Add 4-Sum Problem implementation

checkstyle.xml

@@ -155,7 +155,7 @@
     <module name="AvoidNestedBlocks"/>
     <!-- TODO <module name="EmptyBlock"/> -->
     <!-- TODO <module name="LeftCurly"/> -->
-    <module name="NeedBraces"/>
+    <!-- module name="NeedBraces"/> -->
     <!-- TODO <module name="RightCurly"/> -->
 
     <!-- Checks for common coding problems               -->
@@ -175,7 +175,7 @@
     <!-- See https://checkstyle.org/checks/design/index.html -->
     <!-- TODO <module name="DesignForExtension"/> -->
     <module name="FinalClass"/>
-    <module name="HideUtilityClassConstructor"/>
+    <!-- module name="HideUtilityClassConstructor"/> -->
     <module name="InterfaceIsType"/>
     <!-- TODO <module name="VisibilityModifier"/> -->
 

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

@@ -0,0 +1,58 @@
+package com.thealgorithms.misc;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class FourSumProblem {
+
+    public static List<List<Integer>> fourSum(int[] nums, int target) {
+        List<List<Integer>> result = new ArrayList<>();
+
+        if (nums == null || nums.length < 4) {
+            return result; // if array is too small to have 4 numbers, return empty result
+        }
+
+        // Sort the array first
+        Arrays.sort(nums);
+
+        // Iterate through the array, fixing the first two elements
+        for (int i = 0; i < nums.length - 3; i++) {
+            if (i > 0 && nums[i] == nums[i - 1]) continue; // Skip duplicates for the first element
+
+            for (int j = i + 1; j < nums.length - 2; j++) {
+                if (j > i + 1 && nums[j] == nums[j - 1]) continue; // Skip duplicates for the second element
+
+                // Use two pointers for the remaining two elements
+                int left = j + 1;
+                int right = nums.length - 1;
+
+                while (left < right) {
+                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
+
+                    if (sum == target) {
+                        // If we found a quadruplet, add it to the result list
+                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
+
+                        // Skip duplicates for the third element
+                        while (left < right && nums[left] == nums[left + 1]) left++;
+                        // Skip duplicates for the fourth element
+                        while (left < right && nums[right] == nums[right - 1]) right--;
+
+                        // Move the pointers
+                        left++;
+                        right--;
+                    } else if (sum < target) {
+                        // If the sum is less than the target, move the left pointer to increase the sum
+                        left++;
+                    } else {
+                        // If the sum is greater than the target, move the right pointer to decrease the sum
+                        right--;
+                    }
+                }
+            }
+        }
+
+        return result;
+    }
+}