Add solution #1177

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,167 LeetCode solutions in JavaScript
+# 1,168 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -915,6 +915,7 @@
 1171|[Remove Zero Sum Consecutive Nodes from Linked List](./solutions/1171-remove-zero-sum-consecutive-nodes-from-linked-list.js)|Medium|
 1172|[Dinner Plate Stacks](./solutions/1172-dinner-plate-stacks.js)|Hard|
 1175|[Prime Arrangements](./solutions/1175-prime-arrangements.js)|Easy|
+1177|[Can Make Palindrome from Substring](./solutions/1177-can-make-palindrome-from-substring.js)|Medium|
 1189|[Maximum Number of Balloons](./solutions/1189-maximum-number-of-balloons.js)|Easy|
 1200|[Minimum Absolute Difference](./solutions/1200-minimum-absolute-difference.js)|Easy|
 1206|[Design Skiplist](./solutions/1206-design-skiplist.js)|Hard|

solutions/1177-can-make-palindrome-from-substring.js

@@ -0,0 +1,48 @@
+/**
+ * 1177. Can Make Palindrome from Substring
+ * https://leetcode.com/problems/can-make-palindrome-from-substring/
+ * Difficulty: Medium
+ *
+ * You are given a string s and array queries where queries[i] = [lefti, righti, ki]. We may
+ * rearrange the substring s[lefti...righti] for each query and then choose up to ki of them
+ * to replace with any lowercase English letter.
+ *
+ * If the substring is possible to be a palindrome string after the operations above, the
+ * result of the query is true. Otherwise, the result is false.
+ *
+ * Return a boolean array answer where answer[i] is the result of the ith query queries[i].
+ *
+ * Note that each letter is counted individually for replacement, so if, for example
+ * s[lefti...righti] = "aaa", and ki = 2, we can only replace two of the letters. Also, note
+ * that no query modifies the initial string s.
+ */
+
+/**
+ * @param {string} s
+ * @param {number[][]} queries
+ * @return {boolean[]}
+ */
+var canMakePaliQueries = function(s, queries) {
+  const prefixCounts = new Array(s.length + 1).fill().map(() => new Array(26).fill(0));
+
+  for (let i = 0; i < s.length; i++) {
+    prefixCounts[i + 1] = [...prefixCounts[i]];
+    prefixCounts[i + 1][s.charCodeAt(i) - 97]++;
+  }
+
+  const getOddCount = (left, right) => {
+    let odds = 0;
+    for (let j = 0; j < 26; j++) {
+      const count = prefixCounts[right + 1][j] - prefixCounts[left][j];
+      odds += count % 2;
+    }
+    return odds;
+  };
+
+  const result = queries.map(([left, right, k]) => {
+    const oddCount = getOddCount(left, right);
+    return oddCount <= 2 * k + 1;
+  });
+
+  return result;
+};