Add solution #3495

Author: JoshCrozier

README.md

@@ -2757,6 +2757,7 @@
 3481|[Apply Substitutions](./solutions/3481-apply-substitutions.js)|Medium|
 3487|[Maximum Unique Subarray Sum After Deletion](./solutions/3487-maximum-unique-subarray-sum-after-deletion.js)|Easy|
 3491|[Phone Number Prefix](./solutions/3491-phone-number-prefix.js)|Easy|
+3495|[Minimum Operations to Make Array Elements Zero](./solutions/3495-minimum-operations-to-make-array-elements-zero.js)|Hard|
 3496|[Maximize Score After Pair Deletions](./solutions/3496-maximize-score-after-pair-deletions.js)|Medium|
 3506|[Find Time Required to Eliminate Bacterial Strains](./solutions/3506-find-time-required-to-eliminate-bacterial-strains.js)|Hard|
 3511|[Make a Positive Array](./solutions/3511-make-a-positive-array.js)|Medium|

solutions/3495-minimum-operations-to-make-array-elements-zero.js

@@ -0,0 +1,44 @@
+/**
+ * 3495. Minimum Operations to Make Array Elements Zero
+ * https://leetcode.com/problems/minimum-operations-to-make-array-elements-zero/
+ * Difficulty: Hard
+ *
+ * You are given a 2D array queries, where queries[i] is of the form [l, r]. Each
+ * queries[i] defines an array of integers nums consisting of elements ranging
+ * from l to r, both inclusive.
+ *
+ * In one operation, you can:
+ * - Select two integers a and b from the array.
+ * - Replace them with floor(a / 4) and floor(b / 4).
+ *
+ * Your task is to determine the minimum number of operations required to reduce all
+ * elements of the array to zero for each query. Return the sum of the results for
+ * all queries.
+ */
+
+/**
+ * @param {number[][]} queries
+ * @return {number}
+ */
+var minOperations = function(queries) {
+  let result = 0;
+
+  for (const [start, end] of queries) {
+    let operationsNeeded = 0;
+
+    for (let depth = 1, previousBound = 1; depth < 17; depth++) {
+      const currentBound = previousBound * 4;
+      const intersectionLeft = Math.max(start, previousBound);
+      const intersectionRight = Math.min(end, currentBound - 1);
+
+      if (intersectionRight >= intersectionLeft) {
+        operationsNeeded += (intersectionRight - intersectionLeft + 1) * depth;
+      }
+      previousBound = currentBound;
+    }
+
+    result += Math.ceil(operationsNeeded / 2);
+  }
+
+  return result;
+};