1508. Range Sum of Sorted Subarray Sums.md

1508. Range Sum of Sorted Subarray Sums

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the second one in Biweekly Contest 30.

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Difficulty : Medium

Related Topics : Array、Sort

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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

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Solution

• mine
  • Java
    • Runtime: 81 ms, faster than 66.38%, Memory Usage: 45 MB, less than 100.00% of Java online submissions

      //O(N^4logN)time   N = n
      //O(N^2)space
      public int rangeSum(int[] nums, int n, int left, int right) {
          int[] arr = new int[n * (n + 1) / 2];
          List<Integer> temp = new ArrayList<>();
          int i = 0;
          for (int num : nums) {
              for (int j = 0; j < temp.size(); j++) {
                  temp.set(j, temp.get(j) + num);
                  arr[i++] = temp.get(j);
              }
              temp.add(num);
              arr[i++] = num;
          }
          Arrays.sort(arr);
          int mod = 1000000007;
          int ans = 0;
          for (int j = left - 1; j < right; j++) {
              ans = (ans + arr[j]) % mod;
          }
          return ans;
      }
      

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