Update Readme.md

Sorting/Quick Sort/Readme.md

@@ -1,46 +1,46 @@
 
-QuickSort is a Divide and Conquer algorithm. It picks an element as pivot and partitions the given array around the picked pivot. There are many different versions of quickSort that pick pivot in different ways.
+QuickSort is a Divide and Conquer algorithm. It picks an element as a pivot and partitions the given array around the picked pivot. There are many different versions of quickSort that picks a pivot in different ways.
 
-1. Always pick first element as pivot.
-2. Always pick last element as pivot (implemented below)
-3. Pick a random element as pivot.
-4. Pick median as pivot.
-The key process in quickSort is partition(). Target of partitions is, given an array and an element x of array as pivot, put x at its correct position in sorted array and put all smaller elements (smaller than x) before x, and put all greater elements (greater than x) after x. All this should be done in linear time.
+1. Always pick the first element as the pivot.
+2. Always pick the last element as the pivot (implemented below)
+3. Pick a random element as the pivot.
+4. Pick median as the pivot.
+The key process in quickSort is partition(). The goal of partitions is to, given an array and an element x of an array as the pivot, put x at its correct position in the sorted array and put all the smaller elements (smaller than x) before x, and put all the greater elements (greater than x) after x. All of this should be done in linear time.
 
 Analysis of QuickSort
-Time taken by QuickSort in general can be written as following.
+The time taken by QuickSort, in general, can be written as the following:
 
  T(n) = T(k) + T(n-k-1) + O(n)
-The first two terms are for two recursive calls, the last term is for the partition process. k is the number of elements which are smaller than pivot.
-The time taken by QuickSort depends upon the input array and partition strategy. Following are three cases.
+The first two terms are for two recursive calls, the last term is for the partition process. K is the number of elements that are smaller than the pivot.
+The time taken by QuickSort depends upon the input array and partition strategy. Following are three cases:
 
-Worst Case: The worst case occurs when the partition process always picks greatest or smallest element as pivot. If we consider above partition strategy where last element is always picked as pivot, the worst case would occur when the array is already sorted in increasing or decreasing order. Following is recurrence for worst case.
+Worst Case: The worst case occurs when the partition process always picks the greatest or smallest element as the pivot. If we consider the above partition strategy where the last element is always picked as the pivot, the worst case would occur when the array is already sorted in increasing or decreasing order. The following is a recurrence for the worst case:
 
  T(n) = T(0) + T(n-1) + O(n)
 which is equivalent to  
  T(n) = T(n-1) + O(n)
 The solution of above recurrence is O(n^2).
 
-Best Case: The best case occurs when the partition process always picks the middle element as pivot. Following is recurrence for best case.
+Best Case: The best case occurs when the partition process always picks the middle element as the pivot. The following is a recurrence for the best case.
 
  T(n) = 2T(n/2) + O(n)
-The solution of above recurrence is O(nLogn). It can be solved using case 2 of Master Theorem
+The solution to the above recurrence is O(nLogn). It can be solved using case 2 of the Master Theorem.
 
 Average Case:
-To do average case analysis, we need to consider all possible permutation of array and calculate time taken by every permutation which doesn’t look easy.
-We can get an idea of average case by considering the case when partition puts O(n/9) elements in one set and O(9n/10) elements in other set. Following is recurrence for this case.
+To do the average case analysis, we need to consider all possible permutation of the array and calculate the time taken by every permutation which doesn’t look easy.
+We can get an idea of the average case by considering the case when the partition puts O(n/9) elements in one set and O(9n/10) elements in another set. The following is the recurrence for this case.
 
  T(n) = T(n/9) + T(9n/10) + O(n)
  Solution of above recurrence is also O(nLogn)
 
-Explanation
+Explanation:
 arr[] = {10, 80, 30, 90, 40, 50, 70}
 Indexes:  0   1   2   3   4   5   6 
 
 low = 0, high =  6, pivot = arr[h] = 70
-Initialize index of smaller element, i = -1
+Initialize the index of smaller element, i = -1
 
-Traverse elements from j = low to high-1
+Traverse the elements from j = low to high-1
 j = 0 : Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])
 i = 0 
 arr[] = {10, 80, 30, 90, 40, 50, 70} // No change as i and j 
@@ -63,11 +63,11 @@ j = 5 : Since arr[j] <= pivot, do i++ and swap arr[i] with arr[j]
 i = 3 
 arr[] = {10, 30, 40, 50, 80, 90, 70} // 90 and 50 Swapped 
 
-We come out of loop because j is now equal to high-1.
-Finally we place pivot at correct position by swapping
+We come out of the loop because j is now equal to high-1.
+Finally, we place pivot in the correct position by swapping
 arr[i+1] and arr[high] (or pivot) 
 arr[] = {10, 30, 40, 50, 70, 90, 80} // 80 and 70 Swapped 
 
-Now 70 is at its correct place. All elements smaller than
+Now 70 is in its correct place. All elements smaller than
 70 are before it and all elements greater than 70 are after
 it.