Add solution #1536

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,351 LeetCode solutions in JavaScript
+# 1,352 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1173,6 +1173,7 @@
 1531|[String Compression II](./solutions/1531-string-compression-ii.js)|Hard|
 1534|[Count Good Triplets](./solutions/1534-count-good-triplets.js)|Easy|
 1535|[Find the Winner of an Array Game](./solutions/1535-find-the-winner-of-an-array-game.js)|Medium|
+1536|[Minimum Swaps to Arrange a Binary Grid](./solutions/1536-minimum-swaps-to-arrange-a-binary-grid.js)|Medium|
 1550|[Three Consecutive Odds](./solutions/1550-three-consecutive-odds.js)|Easy|
 1551|[Minimum Operations to Make Array Equal](./solutions/1551-minimum-operations-to-make-array-equal.js)|Medium|
 1566|[Detect Pattern of Length M Repeated K or More Times](./solutions/1566-detect-pattern-of-length-m-repeated-k-or-more-times.js)|Easy|

solutions/1536-minimum-swaps-to-arrange-a-binary-grid.js

@@ -0,0 +1,58 @@
+/**
+ * 1536. Minimum Swaps to Arrange a Binary Grid
+ * https://leetcode.com/problems/minimum-swaps-to-arrange-a-binary-grid/
+ * Difficulty: Medium
+ *
+ * Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and
+ * swap them.
+ *
+ * A grid is said to be valid if all the cells above the main diagonal are zeros.
+ *
+ * Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot
+ * be valid.
+ *
+ * The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var minSwaps = function(grid) {
+  const n = grid.length;
+  const trailingZeros = new Array(n).fill(0);
+
+  for (let i = 0; i < n; i++) {
+    trailingZeros[i] = countTrailingZeros(i);
+  }
+
+  let swaps = 0;
+  for (let i = 0; i < n - 1; i++) {
+    const requiredZeros = n - i - 1;
+    let found = false;
+
+    for (let j = i; j < n; j++) {
+      if (trailingZeros[j] >= requiredZeros) {
+        found = true;
+        for (let k = j; k > i; k--) {
+          [trailingZeros[k], trailingZeros[k - 1]] = [trailingZeros[k - 1], trailingZeros[k]];
+          swaps++;
+        }
+        break;
+      }
+    }
+
+    if (!found) return -1;
+  }
+
+  return swaps;
+
+  function countTrailingZeros(row) {
+    let count = 0;
+    for (let j = n - 1; j >= 0; j--) {
+      if (grid[row][j] === 0) count++;
+      else break;
+    }
+    return count;
+  }
+};