Commit solution with explanation

Author: SartHak-0-Sach

212-30th_July-Min_deletions/approach-explanation.txt

@@ -0,0 +1,12 @@
+Intuition
+Basically, the problem asks you to have all a before b in string s by deleting minimum number of characters from string s.
+The intuition is simple enough. We just need to store how many b come before a at a specific position i in string s.
+This will tell us how many a we have to delete which come after b or how many b we have to delete which come before a.
+In the end, we will count the minimum number of deletions we have to make in order to achieve the string with all a before b.
+Approach
+Create a prefix_b array to store all b from left to right and create a suffix_a array to store all a from right to left.
+Now, at every index i check how many prefix b and suffix a we have to delete by adding prefix_b[i] and suffix_a[i].
+Keep the track of minimum deletions using a variable mini which will be our final answer to return.
+Complexity
+Time complexity: O(n)
+Space complexity: O(n)

212-30th_July-Min_deletions/code.cpp

@@ -0,0 +1,43 @@
+class Solution
+{
+public:
+    int minimumDeletions(string s)
+    {
+        vector<int> prefix_b(s.size() + 2, 0), suffix_a(s.size() + 2, 0);
+        prefix_b[0] = 0;
+        suffix_a[suffix_a.size() - 1] = 0;
+
+        for (int i = 1; i < prefix_b.size() - 1; i++)
+        {
+            if (s[i - 1] == 'b')
+            {
+                prefix_b[i + 1] = prefix_b[i] + 1;
+            }
+            else
+            {
+                prefix_b[i + 1] = prefix_b[i] + 0;
+            }
+        }
+
+        for (int i = s.size(); i >= 1; i--)
+        {
+            if (s[i - 1] == 'a')
+            {
+                suffix_a[i - 1] = suffix_a[i] + 1;
+            }
+            else
+            {
+                suffix_a[i - 1] = suffix_a[i] + 0;
+            }
+        }
+
+        int mini = INT_MAX;
+
+        for (int i = 0; i < s.size() + 2; i++)
+        {
+            mini = min(prefix_b[i] + suffix_a[i], mini);
+        }
+
+        return mini;
+    }
+};