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 ( n 2) 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.

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

 

这道题给了我们一个数组,又给了一个下限和一个上限,让求有多少个不同的区间使得每个区间的和在给定的上下限之间。这道题的难度系数给的是 Hard,的确是一道难度不小的题,题目中也说了 Brute Force 的方法太 Naive 了,只能另想方法了。To be honest,这题超出了博主的能力范围,所以博主也没挣扎了,直接上网搜大神们的解法啦。首先根据前面的那几道类似题 Range Sum Query - MutableRange Sum Query 2D - Immutable 和 Range Sum Query - Immutable 的解法可知类似的区间和的问题一定是要计算累积和数组 sums 的,其中 sum[i] = nums[0] + nums[1] + … + nums[i],对于某个i来说,只有那些满足 lower <= sum[i] - sum[j] <= upper 的j能形成一个区间 [j, i] 满足题意,目标就是来找到有多少个这样的 j (0 =< j < i) 满足 sum[i] - upper =< sum[j] <= sum[i] - lower,可以用 C++ 中由红黑树实现的 multiset 数据结构可以对其中数据排序,然后用 upperbound 和 lowerbound 来找临界值。lower_bound 是找数组中第一个不小于给定值的数(包括等于情况),而 upper_bound 是找数组中第一个大于给定值的数,那么两者相减,就是j的个数,参见代码如下:

 

解法一:

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        int res = 0;
        long long sum = 0;
        multiset<long long> sums;
        sums.insert(0);
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            res += distance(sums.lower_bound(sum - upper), sums.upper_bound(sum - lower));
            sums.insert(sum);
        }
        return res;
    }        
};

 

我们再来看一种方法,这种方法的思路和前一种一样,只是没有 STL 的 multiset 和 lower_bound 和 upper_bound 函数,而是使用了 Merge Sort 来解,在混合的过程中,已经给左半边 [start, mid) 和右半边 [mid, end) 排序了。当遍历左半边,对于每个i,需要在右半边找出k和j,使其满足:

j是第一个满足 sums[j] - sums[i] > upper 的下标

k是第一个满足 sums[k] - sums[i] >= lower 的下标 

那么在 [lower, upper] 之间的区间的个数是 j - k,同时也需要另一个下标t,用来拷贝所有满足 sums[t] < sums[i] 到一个寄存器 Cache 中以完成混合排序的过程,这个步骤是混合排序的精髓所在,实际上这个寄存器的作用就是将 [start, end) 范围内的数字排好序先存到寄存器中,然后再覆盖原数组对应的位置即可,(注意这里 sums 可能会整型溢出,使用长整型 long 代替),参见代码如下:

 

解法二:

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        vector<long> sums(nums.size() + 1, 0);
        for (int i = 0; i < nums.size(); ++i) {
            sums[i + 1] = sums[i] + nums[i];
        }
        return countAndMergeSort(sums, 0, sums.size(), lower, upper);
    }
    int countAndMergeSort(vector<long>& sums, int start, int end, int lower, int upper) {
        if (end - start <= 1) return 0;
        int mid = start + (end - start) / 2;
        int cnt = countAndMergeSort(sums, start, mid, lower, upper) + countAndMergeSort(sums, mid, end, lower, upper);
        int j = mid, k = mid, t = mid;
        vector<int> cache(end - start, 0);
        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++];
            cache[r] = sums[i];
            cnt += j - k;
        }
        copy(cache.begin(), cache.begin() + t - start, sums.begin() + start);
        return cnt;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/327

 

类似题目:

Range Sum Query - Mutable 

Range Sum Query 2D - Immutable

Range Sum Query - Immutable

Reverse Pairs

Count of Smaller Numbers After Self

 

参考资料:

https://leetcode.com/problems/count-of-range-sum/

https://leetcode.com/problems/count-of-range-sum/discuss/77990/Share-my-solution

https://leetcode.com/problems/count-of-range-sum/discuss/78006/Summary-of-the-Divide-and-Conquer-based-and-Binary-Indexed-Tree-based-solutions

https://leetcode.com/problems/count-of-range-sum/discuss/78030/8-line-multiset-C%2B%2B-solution-(100ms)-also-binary-search-tree-(180ms)-%2B-mergesort(52ms)

 

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