Merge branch 'master' into refactor/InfixToPostfix

Author: DenizAltunkapan

src/main/java/com/thealgorithms/misc/RangeInSortedArray.java

@@ -1,25 +1,46 @@
 package com.thealgorithms.misc;
 
+/**
+ * Utility class for operations to find the range of occurrences of a key
+ * in a sorted (non-decreasing) array, and to count elements less than or equal to a given key.
+ */
 public final class RangeInSortedArray {
+
     private RangeInSortedArray() {
     }
 
-    // Get the 1st and last occurrence index of a number 'key' in a non-decreasing array 'nums'
-    // Gives [-1, -1] in case element doesn't exist in array
+    /**
+     * Finds the first and last occurrence indices of the key in a sorted array.
+     *
+     * @param nums sorted array of integers (non-decreasing order)
+     * @param key the target value to search for
+     * @return int array of size two where
+     *         - index 0 is the first occurrence of key,
+     *         - index 1 is the last occurrence of key,
+     *         or [-1, -1] if the key does not exist in the array.
+     */
     public static int[] sortedRange(int[] nums, int key) {
         int[] range = new int[] {-1, -1};
-        alteredBinSearchIter(nums, key, 0, nums.length - 1, range, true);
-        alteredBinSearchIter(nums, key, 0, nums.length - 1, range, false);
+        alteredBinSearchIter(nums, key, 0, nums.length - 1, range, true); // find left boundary
+        alteredBinSearchIter(nums, key, 0, nums.length - 1, range, false); // find right boundary
         return range;
     }
 
-    // Recursive altered binary search which searches for leftmost as well as rightmost occurrence
-    // of 'key'
+    /**
+     * Recursive altered binary search to find either the leftmost or rightmost occurrence of a key.
+     *
+     * @param nums the sorted array
+     * @param key the target to find
+     * @param left current left bound in search
+     * @param right current right bound in search
+     * @param range array to update with boundaries: range[0] for leftmost, range[1] for rightmost
+     * @param goLeft if true, searches for leftmost occurrence; if false, for rightmost occurrence
+     */
     public static void alteredBinSearch(int[] nums, int key, int left, int right, int[] range, boolean goLeft) {
         if (left > right) {
             return;
         }
-        int mid = (left + right) >>> 1;
+        int mid = left + ((right - left) >>> 1);
         if (nums[mid] > key) {
             alteredBinSearch(nums, key, left, mid - 1, range, goLeft);
         } else if (nums[mid] < key) {
@@ -41,11 +62,19 @@ public static void alteredBinSearch(int[] nums, int key, int left, int right, in
         }
     }
 
-    // Iterative altered binary search which searches for leftmost as well as rightmost occurrence
-    // of 'key'
+    /**
+     * Iterative altered binary search to find either the leftmost or rightmost occurrence of a key.
+     *
+     * @param nums the sorted array
+     * @param key the target to find
+     * @param left initial left bound
+     * @param right initial right bound
+     * @param range array to update with boundaries: range[0] for leftmost, range[1] for rightmost
+     * @param goLeft if true, searches for leftmost occurrence; if false, for rightmost occurrence
+     */
     public static void alteredBinSearchIter(int[] nums, int key, int left, int right, int[] range, boolean goLeft) {
         while (left <= right) {
-            final int mid = (left + right) >>> 1;
+            int mid = left + ((right - left) >>> 1);
             if (nums[mid] > key) {
                 right = mid - 1;
             } else if (nums[mid] < key) {
@@ -55,33 +84,48 @@ public static void alteredBinSearchIter(int[] nums, int key, int left, int right
                     if (mid == 0 || nums[mid - 1] != key) {
                         range[0] = mid;
                         return;
-                    } else {
-                        right = mid - 1;
                     }
+                    right = mid - 1;
                 } else {
                     if (mid == nums.length - 1 || nums[mid + 1] != key) {
                         range[1] = mid;
                         return;
-                    } else {
-                        left = mid + 1;
                     }
+                    left = mid + 1;
                 }
             }
         }
     }
 
