Add solution #2780

Author: JoshCrozier

README.md

@@ -1001,6 +1001,7 @@
 2725|[Interval Cancellation](./solutions/2725-interval-cancellation.js)|Easy|
 2726|[Calculator with Method Chaining](./solutions/2726-calculator-with-method-chaining.js)|Easy|
 2727|[Is Object Empty](./solutions/2727-is-object-empty.js)|Easy|
+2780|[Minimum Index of a Valid Split](./solutions/2780-minimum-index-of-a-valid-split.js)|Medium|
 2948|[Make Lexicographically Smallest Array by Swapping Elements](./solutions/2948-make-lexicographically-smallest-array-by-swapping-elements.js)|Medium|
 2965|[Find Missing and Repeated Values](./solutions/2965-find-missing-and-repeated-values.js)|Easy|
 3042|[Count Prefix and Suffix Pairs I](./solutions/3042-count-prefix-and-suffix-pairs-i.js)|Easy|

solutions/2780-minimum-index-of-a-valid-split.js

@@ -0,0 +1,52 @@
+/**
+ * 2780. Minimum Index of a Valid Split
+ * https://leetcode.com/problems/minimum-index-of-a-valid-split/
+ * Difficulty: Medium
+ *
+ * An element x of an integer array arr of length m is dominant if more than half the elements
+ * of arr have a value of x.
+ *
+ * You are given a 0-indexed integer array nums of length n with one dominant element.
+ *
+ * You can split nums at an index i into two arrays nums[0, ..., i] and nums[i + 1, ..., n - 1],
+ * but the split is only valid if:
+ * - 0 <= i < n - 1
+ * - nums[0, ..., i], and nums[i + 1, ..., n - 1] have the same dominant element.
+ *
+ * Here, nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j,
+ * both ends being inclusive. Particularly, if j < i then nums[i, ..., j] denotes an empty subarray.
+ *
+ * Return the minimum index of a valid split. If no valid split exists, return -1.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumIndex = function(nums) {
+  const map = new Map();
+
+  for (const num of nums) {
+    map.set(num, (map.get(num) || 0) + 1);
+  }
+
+  let dominantElement;
+  for (const [num, count] of map) {
+    if (count * 2 > nums.length) {
+      dominantElement = num;
+      break;
+    }
+  }
+
+  let leftCount = 0;
+  for (let i = 0; i < nums.length - 1; i++) {
+    leftCount += nums[i] === dominantElement ? 1 : 0;
+    const rightCount = map.get(dominantElement) - leftCount;
+
+    if (leftCount * 2 > i + 1 && rightCount * 2 > nums.length - i - 1) {
+      return i;
+    }
+  }
+
+  return -1;
+};