Formatted question description: https://leetcode.ca/all/1508.html

1508. Range Sum of Sorted Subarray Sums (Medium)

Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.

Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 10^9 + 7.

 

Example 1:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13 
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 

Example 2:

Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.

Example 3:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50

 

Constraints:

  • 1 <= nums.length <= 10^3
  • nums.length == n
  • 1 <= nums[i] <= 100
  • 1 <= left <= right <= n * (n + 1) / 2

Related Topics:
Array, Sort

Solution 1. Prefix Sum

Compute the prefix sum array using pre. Use pre to get the array of all subarray sums – sum. Sort sum array then add from sum[left - 1] to sum[right - 1].

// OJ: https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/

// Time: O(N^2 * log(N^2))
// Space: O(N^2)
class Solution {
public:
    int rangeSum(vector<int>& A, int N, int left, int right) {
        vector<int> sum;
        long ans = 0, mod = 1e9+7;
        for (int i = 0; i < N; ++i) {
            int pre = 0;
            for (int j = i; j < N; ++j) {
                pre += A[j];
                sum.push_back(pre);
            }
        }
        sort(begin(sum), end(sum));
        for (int i = left; i <= right; ++i) ans = (ans + sum[i - 1]) % mod;
        return ans;
    }
};

Solution 2. Heap

// OJ: https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/

// Time: O(right * logN)
// Space: O(N)
// Ref: https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/discuss/730511/C++-priority_queue-solution
class Solution {
public:
    int rangeSum(vector<int>& A, int N, int left, int right) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q; // partial sum, next index
        for (int i = 0; i < N; ++i) q.emplace(A[i], i + 1);
        long ans = 0, mod = 1e9+7;
        for (int i = 1; i <= right; ++i) {
            auto p = q.top();
            q.pop();
            if (i >= left) ans = (ans + p.first) % mod;
            if (p.second < N) {
                p.first += A[p.second++];
                q.push(p);
            }
        }
        return ans;
    }
};

Solution 3. Binary Search + Sliding Window

// OJ: https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/

// Time: O(Nlog(sum(A)))
// Space: O(N)
// Ref: https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/discuss/733047/Python-Binary-Search-Time-O(NlogSum(A))
class Solution {
    vector<int> B, C; // B is partial_sum array of A. C is partial_sum array of B.
    int N;
    int countSumUnder(int score) { // Use sliding window to cound the partial sums that are smaller or equal to `score`.
        int ans = 0, i = 0;
        for (int j = 0; j < N + 1; ++j) {
            while (B[j] - B[i] > score) ++i;
            ans += j - i;
        }
        return ans;
    }
    int sumKSums(int k) { // Use sliding window to get the sum of the k smallest items in all the subarray sums.
        int score = kthScore(k), ans = 0, i = 0;
        for (int j = 0; j < N + 1; ++j) {
            while (B[j] - B[i] > score) ++i;
            ans += B[j] * (j - i + 1) - (C[j] - (i ? C[i - 1] : 0));
        }
        return ans - (countSumUnder(score) - k) * score;
    }
    int kthScore(int k) { // use binary search to find the kth smallest sum in all subarray sums.
        int L = 0, R = B[N];
        while (L < R) {
            int M = (L + R) / 2;
            if (countSumUnder(M) < k) ++L;
            else R = M;
        }
        return L;
    }
public:
    int rangeSum(vector<int>& A, int n, int left, int right) {
        N = n;
        B.assign(N + 1, 0);
        C.assign(N + 1, 0);
        for (int i = 0; i < N; ++i) {
            B[i + 1] = B[i] + A[i];
            C[i + 1] = C[i] + B[i + 1];
        }
        return sumKSums(right) - sumKSums(left - 1);
    }
};

Java

class Solution {
    public int rangeSum(int[] nums, int n, int left, int right) {
        final int MODULO = 1000000007;
        int length = nums.length;
        int sumsLength = length * (length + 1) / 2;
        int[] sums = new int[sumsLength];
        int index = 0;
        for (int i = 0; i < length; i++) {
            int sum = 0;
            for (int j = i; j < length; j++) {
                sum += nums[j];
                sums[index++] = sum;
            }
        }
        Arrays.sort(sums);
        int rangeSum = 0;
        for (int i = left - 1; i < right; i++)
            rangeSum = (rangeSum + sums[i]) % MODULO;
        return rangeSum;
    }
}

All Problems

All Solutions