2017-09-14 LeetCode_327 Count of Range Sum

327. Count of Range Sum

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

.
Credits:

Special thanks to @dietpepsi for adding this problem and creating all test cases.
Solution:

class Solution {

2

public:

3

void sort1(vector<long long>& sums, vector<long long>& nums, long long start, long long end) {

4

if (start >= end) return;

5

long long m1 = sums[start], m2 = nums[start], i = start, j = end;

6

while (i < j) {

7

while (i < j && sums[j]-nums[j] >= m1-m2) j--;

8

sums[i] = sums[j]; nums[i] = nums[j];

9

while (i < j && sums[i]-nums[i] <= m1-m2) i++;

10

sums[j] = sums[i]; nums[j] = nums[i];

11

}

12

sums[i] = m1; nums[i] = m2;

13

sort1(sums, nums, start, i-1);

14

sort1(sums, nums, i+1, end);

15

}

16

void sort2(vector<long long>& sums, vector<long long>& nums, long long start, long long end) {

17

if (start >= end) return;

18

long long m1 = sums[start], m2 = nums[start], i = start, j = end;

19

while (i < j) {

20

while (i < j && sums[j] >= m1) j--;

21

sums[i] = sums[j]; nums[i] = nums[j];

22

while (i < j && sums[i] <= m1) i++;

23

sums[j] = sums[i]; nums[j] = nums[i];

24

}

25

sums[i] = m1; nums[i] = m2;

26

sort2(sums, nums, start, i-1);

27

sort2(sums, nums, i+1, end);

28

}

29

long long divides(vector<long long>& sums, vector<long long>& nums, long long start, long long end, int lower, int upper) {

30

if (start == end) return (nums[start] <= upper && nums[start] >= lower);

31

long long count = divides(sums, nums, start, (start+end)/2, lower, upper)+divides(sums, nums, (start+end)/2+1, end, lower, upper);

32

sort1(sums, nums, start, (start+end)/2);

33

sort2(sums, nums, (start+end)/2+1, end);

34

35

for (int left = start, right0 = (start+end)/2+1, right1 = (start+end)/2+1; left <= (start+end)/2; left++) {

36

    while (right0 <= end && sums[right0]-sums[left]+nums[left] < lower) right0++;

37

    while (right1 <= end && sums[right1]-sums[left]+nums[left] <= upper) right1++;

38

    count += right1-right0;

39

}

40

return count;

41

}

42

   int countRangeSum(vector<int>& nums, int lower, int upper) {

43

       if (nums.size() == 0) return 0;

44

       vector<long long> sums, mynums; sums.push_back(nums[0]); mynums.push_back(nums[0]);

45

       for (int i = 1; i < nums.size(); i++) {

46

           mynums.push_back(nums[i]);

47

           sums.push_back(sums[i-1]+nums[i]);

48

}

49

       return divides(sums, mynums, 0, nums.size()-1, lower, upper);

50

   }

51

};