Add solution #2598

JoshCrozier · JoshCrozier · commit 0000c47f5f85 · 2025-06-05T23:36:49.000-05:00
diff --git a/README.md b/README.md
@@ -1942,6 +1942,7 @@
 2595|[Number of Even and Odd Bits](./solutions/2595-number-of-even-and-odd-bits.js)|Easy|
 2596|[Check Knight Tour Configuration](./solutions/2596-check-knight-tour-configuration.js)|Medium|
 2597|[The Number of Beautiful Subsets](./solutions/2597-the-number-of-beautiful-subsets.js)|Medium|
+2598|[Smallest Missing Non-negative Integer After Operations](./solutions/2598-smallest-missing-non-negative-integer-after-operations.js)|Medium|
 2600|[K Items With the Maximum Sum](./solutions/2600-k-items-with-the-maximum-sum.js)|Easy|
 2601|[Prime Subtraction Operation](./solutions/2601-prime-subtraction-operation.js)|Medium|
 2605|[Form Smallest Number From Two Digit Arrays](./solutions/2605-form-smallest-number-from-two-digit-arrays.js)|Easy|
diff --git a/solutions/2598-smallest-missing-non-negative-integer-after-operations.js b/solutions/2598-smallest-missing-non-negative-integer-after-operations.js
@@ -0,0 +1,40 @@
+/**
+ * 2598. Smallest Missing Non-negative Integer After Operations
+ * https://leetcode.com/problems/smallest-missing-non-negative-integer-after-operations/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums and an integer value.
+ *
+ * In one operation, you can add or subtract value from any element of nums.
+ * - For example, if nums = [1,2,3] and value = 2, you can choose to subtract value from nums[0]
+ *   to make nums = [-1,2,3].
+ *
+ * The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it.
+ * - For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2.
+ *
+ * Return the maximum MEX of nums after applying the mentioned operation any number of times.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} value
+ * @return {number}
+ */
+var findSmallestInteger = function(nums, value) {
+  const remainderCount = new Map();
+
+  for (const num of nums) {
+    const remainder = ((num % value) + value) % value;
+    remainderCount.set(remainder, (remainderCount.get(remainder) || 0) + 1);
+  }
+
+  for (let mex = 0; mex <= nums.length; mex++) {
+    const remainder = mex % value;
+    if (!remainderCount.has(remainder) || remainderCount.get(remainder) === 0) {
+      return mex;
+    }
+    remainderCount.set(remainder, remainderCount.get(remainder) - 1);
+  }
+
+  return nums.length;
+};
