327. Count of Range Sum(inplace_marge)

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:

Input: nums = [2, 5, -1], lower = -2, upper = 2,
Output: 3 
Explanation: The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

 

Approach #1: C++.

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        int len = nums.size();
        if (len == 0) return 0;
        vector<long> sum(len+1, 0);
        for (int i = 0; i < len; ++i)
            sum[i+1] += sum[i] + nums[i];
        return mergeSort(sum, lower, upper, 0, len+1);
    }
    
private:
    int mergeSort(vector<long>& sum, int lower, int upper, int left, int right) {
        if (right - left <= 1) return 0;
        int mid = left + (right - left) / 2;
        int m = mid, n = mid, count = 0;
        count = mergeSort(sum, lower, upper, left, mid) + mergeSort(sum, lower, upper, mid, right);
        for (int i = left; i < mid; ++i) {
            while (m < right && sum[m] - sum[i] < lower) m++;
            while (n < right && sum[n] - sum[i] <= upper) n++;
            count += n - m;
        }
        inplace_merge(sum.begin()+left, sum.begin()+mid, sum.begin()+right);
        return count;
    }
};

　　

Approach #2: Java.

class Solution {
    public int countRangeSum(int[] nums, int lower, int upper) {
        if (nums == null || nums.length == 0) return 0;
        long[] sums = new long[nums.length];
        long sum = 0;
        for (int i = 0; i < nums.length; ++i) {
            sum += nums[i];
            sums[i] += sum;
        }
        return mergeSort(sums, lower, upper, 0, nums.length-1);
    }
    
    private int mergeSort(long[] sums, int lower, int upper, int left, int right) {
        if (right < left) return 0;
        else if (left == right) {
            if (sums[left] >= lower && sums[right] <= upper) return 1;
            else return 0;
        }
        int mid = left + (right - left) / 2;
        int count = mergeSort(sums, lower, upper, left, mid) + mergeSort(sums, lower, upper, mid+1, right);
        int m = mid+1, n = mid+1;
        for (int i = left; i <= mid; ++i) {
            while (m <= right && sums[m] - sums[i] < lower) m++;
            while (n <= right && sums[n] - sums[i] <= upper) n++;
            count += n - m;
        }
        mergeHelper(sums, left, mid, right);
        return count;
    }
    
    private void mergeHelper(long[] sums, int left, int mid, int right) {
        int i = left;
        int j = mid + 1;
        long[] copy = new long[right-left+1];
        int p = 0;
        while (i <= mid && j <= right) {
            if (sums[i] < sums[j]) {
                copy[p++] = sums[i++];
            } else {
                copy[p++] = sums[j++];
            }
        }
        
        while (i <= mid) {
            copy[p++] = sums[i++];
        }
        
        while (j <= right) {
            copy[p++] = sums[j++];
        }
        
        System.arraycopy(copy, 0, sums, left, right-left+1);
    }
}

　　

Approach #3: Python.

class Solution(object):
    def countRangeSum(self, nums, lower, upper):
        """
        :type nums: List[int]
        :type lower: int
        :type upper: int
        :rtype: int
        """
        first = [0]
        for num in nums:
            first.append(first[-1] + num)
            
        def sort(lo, hi):
            mid = (lo + hi) / 2
            if mid == lo:
                return 0
            count = sort(lo, mid) + sort(mid, hi)
            i = j = mid
            for left in first[lo:mid]:
                while i < hi and first[i] - left < lower: i += 1
                while j < hi and first[j] - left <= upper: j += 1
                count += j - i
            first[lo:hi] = sorted(first[lo:hi])
            return count
        return sort(0, len(first))

　　

Notes:

C++ -----> inplace_merge

default (1)	template <class BidirectionalIterator> void inplace_merge (BidirectionalIterator first, BidirectionalIterator middle, BidirectionalIterator last);
custom (2)	template <class BidirectionalIterator, class Compare> void inplace_merge (BidirectionalIterator first, BidirectionalIterator middle, BidirectionalIterator last, Compare comp);

Merge consecutive sorted ranges

Merges two consecutive sorted ranges: [first,middle) and [middle,last), putting the result into the combined sorted range [first,last).

The elements are compared using operator< for the first version, and comp for the second. The elements in both ranges shall already be ordered according to this same criterion (operator< or comp). The resulting range is also sorted according to this.

The function preserves the relative order of elements with equivalent values, with the elements in the first range preceding those equivalent in the second.

for example:

// inplace_merge example
#include <iostream>     // std::cout
#include <algorithm>    // std::inplace_merge, std::sort, std::copy
#include <vector>       // std::vector

int main () {
  int first[] = {5,10,15,20,25};
  int second[] = {50,40,30,20,10};
  std::vector<int> v(10);
  std::vector<int>::iterator it;

  std::sort (first,first+5);
  std::sort (second,second+5);

  it=std::copy (first, first+5, v.begin());
     std::copy (second,second+5,it);

  std::inplace_merge (v.begin(),v.begin()+5,v.end());

  std::cout << "The resulting vector contains:";
  for (it=v.begin(); it!=v.end(); ++it)
    std::cout << ' ' << *it;
  std::cout << '
';

  return 0;
}

　　

output:

The resulting vector contains: 5 10 10 15 20 20 25 30 40 50