Merge pull request #588 from sumitvajarinkar/master

Left rotate array by D in C++

Author: smv1999

Arrays/left_rotate_by_d_brute_force.cpp

@@ -0,0 +1,52 @@
+//Left rotate an Array by D - brute force
+
+//A left rotation operation on an array of size n
+//shifts each of the array's elements 1 unit to the left.
+//Given an integer d, rotate the array that many steps left and return the result.
+
+//O(nD) O(1)
+//approach 1
+
+//we simply call left rotate by one for D times
+
+//i/p : n=5, 1 2 3 4 5, d=2
+//o/p : 3 4 5 1 2
+
+#include<bits/stdc++.h>
+using namespace std;
+
+void leftrotatebyone(int arr[],int n)
+{
+     //store 0th element in temp
+    int temp=arr[0];
+    for(int i=1;i<n;i++)
+         {
+             //move other element back bye one
+         	arr[i-1]=arr[i];
+		 }
+         //insert temp element at end
+       arr[n-1]=temp;
+}
+
+void leftRotateByD(int arr[],int n,int d)
+{
+	for(int i=0;i<d;i++)
+		leftrotatebyone(arr,n);
+}
+
+int main()
+{
+    int n,d;
+    cout<<"Enter the number of elements\n";
+    cin>>n;
+    int arr[n];
+    cout<<"Enter the elements\n";
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+    cout<<"Enter the number of rotations\n";
+    cin>>d;
+    leftRotateByD(arr,n,d);
+      for(int i=0;i<n;i++)
+        cout<<arr[i]<<" ";
+    return 0;
+}

Arrays/left_rotate_by_d_extra_space.cpp

@@ -0,0 +1,52 @@
+//Left rotate an Array by D - extra space
+
+//A left rotation operation on an array of size n
+//shifts each of the array's elements 1 unit to the left.
+//Given an integer d, rotate the array that many steps left and return the result.
+
+//O(n) O(d)
+//approach 2
+
+//i/p : n=5, 1 2 3 4 5, d=2
+//o/p : 3 4 5 1 2
+
+//1st we copy D elements in arr[D]
+//2nd we move arr[i] one by one back
+//3rd insert arr[D] in arr[i] [n-d+i]
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int leftRotateByD(int arr[],int n,int d)
+{
+    int temp[d];
+    for(int i=0;i<d;i++) //O(D)
+        temp[i]=arr[i];
+
+    for(int i=d;i<n;i++) //O(n-D)
+        arr[i-d]=arr[i]; //i-d will place element at 0th
+
+    for(int i=0;i<d;i++) //O(D)
+        arr[n-d+i]=temp[i]; //5-3+1 =3, 4 ,5
+
+        // so here we have, d + n - d + d = n + d = n (d is negligible)
+}
+int main()
+{
+    int n,d;
+    cout<<"Enter the number of elements\n";
+    cin>>n;
+    int arr[n];
+    cout<<"Enter the elements\n";
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+    cout<<"Enter the number of rotations\n";
+    cin>>d;
+    leftRotateByD(arr,n,d);
+      for(int i=0;i<n;i++)
+        cout<<arr[i]<<" ";
+    return 0;
+}
+
+

Arrays/left_rotate_by_d_no_extra_space.cpp

@@ -0,0 +1,60 @@
+//Left rotate an Array by D - no extra space
+
+//A left rotation operation on an array of size n
+//shifts each of the array's elements 1 unit to the left.
+//Given an integer d, rotate the array that many steps left and return the result.
+
+
+//O(n) O(1)
+//approach 3
+
+//i/p : n=5, 1 2 3 4 5, d=2
+//o/p : 3 4 5 1 2
+
+// 1st we reverse D
+//then reverse D to N
+// and finally we reverse whole the array
+// so we get as desire output
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+void reverse(int arr[],int low,int high)
+{
+    while(low<high)
+    {
+        swap(arr[low],arr[high]); //reverse till low and high overlap
+        low++;
+        high--;
+    }
+}
+
+void leftRotateByD(int arr[],int n,int d)
+{                            //e.g.n=5, 1 2 3 4 5 d=2
+
+        reverse(arr,0,d-1); //call to reverse start to D th element  2 1 3 4 5
+
+        reverse(arr,d,n-1); //call to reverse for D th to N-1 number 2 1 5 4 3
+
+        reverse(arr,0,n-1); //call to reverse for 0th to N-1 number  3 4 5 1 2
+
+}
+
+int main()
+{
+    int n,d;
+    cout<<"Enter the number of elements\n";
+    cin>>n;
+    int arr[n];
+    cout<<"Enter the elements\n";
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+    cout<<"Enter the number of rotations\n";
+    cin>>d;
+    leftRotateByD(arr,n,d);
+      for(int i=0;i<n;i++)
+        cout<<arr[i]<<" ";
+    return 0;
+}
+