Add solution #3355

JoshCrozier · JoshCrozier · commit 2ed286a05823 · 2025-05-19T19:06:05.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,883 LeetCode solutions in JavaScript
+# 1,884 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1877,6 +1877,7 @@
 3341|[Find Minimum Time to Reach Last Room I](./solutions/3341-find-minimum-time-to-reach-last-room-i.js)|Medium|
 3342|[Find Minimum Time to Reach Last Room II](./solutions/3342-find-minimum-time-to-reach-last-room-ii.js)|Medium|
 3343|[Count Number of Balanced Permutations](./solutions/3343-count-number-of-balanced-permutations.js)|Hard|
+3355|[Zero Array Transformation I](./solutions/3355-zero-array-transformation-i.js)|Medium|
 3356|[Zero Array Transformation II](./solutions/3356-zero-array-transformation-ii.js)|Medium|
 3375|[Minimum Operations to Make Array Values Equal to K](./solutions/3375-minimum-operations-to-make-array-values-equal-to-k.js)|Easy|
 3392|[Count Subarrays of Length Three With a Condition](./solutions/3392-count-subarrays-of-length-three-with-a-condition.js)|Easy|
diff --git a/solutions/3355-zero-array-transformation-i.js b/solutions/3355-zero-array-transformation-i.js
@@ -0,0 +1,43 @@
+/**
+ * 3355. Zero Array Transformation I
+ * https://leetcode.com/problems/zero-array-transformation-i/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums of length n and a 2D array queries, where
+ * queries[i] = [li, ri].
+ *
+ * For each queries[i]:
+ * - Select a subset of indices within the range [li, ri] in nums.
+ * - Decrement the values at the selected indices by 1.
+ *
+ * A Zero Array is an array where all elements are equal to 0.
+ *
+ * Return true if it is possible to transform nums into a Zero Array after processing all the
+ * queries sequentially, otherwise return false.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[][]} queries
+ * @return {boolean}
+ */
+var isZeroArray = function(nums, queries) {
+  const maxDecrements = new Array(nums.length).fill(0);
+
+  for (const [left, right] of queries) {
+    maxDecrements[left]++;
+    if (right + 1 < nums.length) {
+      maxDecrements[right + 1]--;
+    }
+  }
+
+  let currentDecrements = 0;
+  for (let i = 0; i < nums.length; i++) {
+    currentDecrements += maxDecrements[i];
+    if (nums[i] > currentDecrements) {
+      return false;
+    }
+  }
+
+  return true;
+};
