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@ -0,0 +1,88 @@
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// ====================================================
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// Data-Structures-with-Go Copyright(C) 2017 Furkan Türkal
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// This program comes with ABSOLUTELY NO WARRANTY; This is free software,
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// and you are welcome to redistribute it under certain conditions; See
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// file LICENSE, which is part of this source code package, for details.
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// ====================================================
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/*** INFO
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-Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements
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-1 2 3 4 5 6 7
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-Rotation of the above array by 2 will make array
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-3 4 5 6 7 1 2
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-METHOD 1 (Use temp array) :
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Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
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1) Store d elements in a temp array
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temp[] = [1, 2]
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2) Shift rest of the arr[]
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arr[] = [3, 4, 5, 6, 7, 6, 7]
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3) Store back the d elements
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arr[] = [3, 4, 5, 6, 7, 1, 2]
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-Time complexity O(n)
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-Auxiliary Space: O(d)
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-METHOD 2 (Rotate one by one) :
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leftRotate(arr[], d, n)
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start
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For i = 0 to i < d
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Left rotate all elements of arr[] by one
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end
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To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]
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Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
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Rotate arr[] by one 2 times
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We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.
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-Time complexity: O(n*d)
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-Auxiliary Space: O(1)
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***/
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package main
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import "fmt"
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func leftRotate(arr []int, d int, n int){
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for i := 0; i < d; i++ {
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leftRotateByOne(arr, n)
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}
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}
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func leftRotateByOne(arr []int, n int){
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var i, temp int
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temp = arr[0]
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for i = 0; i < n - 1; i++ {
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arr[i] = arr[i + 1]
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}
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arr[i] = temp
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}
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func printArray(arr []int, size int){
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if(len(arr) < size){
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fmt.Println("[ArrayRotation::printArray()] -> Index out of range. Max : ", len(arr))
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return;
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}
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for i := 0; i < size; i++ {
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fmt.Print(arr[i])
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}
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}
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func main() {
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arr := []int{1, 2, 3, 4, 5, 6, 7}
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fmt.Println("-INPUT-")
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printArray(arr, len(arr) - 1)
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leftRotate(arr, 2 , 7)
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fmt.Println()
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fmt.Println("-OUTPUT-")
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printArray(arr, 7)
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}
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