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Problem solution · Java

Count of Range Sum

Count of Range Sum: a Java solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
60 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Count of Range Sum, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 60 lines of Java from the credited upstream file 327.java.
  • The implementation visibly relies on sequence storage.
  • 7 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount of Range Sum · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int countRangeSum(int[] nums, int lower, int upper) {    final int n = nums.length;    long[] prefix = new long[n + 1];     for (int i = 0; i < n; ++i)      prefix[i + 1] = (long) nums[i] + prefix[i];     mergeSort(prefix, 0, n, lower, upper);    return ans;  }   private int ans = 0;   private void mergeSort(long[] prefix, int l, int r, int lower, int upper) {    if (l >= r)      return;     final int m = (l + r) / 2;    mergeSort(prefix, l, m, lower, upper);    mergeSort(prefix, m + 1, r, lower, upper);    merge(prefix, l, m, r, lower, upper);  }   private void merge(long[] prefix, int l, int m, int r, int lower, int upper) {    int lo = m + 1; // the first index s.t. prefix[lo] - prefix[i] >= lower    int hi = m + 1; // the first index s.t. prefix[hi] - prefix[i] > upper     // For each index i in range [l, m], add hi - lo to `ans`.    for (int i = l; i <= m; ++i) {      while (lo <= r && prefix[lo] - prefix[i] < lower)        ++lo;      while (hi <= r && prefix[hi] - prefix[i] <= upper)        ++hi;      ans += hi - lo;    }     long[] sorted = new long[r - l + 1];    int k = 0;     // sorted's index    int i = l;     // left's index    int j = m + 1; // right's index     while (i <= m && j <= r)      if (prefix[i] < prefix[j])        sorted[k++] = prefix[i++];      else        sorted[k++] = prefix[j++];     // Put the possible remaining left part into the sorted array.    while (i <= m)      sorted[k++] = prefix[i++];     // Put the possible remaining right part into the sorted array.    while (j <= r)      sorted[k++] = prefix[j++];     System.arraycopy(sorted, 0, prefix, l, sorted.length);  }}

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