java 43 lines · 8 steps

The sliding window technique in Java

Find the best fixed-length subarray in one pass by adding and removing one element at a time instead of resumming.

Explained by highlit

Intermediate sliding-window arrays algorithms optimization incremental-update

1import java.util.Arrays;
2 
3public class SlidingWindow {
  // Returns the maximum sum of any contiguous subarray of length k.
  public static long maxSubarraySum(int[] nums, int k) {
      if (nums == null || k <= 0 || k > nums.length) {
          throw new IllegalArgumentException(
              "k must be between 1 and " + (nums == null ? 0 : nums.length));
      }
10 
      long windowSum = 0;
      for (int i = 0; i < k; i++) {
          windowSum += nums[i];
      }
15 
      long maxSum = windowSum;
      for (int end = k; end < nums.length; end++) {
          // Slide the window: add the entering element, drop the leaving one.
          windowSum += nums[end] - nums[end - k];
          maxSum = Math.max(maxSum, windowSum);
      }
      return maxSum;
  }
24 
  // Variant: also report where the best window starts.
  public static int[] maxSubarrayRange(int[] nums, int k) {
      long windowSum = 0;
      for (int i = 0; i < k; i++) {
          windowSum += nums[i];
      }
31 
      long maxSum = windowSum;
      int bestStart = 0;
      for (int end = k; end < nums.length; end++) {
          windowSum += nums[end] - nums[end - k];
          if (windowSum > maxSum) {
              maxSum = windowSum;
              bestStart = end - k + 1;
          }
      }
      return Arrays.copyOfRange(nums, bestStart, bestStart + k);
  }
43}

01 / 01

STEP 01

Walkthrough

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Three takeaways

1Maintaining a running sum and updating it incrementally turns an O(n*k) brute force into a single O(n) pass.
2Validating arguments up front keeps the core loop simple and surfaces misuse clearly.
3Tracking the index where the best result occurs costs almost nothing and lets you return the window itself, not just its score.

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