Add solution #1769

Author: JoshCrozier

README.md

@@ -513,6 +513,7 @@
 1764|[Form Array by Concatenating Subarrays of Another Array](./1764-form-array-by-concatenating-subarrays-of-another-array.js)|Medium|
 1765|[Map of Highest Peak](./1765-map-of-highest-peak.js)|Medium|
 1768|[Merge Strings Alternately](./1768-merge-strings-alternately.js)|Easy|
+1769|[Minimum Number of Operations to Move All Balls to Each Box](./1769-minimum-number-of-operations-to-move-all-balls-to-each-box.js)|Medium|
 1780|[Check if Number is a Sum of Powers of Three](./1780-check-if-number-is-a-sum-of-powers-of-three.js)|Medium|
 1790|[Check if One String Swap Can Make Strings Equal](./1790-check-if-one-string-swap-can-make-strings-equal.js)|Easy|
 1791|[Find Center of Star Graph](./1791-find-center-of-star-graph.js)|Easy|

solutions/1769-minimum-number-of-operations-to-move-all-balls-to-each-box.js

@@ -0,0 +1,37 @@
+/**
+ * 1769. Minimum Number of Operations to Move All Balls to Each Box
+ * https://leetcode.com/problems/minimum-number-of-operations-to-move-all-balls-to-each-box/
+ * Difficulty: Medium
+ *
+ * You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if
+ * the ith box is empty, and '1' if it contains one ball.
+ *
+ * In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to
+ * box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some
+ * boxes.
+ *
+ * Return an array answer of size n, where answer[i] is the minimum number of operations needed
+ * to move all the balls to the ith box.
+ *
+ * Each answer[i] is calculated considering the initial state of the boxes.
+ */
+
+/**
+ * @param {string} boxes
+ * @return {number[]}
+ */
+var minOperations = function(boxes) {
+  const result = new Array(boxes.length).fill(0);
+
+  for (let i = 0; i < boxes.length; i++) {
+    let total = 0;
+    for (let j = 0; j < boxes.length; j++) {
+      if (boxes[j] === '1') {
+        total += Math.abs(j - i);
+      }
+    }
+    result[i] = total;
+  }
+
+  return result;
+};