Add solution #1284

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,208 LeetCode solutions in JavaScript
+# 1,209 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -969,6 +969,7 @@
 1276|[Number of Burgers with No Waste of Ingredients](./solutions/1276-number-of-burgers-with-no-waste-of-ingredients.js)|Medium|
 1277|[Count Square Submatrices with All Ones](./solutions/1277-count-square-submatrices-with-all-ones.js)|Medium|
 1278|[Palindrome Partitioning III](./solutions/1278-palindrome-partitioning-iii.js)|Hard|
+1284|[Minimum Number of Flips to Convert Binary Matrix to Zero Matrix](./solutions/1284-minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.js)|Hard|
 1287|[Element Appearing More Than 25% In Sorted Array](./solutions/1287-element-appearing-more-than-25-in-sorted-array.js)|Easy|
 1290|[Convert Binary Number in a Linked List to Integer](./solutions/1290-convert-binary-number-in-a-linked-list-to-integer.js)|Easy|
 1291|[Sequential Digits](./solutions/1291-sequential-digits.js)|Medium|

solutions/1284-minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.js

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+/**
+ * 1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
+ * https://leetcode.com/problems/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix/
+ * Difficulty: Hard
+ *
+ * Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the
+ * four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are
+ * called neighbors if they share one edge.
+ *
+ * Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.
+ *
+ * A binary matrix is a matrix with all cells equal to 0 or 1 only.
+ *
+ * A zero matrix is a matrix with all cells equal to 0.
+ */
+
+/**
+ * @param {number[][]} mat
+ * @return {number}
+ */
+var minFlips = function(mat) {
+  const rows = mat.length;
+  const cols = mat[0].length;
+  const target = 0;
+  const start = mat.flat().reduce((acc, val, idx) => acc | (val << idx), 0);
+  const queue = [[start, 0]];
+  const seen = new Set([start]);
+  const directions = [[0, 0], [0, 1], [0, -1], [1, 0], [-1, 0]];
+
+  while (queue.length) {
+    const [state, steps] = queue.shift();
+
+    if (state === target) return steps;
+
+    for (let i = 0; i < rows; i++) {
+      for (let j = 0; j < cols; j++) {
+        let nextState = state;
+        for (const [di, dj] of directions) {
+          const ni = i + di;
+          const nj = j + dj;
+          if (ni >= 0 && ni < rows && nj >= 0 && nj < cols) {
+            const pos = ni * cols + nj;
+            nextState ^= (1 << pos);
+          }
+        }
+        if (!seen.has(nextState)) {
+          seen.add(nextState);
+          queue.push([nextState, steps + 1]);
+        }
+      }
+    }
+  }
+
+  return -1;
+};