@@ -158,3 +158,4 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Longest Substring Witout Repeats](problems/src/two_pointers/LongestSubstringWitoutRepeats.java) (Medium)
 - [Three Sum](problems/src/two_pointers/ThreeSum.java) (Medium)
 - [Trapping Rain Water](problems/src/two_pointers/TrappingRainWater.java) (Hard)
+- [3Sum Closest](problems/src/two_pointers/ThreeSumClosest.java) (Hard)
@@ -0,0 +1,57 @@
+package two_pointers;
+
+import java.util.Arrays;
+
+/**
+ * Created by gouthamvidyapradhan on 13/06/2017.
+ * Accepted
+ Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
+
+ For example, given array S = {-1 2 1 -4}, and>
+
+ The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
+ */
+public class ThreeSumClosest {
+
+    public static void main(String[] args) {
+        int[] a = {-1, 2, 1, -4};
+        System.out.println(new ThreeSumClosest().threeSumClosest(a, 1));
+    }
+
+    public int threeSumClosest(int[] a, int target) {
+        Arrays.sort(a);
+        int min = Integer.MAX_VALUE, ans = -1;
+        for(int i = 0, l = a.length; i < l - 2; i ++) {
+            if(i == 0 || ! (a[i] == a[i - 1])) {
+                int j = i + 1, k = l - 1;
+                while(k > j) {
+                    if(j != i + 1 && (a[j] == a[j - 1])) {
+                        j ++;
+                        continue;
+                    }
+                    int sum = a[i] + a[j] + a[k];
+                    if(sum < target) {
+                        int diff = Math.abs(sum - target);
+                        if(diff < min) {
+                            min = diff;
+                            ans = sum;
+                        }
+                        j++;
+                    }
+                    else if(sum > target) {
+                        int diff = Math.abs(sum - target);
+                        if(diff < min) {
+                            min = diff;
+                            ans = sum;
+                        }
+                        k--;
+                    }
+                    else{
+                        return sum;
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+}