Merge branch 'master' into test/EditDistance

Author: siriak

src/main/java/com/thealgorithms/dynamicprogramming/LongestAlternatingSubsequence.java

@@ -1,73 +1,73 @@
 package com.thealgorithms.dynamicprogramming;
 
-/*
-
- * Problem Statement: -
- * Find Longest Alternating Subsequence
-
- * A sequence {x1, x2, .. xn} is alternating sequence if its elements satisfy one of the following
- relations :
-
-   x1 < x2 > x3 < x4 > x5 < …. xn or
-   x1 > x2 < x3 > x4 < x5 > …. xn
+/**
+ * Class for finding the length of the longest alternating subsequence in an array.
+ *
+ * <p>An alternating sequence is a sequence of numbers where the elements alternate
+ * between increasing and decreasing. Specifically, a sequence is alternating if its elements
+ * satisfy one of the following relations:
+ *
+ * <ul>
+ *   <li>{@code x1 < x2 > x3 < x4 > x5 < ... < xn}</li>
+ *   <li>{@code x1 > x2 < x3 > x4 < x5 > ... > xn}</li>
+ * </ul>
+ *
+ * <p>This class provides a method to compute the length of the longest such subsequence
+ * from a given array of integers.
  */
 public final class LongestAlternatingSubsequence {
     private LongestAlternatingSubsequence() {
     }
 
-    /* Function to return longest alternating subsequence length*/
+    /**
+     * Finds the length of the longest alternating subsequence in the given array.
+     *
+     * @param arr an array of integers where the longest alternating subsequence is to be found
+     * @param n the length of the array {@code arr}
+     * @return the length of the longest alternating subsequence
+     *
+     * <p>The method uses dynamic programming to solve the problem. It maintains a 2D array
+     * {@code las} where:
+     * <ul>
+     *   <li>{@code las[i][0]} represents the length of the longest alternating subsequence
+     *   ending at index {@code i} with the last element being greater than the previous element.</li>
+     *   <li>{@code las[i][1]} represents the length of the longest alternating subsequence
+     *   ending at index {@code i} with the last element being smaller than the previous element.</li>
+     * </ul>
+     *
+     * <p>The method iterates through the array and updates the {@code las} array based on
+     * whether the current element is greater or smaller than the previous elements.
+     * The result is the maximum value found in the {@code las} array.
+     */
     static int alternatingLength(int[] arr, int n) {
-        /*
-
-                las[i][0] = Length of the longest
-                        alternating subsequence ending at
-                        index i and last element is
-                        greater than its previous element
-
-                las[i][1] = Length of the longest
-                        alternating subsequence ending at
-                        index i and last element is
-                        smaller than its previous
-                        element
-
-         */
         int[][] las = new int[n][2]; // las = LongestAlternatingSubsequence
 
+        // Initialize the dp array
         for (int i = 0; i < n; i++) {
             las[i][0] = 1;
             las[i][1] = 1;
         }
 
         int result = 1; // Initialize result
 
-        /* Compute values in bottom up manner */
+        // Compute values in a bottom-up manner
         for (int i = 1; i < n; i++) {
-            /* Consider all elements as previous of arr[i]*/
             for (int j = 0; j < i; j++) {
-                /* If arr[i] is greater, then check with las[j][1] */
+                // If arr[i] is greater than arr[j], update las[i][0]
                 if (arr[j] < arr[i] && las[i][0] < las[j][1] + 1) {
                     las[i][0] = las[j][1] + 1;
                 }
 
-                /* If arr[i] is smaller, then check with las[j][0]*/
+                // If arr[i] is smaller than arr[j], update las[i][1]
                 if (arr[j] > arr[i] && las[i][1] < las[j][0] + 1) {
                     las[i][1] = las[j][0] + 1;
                 }
             }
 
-            /* Pick maximum of both values at index i */
-            if (result < Math.max(las[i][0], las[i][1])) {
-                result = Math.max(las[i][0], las[i][1]);
-            }
+            // Pick the maximum of both values at index i
+            result = Math.max(result, Math.max(las[i][0], las[i][1]));
         }
 
         return result;
     }
-
-    public static void main(String[] args) {
-        int[] arr = {10, 22, 9, 33, 49, 50, 31, 60};
-        int n = arr.length;
-        System.out.println("Length of Longest "
-            + "alternating subsequence is " + alternatingLength(arr, n));
-    }
 }

src/main/java/com/thealgorithms/others/LineSweep.java

@@ -1,55 +1,62 @@
 package com.thealgorithms.others;
+
 import java.util.Arrays;
 import java.util.Comparator;
 
-/* Line Sweep algorithm can be used to solve range problems by first sorting the list of ranges
- * by the start value of the range in non-decreasing order and doing a "sweep" through the number
- * line(x-axis) by incrementing the start point by 1 and decrementing the end point+1 by 1 on the
- * number line.
- * An overlapping range is defined as (StartA <= EndB) AND (EndA >= StartB)
- * References
- * https://en.wikipedia.org/wiki/Sweep_line_algorithm
- * https://en.wikipedia.org/wiki/De_Morgan%27s_laws>
+/**
+ * The Line Sweep algorithm is used to solve range problems efficiently. It works by:
+ * 1. Sorting a list of ranges by their start values in non-decreasing order.
+ * 2. Sweeping through the number line (x-axis) while updating a count for each point based on the ranges.
+ *
+ * An overlapping range is defined as:
+ * - (StartA <= EndB) AND (EndA >= StartB)
+ *
+ * References:
+ * - https://en.wikipedia.org/wiki/Sweep_line_algorithm
+ * - https://en.wikipedia.org/wiki/De_Morgan%27s_laws
  */
 public final class LineSweep {
     private LineSweep() {
     }
 
     /**
-     * Find Maximum end point
-     *   param = ranges : Array of range[start,end]
-     *   return Maximum Endpoint
+     * Finds the maximum endpoint from a list of ranges.
+     *
+     * @param ranges a 2D array where each element is a range represented by [start, end]
+     * @return the maximum endpoint among all ranges
      */
     public static int findMaximumEndPoint(int[][] ranges) {
-        Arrays.sort(ranges, Comparator.comparingInt(a -> a[1]));
+        Arrays.sort(ranges, Comparator.comparingInt(range -> range[1]));
         return ranges[ranges.length - 1][1];
     }
 
     /**
-     * Find if any ranges overlap
-     *   param = ranges : Array of range[start,end]
-     *   return true if overlap exists false otherwise.
+     * Determines if any of the given ranges overlap.
+     *
+     * @param ranges a 2D array where each element is a range represented by [start, end]
+     * @return true if any ranges overlap, false otherwise
      */
     public static boolean isOverlap(int[][] ranges) {
+        if (ranges == null || ranges.length == 0) {
+            return false;
+        }
 
         int maximumEndPoint = findMaximumEndPoint(ranges);
-        Arrays.sort(ranges, Comparator.comparingInt(a -> a[0]));
         int[] numberLine = new int[maximumEndPoint + 2];
         for (int[] range : ranges) {
-
             int start = range[0];
             int end = range[1];
 
             numberLine[start] += 1;
             numberLine[end + 1] -= 1;
         }
 
-        int current = 0;
-        int overlaps = 0;
-        for (int num : numberLine) {
-            current += num;
-            overlaps = Math.max(overlaps, current);
+        int currentCount = 0;
+        int maxOverlaps = 0;
+        for (int count : numberLine) {
+            currentCount += count;
+            maxOverlaps = Math.max(maxOverlaps, currentCount);
         }
-        return overlaps > 1;
+        return maxOverlaps > 1;
     }
 }