Add solution #960

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,045 LeetCode solutions in JavaScript
+# 1,046 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -769,6 +769,7 @@
 957|[Prison Cells After N Days](./solutions/0957-prison-cells-after-n-days.js)|Medium|
 958|[Check Completeness of a Binary Tree](./solutions/0958-check-completeness-of-a-binary-tree.js)|Medium|
 959|[Regions Cut By Slashes](./solutions/0959-regions-cut-by-slashes.js)|Medium|
+960|[Delete Columns to Make Sorted III](./solutions/0960-delete-columns-to-make-sorted-iii.js)|Hard|
 966|[Vowel Spellchecker](./solutions/0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./solutions/0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./solutions/0976-largest-perimeter-triangle.js)|Easy|

solutions/0960-delete-columns-to-make-sorted-iii.js

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+/**
+ * 960. Delete Columns to Make Sorted III
+ * https://leetcode.com/problems/delete-columns-to-make-sorted-iii/
+ * Difficulty: Hard
+ *
+ * You are given an array of n strings strs, all of the same length.
+ *
+ * We may choose any deletion indices, and we delete all the characters in those indices
+ * for each string.
+ *
+ * For example, if we have strs = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then
+ * the final array after deletions is ["bef", "vyz"].
+ *
+ * Suppose we chose a set of deletion indices answer such that after deletions, the final
+ * array has every string (row) in lexicographic order. (i.e.,
+ * (strs[0][0] <= strs[0][1] <= ... <= strs[0][strs[0].length - 1]),
+ * and (strs[1][0] <= strs[1][1] <= ... <= strs[1][strs[1].length - 1]),
+ * and so on). Return the minimum possible value of answer.length.
+ */
+
+/**
+ * @param {string[]} strs
+ * @return {number}
+ */
+var minDeletionSize = function(strs) {
+  const isValid = (input, a, b) => input.every(row => row[a] <= row[b]);
+  const cols = strs[0].length;
+  const dp = new Array(cols).fill(1);
+
+  for (let i = 1; i < cols; i++) {
+    for (let j = 0; j < i; j++) {
+      if (isValid(strs, j, i)) {
+        dp[i] = Math.max(dp[i], dp[j] + 1);
+      }
+    }
+  }
+
+  return cols - Math.max(...dp);
+};