sorted[k] = prefix[j]
k += 1
j += 1
# Put possible remaining left part to the sorted array
while i <= m:
sorted[k] = prefix[i]
k += 1
i += 1
# Put possible remaining right part to the sorted array
while j <= r:
sorted[k] = prefix[j]
k += 1
j += 1
prefix[l:l + len(sorted)] = sorted