Smallest subarray sorting which sorts array

Author: kumar91gopi

Diff for: arrays/UnsortedSubArray.rb

@@ -0,0 +1,70 @@
+#Given an unsorted array,find the minimum length of subarraay,sorting which sorts the whole array.
+#Time-complexity: O(n),Auxiliary-space:O(1)
+=begin
+Algorithm: Scan left to right and find first index where next element is less than current,let it be l
+(if l==a.length-1 it is already sorted),similarly scan right to left and search first index
+where previous element is larger than current,let it be r.Now search for min and max in l to r,
+now from 0 to l-1 find first element which is greater than min(let it be i) and 
+from r+1 to a.length-1 find first element which is less than max.(let it be j)
+set l= i and r=j
+minimum length= r-l+1
+start=l
+end=r
+=end
+def unsorted_sub(a)
+    n=a.length
+    l=0
+    r=n-1
+    for i in 0...n-1
+        if a[i]>a[i+1]
+            l=i
+            break
+        else
+          l+=1
+        end
+    end
+    return "The complete array is sorted" if l==n-1
+    
+    for i in (n-1).downto(1)
+        if a[i]<a[i-1]
+            r=i
+            break
+        else
+          	r-=1
+        end
+    end
+    
+    min_i,max_i=find_minmax(a,l,r)
+    
+    for i in 0...l
+       if a[i]>a[min_i]
+           l=i
+           break
+       end
+   end
+   
+    for i in (n-1).downto(r+1)
+        if a[i]<a[max_i]
+            r=i
+            break
+        end
+    end
+  return "Length: #{r-l+1}, Starting index: #{l}, Ending index: #{r}"
+end
+
+def find_minmax(a,lo,hi)
+    min=max=lo
+    for i in (lo+1)..hi
+        if a[i]>a[max]
+            max=i
+        elsif a[i]<min
+            min=i
+        else
+            next
+        end
+    end
+    return min,max
+end
+                
+unsorted_sub([15, 16, 21, 30, 25, 41, 28, 39, 58])   # => Length: 5, Starting index: 3, Ending index: 7  
+unsorted_sub([1,2,3,4,5,6])   # => The complete array is sorted