Add solution #955

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,040 LeetCode solutions in JavaScript
+# 1,041 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -764,6 +764,7 @@
 952|[Largest Component Size by Common Factor](./solutions/0952-largest-component-size-by-common-factor.js)|Hard|
 953|[Verifying an Alien Dictionary](./solutions/0953-verifying-an-alien-dictionary.js)|Easy|
 954|[Array of Doubled Pairs](./solutions/0954-array-of-doubled-pairs.js)|Medium|
+955|[Delete Columns to Make Sorted II](./solutions/0955-delete-columns-to-make-sorted-ii.js)|Medium|
 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/0955-delete-columns-to-make-sorted-ii.js

@@ -0,0 +1,47 @@
+/**
+ * 955. Delete Columns to Make Sorted II
+ * https://leetcode.com/problems/delete-columns-to-make-sorted-ii/
+ * Difficulty: Medium
+ *
+ * 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 its elements in lexicographic order (i.e., strs[0] <= strs[1] <= strs[2]
+ * <= ... <= strs[n - 1]). Return the minimum possible value of answer.length.
+ */
+
+/**
+ * @param {string[]} strs
+ * @return {number}
+ */
+var minDeletionSize = function(strs) {
+  let previousSorted = new Array(strs.length).fill('');
+  let result = 0;
+
+  for (let col = 0; col < strs[0].length; col++) {
+    const currentSorted = previousSorted.slice();
+    let isValid = true;
+
+    for (let row = 0; row < strs.length; row++) {
+      currentSorted[row] += strs[row][col];
+      if (row > 0 && currentSorted[row] < currentSorted[row - 1]) {
+        isValid = false;
+        break;
+      }
+    }
+
+    if (isValid) {
+      previousSorted = currentSorted;
+    } else {
+      result++;
+    }
+  }
+
+  return result;
+};