diff --git a/Sorts/SelectionSort.js b/Sorts/SelectionSort.js
index 9170e6f..fdd6266 100644
--- a/Sorts/SelectionSort.js
+++ b/Sorts/SelectionSort.js
@@ -1,24 +1,30 @@
-export default SelectionSort = function (arr) {
-  let n = arr.length;
+/*
 
-  for (let i = 0; i < n; i++) {
-    // Set the minimum to this position.
-    min_index = i;
+Selection sort is fairly simple. Assume the left part of your array is
+sorted and right part is unsorted. Inititally the sorted(left) part is
+empty. Now select the smallest element from the unsorted(right) part and
+swap it with the first element of the unsorted(right) part. Now this element
+is sorted, move to the next iteration and repeat without touch the sorted(left)
+part.
 
-    // Looking for a smaller number in the unsorted subarray.
-    for (let j = i + 1; j < n; j++) {
-      // If find one, set the minimum to this position.
-      if (arr[j] < arr[min_index]) {
-        min_index = j;
+*/
+function selectionSort(arr) {
+  for (let i = 0; i < arr.length; i++) {
+    let firstOfUnsorted = arr[i];
+
+    // find the index of the smallest element in the unsorted part
+    let minIndex = i;
+    for (let j = i + 1; j < arr.length; j++) {
+      if (arr[minIndex] > arr[j]) {
+        minIndex = j;
       }
     }
 
-    // If the current minimum index is not the minimum index you started with, swap the elements.
-    if (min_index != i) {
-      let temp = arr[i];
-      arr[i] = arr[min_index];
-      arr[min_index] = temp;
-    }
+    // swapping the minimum with first element of unsorted part
+    arr[i] = arr[minIndex];
+    arr[minIndex] = firstOfUnsorted;
   }
   return arr;
-};
+}
+
+// This has the time complexity of O(n^2)