Added FourSumProblem.java #5786

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

@@ -0,0 +1,63 @@
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+class Solution {
+    public List<List<Integer>> fourSum(int[] nums, int target) {
+        int n = nums.length;
+
+        List<List<Integer>> ans = new ArrayList<>();
+        Arrays.sort(nums);
+
+        // select the first element of the quadruplet
+        for (int i = 0; i < n; i++) {
+            // Skip duplicate 
+            if (i > 0 && nums[i] == nums[i - 1]) continue;
+
+            // select the second element 
+            for (int j = i + 1; j < n; j++) {
+                // Skip duplicate 
+                if (j != i + 1 && nums[j] == nums[j - 1]) continue;
+                //assign two pointers
+                int k = j + 1;
+                int l = n - 1;
+
+                // Two-pointer approach to find the remaining two elements
+                while (k < l) {
+                    // Calculate the sum of the current quadruplet
+                    long sum = nums[i];
+                    sum += nums[j];
+                    sum += nums[k];
+                    sum += nums[l];
+
+                    if (sum == target) {
+                        //if the sum matches the target store it in ans
+                        List<Integer> temp = new ArrayList<>();
+                        temp.add(nums[i]);
+                        temp.add(nums[j]);
+                        temp.add(nums[k]);
+                        temp.add(nums[l]);
+                        ans.add(temp);
+
+                        // Move both pointers 
+                        k++;
+                        l--;
+
+                        // skip duplicates for the third element & forth ele
+                        while (k < l && nums[k] == nums[k - 1]) k++;
+                        while (k < l && nums[l] == nums[l + 1]) l--;
+                    } 
+                    else if (sum < target)
+                        // If the sum < target, move the left pointer (k) to increase the sum
+                        //bcz array is in sorting order so next element always greater 
+                        k++;
+                    else 
+                        // If the sum > the target, move the right pointer (l) to decrease the sum
+                        //bcz array is in sorting order so previous element of 'l' pointer always smaller
+                        l--;
+                }
+            }
+        }
+        return ans;
+    }
+}