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Problem solution · Java

Find if Array Can Be Sorted

Find if Array Can Be Sorted: 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
30 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 Find if Array Can Be Sorted, 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

  • 30 lines of Java from the credited upstream file 3011.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 codeFind if Array Can Be Sorted · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public boolean canSortArray(int[] nums) {    // Divide the array into distinct segments where each segment is comprised    // of consecutive elements sharing an equal number of set bits. Ensure that    // for each segment, when moving from left to right, the maximum of a    // preceding segment is less than the minimum of the following segment.    int prevSetBits = 0;    int prevMax = Integer.MIN_VALUE; // the maximum of the previous segment    int currMax = Integer.MIN_VALUE; // the maximum of the current segment    int currMin = Integer.MAX_VALUE; // the minimum of the current segment     for (final int num : nums) {      final int setBits = Integer.bitCount(num);      if (setBits != prevSetBits) { // Start a new segment.        if (prevMax > currMin)          return false;        prevSetBits = setBits;        prevMax = currMax;        currMax = num;        currMin = num;      } else { // Continue with the current segment.        currMax = Math.max(currMax, num);        currMin = Math.min(currMin, num);      }    }     return prevMax <= currMin;  }}

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