Add solution #1727

JoshCrozier · JoshCrozier · commit ebe130696cd4 · 2025-04-27T14:49:29.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,503 LeetCode solutions in JavaScript
+# 1,504 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1331,6 +1331,7 @@
 1723|[Find Minimum Time to Finish All Jobs](./solutions/1723-find-minimum-time-to-finish-all-jobs.js)|Hard|
 1725|[Number Of Rectangles That Can Form The Largest Square](./solutions/1725-number-of-rectangles-that-can-form-the-largest-square.js)|Easy|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|
+1727|[Largest Submatrix With Rearrangements](./solutions/1727-largest-submatrix-with-rearrangements.js)|Medium|
 1732|[Find the Highest Altitude](./solutions/1732-find-the-highest-altitude.js)|Easy|
 1748|[Sum of Unique Elements](./solutions/1748-sum-of-unique-elements.js)|Easy|
 1749|[Maximum Absolute Sum of Any Subarray](./solutions/1749-maximum-absolute-sum-of-any-subarray.js)|Medium|
diff --git a/solutions/1727-largest-submatrix-with-rearrangements.js b/solutions/1727-largest-submatrix-with-rearrangements.js
@@ -0,0 +1,37 @@
+/**
+ * 1727. Largest Submatrix With Rearrangements
+ * https://leetcode.com/problems/largest-submatrix-with-rearrangements/
+ * Difficulty: Medium
+ *
+ * You are given a binary matrix matrix of size m x n, and you are allowed to rearrange the
+ * columns of the matrix in any order.
+ *
+ * Return the area of the largest submatrix within matrix where every element of the submatrix
+ * is 1 after reordering the columns optimally.
+ */
+
+/**
+ * @param {number[][]} matrix
+ * @return {number}
+ */
+var largestSubmatrix = function(matrix) {
+  const rows = matrix.length;
+  const cols = matrix[0].length;
+  let result = 0;
+
+  for (let row = 0; row < rows; row++) {
+    for (let col = 0; col < cols; col++) {
+      if (row > 0 && matrix[row][col] === 1) {
+        matrix[row][col] += matrix[row - 1][col];
+      }
+    }
+
+    const heights = matrix[row].slice().sort((a, b) => b - a);
+    for (let i = 0; i < cols; i++) {
+      if (heights[i] === 0) break;
+      result = Math.max(result, heights[i] * (i + 1));
+    }
+  }
+
+  return result;
+};
