[6kyu][algorithms][my-solution] Find The Parity Outlier solution

You are given an array (which will have a length of at least 3, but
could be very large) containing integers. The array is either entirely
comprised of odd integers or entirely comprised of even integers except
for a single integer N. Write a method that takes the array as an
argument and returns this "outlier" N.

lib/findOUtlier/findOUtlier.ts

@@ -10,10 +10,47 @@
 //
 
 
+// export function findOUtlier(integers: number[]): number {
+
+//   const isEvenNum = integers.filter(num => num % 2 == 0);
+//   const isOddNum = integers.filter(num => num % 2 != 0);
+
+//   return isEvenNum.length == 1 ? Number(isEvenNum) : Number(isOddNum);
+// };
+
 export function findOUtlier(integers: number[]): number {
+  const evenNum: number[] = [];
+  const oddNum: number[] = [];
+  const middle = Math.trunc(integers.length / 2);
+  let end = integers.length - 1;
+  let start = 0;
 
-  const isEvenNum = integers.filter(num => num % 2 == 0);
-  const isOddNum = integers.filter(num => num % 2 != 0);
+  while (start < middle || end >= middle) {
 
-  return isEvenNum.length == 1 ? Number(isEvenNum) : Number(isOddNum);
+    if (start !== middle) {
+      if (!(integers[start] % 2)) {
+        evenNum.push(integers[start])
+      } else {
+        oddNum.push(integers[start])
+      }
+    }
+
+
+    if (!(integers[end] % 2)) {
+      evenNum.push(integers[end])
+    } else {
+      oddNum.push(integers[end])
+    }
+
+    if (evenNum.length === 1 && oddNum.length > 1 || oddNum.length === 1 && evenNum.length > 1) {
+      break;
+    }
+
+    start++;
+    end--;
+  }
+
+
+  return evenNum.length === 1 ? evenNum[0] : oddNum[0];
 };
+