Add solution #259

JoshCrozier · JoshCrozier · commit 34bce6d304df · 2025-06-10T22:33:01.000-05:00
diff --git a/README.md b/README.md
@@ -248,6 +248,7 @@
 256|[Paint House](./solutions/0256-paint-house.js)|Medium|
 257|[Binary Tree Paths](./solutions/0257-binary-tree-paths.js)|Easy|
 258|[Add Digits](./solutions/0258-add-digits.js)|Easy|
+259|[3Sum Smaller](./solutions/0259-3sum-smaller.js)|Medium|
 260|[Single Number III](./solutions/0260-single-number-iii.js)|Medium|
 263|[Ugly Number](./solutions/0263-ugly-number.js)|Easy|
 264|[Ugly Number II](./solutions/0264-ugly-number-ii.js)|Medium|
diff --git a/solutions/0259-3sum-smaller.js b/solutions/0259-3sum-smaller.js
@@ -0,0 +1,36 @@
+/**
+ * 259. 3Sum Smaller
+ * https://leetcode.com/problems/3sum-smaller/
+ * Difficulty: Medium
+ *
+ * Given an array of n integers nums and an integer target, find the number of index triplets
+ * i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var threeSumSmaller = function(nums, target) {
+  nums.sort((a, b) => a - b);
+  const n = nums.length;
+  let result = 0;
+
+  for (let i = 0; i < n - 2; i++) {
+    let left = i + 1;
+    let right = n - 1;
+
+    while (left < right) {
+      const sum = nums[i] + nums[left] + nums[right];
+      if (sum < target) {
+        result += right - left;
+        left++;
+      } else {
+        right--;
+      }
+    }
+  }
+
+  return result;
+};