+    /**
+     * Counts the number of elements strictly less than the given key.
+     *
+     * @param nums sorted array
+     * @param key the key to compare
+     * @return the count of elements less than the key
+     */
     public static int getCountLessThan(int[] nums, int key) {
         return getLessThan(nums, key, 0, nums.length - 1);
     }
 
+    /**
+     * Helper method using binary search to count elements less than or equal to the key.
+     *
+     * @param nums sorted array
+     * @param key the key to compare
+     * @param left current left bound
+     * @param right current right bound
+     * @return count of elements less than or equal to the key
+     */
     public static int getLessThan(int[] nums, int key, int left, int right) {
         int count = 0;
         while (left <= right) {
-            final int mid = (left + right) >>> 1;
+            int mid = left + ((right - left) >>> 1);
             if (nums[mid] > key) {
                 right = mid - 1;
-            } else if (nums[mid] <= key) {
-                count = mid + 1; // At least mid+1 elements exist which are <= key
+            } else {
+                // nums[mid] <= key
+                count = mid + 1; // all elements from 0 to mid inclusive are <= key
                 left = mid + 1;
             }
         }

src/main/java/com/thealgorithms/stacks/LargestRectangle.java

@@ -3,36 +3,50 @@
 import java.util.Stack;
 
 /**
+ * Utility class to calculate the largest rectangle area in a histogram.
+ * Each bar's width is assumed to be 1 unit.
  *
- * @author mohd rameez github.com/rameez471
+ * <p>This implementation uses a monotonic stack to efficiently calculate
+ * the area of the largest rectangle that can be formed from the histogram bars.</p>
+ *
+ * <p>Example usage:
+ * <pre>{@code
+ *     int[] heights = {2, 1, 5, 6, 2, 3};
+ *     String area = LargestRectangle.largestRectangleHistogram(heights);
+ *     // area is "10"
+ * }</pre>
  */
-
 public final class LargestRectangle {
+
     private LargestRectangle() {
     }
 
+    /**
+     * Calculates the largest rectangle area in the given histogram.
+     *
+     * @param heights an array of non-negative integers representing bar heights
+     * @return the largest rectangle area as a {@link String}
+     */
     public static String largestRectangleHistogram(int[] heights) {
-        int n = heights.length;
         int maxArea = 0;
-        Stack<int[]> st = new Stack<>();
-        for (int i = 0; i < n; i++) {
+        Stack<int[]> stack = new Stack<>();
+
+        for (int i = 0; i < heights.length; i++) {
             int start = i;
-            while (!st.isEmpty() && st.peek()[1] > heights[i]) {
-                int[] tmp = st.pop();
-                maxArea = Math.max(maxArea, tmp[1] * (i - tmp[0]));
-                start = tmp[0];
+            while (!stack.isEmpty() && stack.peek()[1] > heights[i]) {
+                int[] popped = stack.pop();
+                maxArea = Math.max(maxArea, popped[1] * (i - popped[0]));
+                start = popped[0];
             }
-            st.push(new int[] {start, heights[i]});
+            stack.push(new int[] {start, heights[i]});
         }
-        while (!st.isEmpty()) {
-            int[] tmp = st.pop();
-            maxArea = Math.max(maxArea, tmp[1] * (n - tmp[0]));
+
+        int totalLength = heights.length;
+        while (!stack.isEmpty()) {
+            int[] remaining = stack.pop();
+            maxArea = Math.max(maxArea, remaining[1] * (totalLength - remaining[0]));
         }
-        return Integer.toString(maxArea);
-    }
 
-    public static void main(String[] args) {
-        assert largestRectangleHistogram(new int[] {2, 1, 5, 6, 2, 3}).equals("10");
-        assert largestRectangleHistogram(new int[] {2, 4}).equals("4");
+        return Integer.toString(maxArea);
     }
 }

src/main/java/com/thealgorithms/strings/Isomorphic.java

@@ -5,35 +5,54 @@
 import java.util.Map;
 import java.util.Set;
 
