group anagrams solution

Author: robinyoon-dev

group-anagrams/robinyoon-dev.js

@@ -0,0 +1,45 @@
+/**
+ * @param {string[]} strs
+ * @return {string[][]}
+ */
+
+// 문제는 풀었으나 시간 복잡도 측면에서 효율이 너무 떨어지는 풀이 방법....
+var groupAnagrams = function (strs) {
+    let outputArr = [];
+    let countArr = [];
+
+    const A_ASCII = 'a'.charCodeAt(0);
+    const Z_ASCII = 'z'.charCodeAt(0);
+
+    let charCounts = Z_ASCII - A_ASCII + 1;
+    let charCountArr = new Array(charCounts).fill(0); //인덱스가 알파벳을 나타냄.
+
+    for (str of strs) {
+        let strCountString = getStrCountString(str);
+
+        let hasSameCountIndex = countArr.findIndex((item) => item === strCountString);
+
+        if (hasSameCountIndex !== -1) {
+            outputArr[hasSameCountIndex].push(str);
+        } else {
+            countArr.push(strCountString);
+
+            outputArr.push([str]);
+        }
+    }
+
+    return outputArr;
+
+    function getStrCountString(str) {
+        let tempArr = [...charCountArr];
+
+        for (char of str) {
+            let charAscii = char.charCodeAt(0);
+            let charIndex = charAscii - A_ASCII;
+            tempArr[charIndex] += 1;
+        }
+        return tempArr.join(',');
+    }
+};
+
+