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Problem solution · Java

Count of Smaller Numbers After Self

Count of Smaller Numbers After Self: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
55 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

Sliding window or two pointers

For Count of Smaller Numbers After Self, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 55 lines of Java from the credited upstream file 315-4.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 Smaller Numbers After Self · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public List<Integer> countSmaller(int[] nums) {    final int n = nums.length;    int[] ans = new int[n];    Item[] items = new Item[n];     for (int i = 0; i < n; ++i)      items[i] = new Item(nums[i], i);     mergeSort(items, 0, n - 1, ans);    return Arrays.stream(ans).boxed().collect(Collectors.toList());  }   private record Item(int num, int index) {}   private void mergeSort(Item[] items, int l, int r, int[] ans) {    if (l >= r)      return;     final int m = (l + r) / 2;    mergeSort(items, l, m, ans);    mergeSort(items, m + 1, r, ans);    merge(items, l, m, r, ans);  }   private void merge(Item[] items, int l, int m, int r, int[] ans) {    Item[] sorted = new Item[r - l + 1];    int k = 0;          // sorted's index    int i = l;          // left's index    int j = m + 1;      // right's index    int rightCount = 0; // the number of `nums` < items[i].num     while (i <= m && j <= r)      if (items[i].num > items[j].num) {        ++rightCount;        sorted[k++] = items[j++];      } else {        ans[items[i].index] += rightCount;        sorted[k++] = items[i++];      }     // Put the possible remaining left part into the sorted array.    while (i <= m) {      ans[items[i].index] += rightCount;      sorted[k++] = items[i++];    }     // Put the possible remaining right part into the sorted array.    while (j <= r)      sorted[k++] = items[j++];     System.arraycopy(sorted, 0, items, l, sorted.length);  }}

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