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Problem solution · Java

Earliest Second to Mark Indices I

Earliest Second to Mark Indices I: 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
46 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 Earliest Second to Mark Indices I, 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

  • 46 lines of Java from the credited upstream file 3048.java.
  • The implementation visibly relies on sequence storage.
  • 3 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 codeEarliest Second to Mark Indices I · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {    int l = 0;    int r = changeIndices.length + 1;     while (l < r) {      final int m = (l + r) / 2;      if (canMark(nums, changeIndices, m))        r = m;      else        l = m + 1;    }     return l <= changeIndices.length ? l : -1;  }   // Returns true if all indices of `nums` can be marked within `second`.  private boolean canMark(int[] nums, int[] changeIndices, int second) {    int numMarked = 0;    int decrement = 0;    // indexToLastSecond[i] := the last second to mark the index i    int[] indexToLastSecond = new int[nums.length];    Arrays.fill(indexToLastSecond, -1);     for (int i = 0; i < second; ++i)      indexToLastSecond[changeIndices[i] - 1] = i;     for (int i = 0; i < second; ++i) {      final int index = changeIndices[i] - 1; // Convert to 0-indexed.      if (i == indexToLastSecond[index]) {        // Reach the last occurrence of the number.        // So, the current second will be used to mark the index.        if (nums[index] > decrement)          // The decrement is less than the number to be marked.          return false;        decrement -= nums[index];        ++numMarked;      } else {        ++decrement;      }    }     return numMarked == nums.length;  }}

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