+/**
+ * Utility class to check if two strings are isomorphic.
+ *
+ * <p>
+ * Two strings {@code s} and {@code t} are isomorphic if the characters in {@code s}
+ * can be replaced to get {@code t}, while preserving the order of characters.
+ * Each character must map to exactly one character, and no two characters can map to the same character.
+ * </p>
+ *
+ * @see <a href="https://en.wikipedia.org/wiki/Isomorphism_(computer_science)">Isomorphic Strings</a>
+ */
 public final class Isomorphic {
+
     private Isomorphic() {
     }
 
-    public static boolean checkStrings(String s, String t) {
+    /**
+     * Checks if two strings are isomorphic.
+     *
+     * @param s the first input string
+     * @param t the second input string
+     * @return {@code true} if {@code s} and {@code t} are isomorphic; {@code false} otherwise
+     */
+    public static boolean areIsomorphic(String s, String t) {
         if (s.length() != t.length()) {
             return false;
         }
 
-        // To mark the characters of string using MAP
-        // character of first string as KEY and another as VALUE
-        // now check occurence by keeping the track with SET data structure
-        Map<Character, Character> characterMap = new HashMap<>();
-        Set<Character> trackUniqueCharacter = new HashSet<>();
+        Map<Character, Character> map = new HashMap<>();
+        Set<Character> usedCharacters = new HashSet<>();
 
         for (int i = 0; i < s.length(); i++) {
-            if (characterMap.containsKey(s.charAt(i))) {
-                if (t.charAt(i) != characterMap.get(s.charAt(i))) {
+            char sourceChar = s.charAt(i);
+            char targetChar = t.charAt(i);
+
+            if (map.containsKey(sourceChar)) {
+                if (map.get(sourceChar) != targetChar) {
                     return false;
                 }
             } else {
-                if (trackUniqueCharacter.contains(t.charAt(i))) {
+                if (usedCharacters.contains(targetChar)) {
                     return false;
                 }
-
-                characterMap.put(s.charAt(i), t.charAt(i));
+                map.put(sourceChar, targetChar);
+                usedCharacters.add(targetChar);
             }
-            trackUniqueCharacter.add(t.charAt(i));
         }
+
         return true;
     }
 }

src/main/java/com/thealgorithms/strings/LongestCommonPrefix.java

@@ -2,21 +2,41 @@
 
 import java.util.Arrays;
 
+/**
+ * Utility class for string operations.
+ * <p>
+ * This class provides a method to find the longest common prefix (LCP)
+ * among an array of strings.
+ * </p>
+ *
+ * @see <a href="https://en.wikipedia.org/wiki/Longest_common_prefix">Longest Common Prefix - Wikipedia</a>
+ */
 public final class LongestCommonPrefix {
-    public String longestCommonPrefix(String[] strs) {
+
+    private LongestCommonPrefix() {
+    }
+
+    /**
+     * Finds the longest common prefix among a list of strings using lexicographical sorting.
+     * The prefix is common to the first and last elements after sorting the array.
+     *
+     * @param strs array of input strings
+     * @return the longest common prefix, or empty string if none exists
+     */
+    public static String longestCommonPrefix(String[] strs) {
         if (strs == null || strs.length == 0) {
             return "";
         }
 
         Arrays.sort(strs);
-        String shortest = strs[0];
-        String longest = strs[strs.length - 1];
+        String first = strs[0];
+        String last = strs[strs.length - 1];
 
         int index = 0;
-        while (index < shortest.length() && index < longest.length() && shortest.charAt(index) == longest.charAt(index)) {
+        while (index < first.length() && index < last.length() && first.charAt(index) == last.charAt(index)) {
             index++;
         }
 
-        return shortest.substring(0, index);
+        return first.substring(0, index);
     }
 }