Given two arrays arr1[] and arr2[] both of size N, the task is to shift each element of the cell in Z form such that arr2[0] is in arr1[0], arr1[0] is in arr2[1], arr2[1] is in arr1[1] and so on and arr1[N-1] is in arr2[0].
Examples:
Input: arr1[] = {61, 45, 19, 33, 59, 7, 42, 24, 98, 77}
arr2[] = {86, 52, 10, 36, 22, 5, 98, 91, 13, 6}
Output: arr1[] = {86, 52, 10, 36, 22, 5, 98, 91, 13, 6}
arr2[] = {77, 61, 45, 19, 33, 59, 7, 42, 24, 98}Explanation:
Input: arr1[] = {6, 24, 39, 99, 67}
arr2[] = {12, 84, 9, 13, 5}
Output: arr1[] = {12, 84, 9, 13, 5}
arr2[] = {67, 6, 24, 39, 99}
Approach: Let’s understand how the function zshift works step by step:
- The variable
tis assigned the last element ofarr1, which will be used for swapping later. - The loop starts from the last index (
N-1) and iterates down to the second index (1). - Inside the loop, the current element of
arr2is assigned to the corresponding index inarr1, and the previous element ofarr1is assigned the current element ofarr2. This process effectively shifts the elements in a Z-form between the arrays. - After the loop, the first element of
arr2is assigned to the first element ofarr1, and the original last element ofarr1(stored int) is assigned to the first element ofarr2. This finalizes the shifting process.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to perform Z-form shifting// between two arraysvoid zshift(int* arr1, int* arr2, int N){ // Store the last element of arr1 in t int t = arr1[N - 1]; for (int i = N - 1; i >= 1; i--) { // Shift elements of arr2[i] to arr1[i] arr1[i] = arr2[i]; // Shift previous elements of arr1 to current index // of arr2 arr2[i] = arr1[i - 1]; } // Shift the first element of arr2 to arr1 arr1[0] = arr2[0]; // Shift the original last element of arr1 to arr2 arr2[0] = t;}// Drivers codeint main(){ int arr1[] = { 61, 45, 19, 33, 59, 7, 42, 24, 98, 77 }; int arr2[] = { 86, 52, 10, 36, 22, 5, 98, 91, 13, 6 }; int N = sizeof(arr1) / sizeof(arr1[0]); zshift(arr1, arr2, N); /// Print the elements of arr1 for (int i = 0; i < N; i++) cout << arr1[i] << " "; cout << endl; // Print the elements of arr2 for (int i = 0; i < N; i++) cout << arr2[i] << " "; return 0;} |
Java
// Java program for the above approachimport java.util.Arrays;class GFG { // Function to perform Z-form shifting // between two arrays public static void zshift(int[] arr1, int[] arr2, int N) { // Store the last element of arr1 in t int t = arr1[N - 1]; for (int i = N - 1; i >= 1; i--) { // Shift elements of arr2[i] to arr1[i] arr1[i] = arr2[i]; // Shift previous elements of arr1 to current // index of arr2 arr2[i] = arr1[i - 1]; } // Shift the first element of arr2 to arr1 arr1[0] = arr2[0]; // Shift the original last element of arr1 to arr2 arr2[0] = t; } // Drivers code public static void main(String[] args) { int arr1[] = { 61, 45, 19, 33, 59, 7, 42, 24, 98, 77 }; int arr2[] = { 86, 52, 10, 36, 22, 5, 98, 91, 13, 6 }; int N = arr1.length; zshift(arr1, arr2, N); // Print the elements of arr1 for (int i = 0; i < N; i++) System.out.print(arr1[i] + " "); System.out.println(); // Print the elements of arr2 for (int i = 0; i < N; i++) System.out.print(arr2[i] + " "); System.out.println(); }}// This code is contributed by Abhishek Kumar |
Python3
def zshift(arr1, arr2, N): # Store the last element of arr1 in t t = arr1[N - 1] for i in range(N - 1, 0, -1): # Shift elements of arr2[i] to arr1[i] arr1[i] = arr2[i] # Shift previous elements of arr1 to current index of arr2 arr2[i] = arr1[i - 1] # Shift the first element of arr2 to arr1 arr1[0] = arr2[0] # Shift the original last element of arr1 to arr2 arr2[0] = t# Drivers codearr1 = [61, 45, 19, 33, 59, 7, 42, 24, 98, 77]arr2 = [86, 52, 10, 36, 22, 5, 98, 91, 13, 6]N = len(arr1)zshift(arr1, arr2, N)# Print the elements of arr1for i in range(0, N, 1): print(arr1[i], end=" ")print()# Print the elements of arr2for i in range(0, N, 1): print(arr2[i], end=" ")# This code is contributed by Abhishek Kumar |
C#
using System;public class Program { // Function to perform Z-form shifting between two // arrays static void ZShift(int[] arr1, int[] arr2, int N) { // Store the last element of arr1 in t int t = arr1[N - 1]; for (int i = N - 1; i >= 1; i--) { // Shift elements of arr2[i] to arr1[i] arr1[i] = arr2[i]; // Shift previous elements of arr1 to current // index of arr2 arr2[i] = arr1[i - 1]; } // Shift the first element of arr2 to arr1 arr1[0] = arr2[0]; // Shift the original last element of arr1 to arr2 arr2[0] = t; } public static void Main(string[] args) { int[] arr1 = { 61, 45, 19, 33, 59, 7, 42, 24, 98, 77 }; int[] arr2 = { 86, 52, 10, 36, 22, 5, 98, 91, 13, 6 }; int N = arr1.Length; ZShift(arr1, arr2, N); // Print the elements of arr1 foreach(int num in arr1) { Console.Write(num + " "); } Console.WriteLine(); // Print the elements of arr2 foreach(int num in arr2) { Console.Write(num + " "); } }} |
Javascript
function zShift(arr1, arr2, N) { // Store the last element of arr1 in t const t = arr1[N - 1]; for (let i = N - 1; i >= 1; i--) { // Shift elements of arr2[i] to arr1[i] arr1[i] = arr2[i]; // Shift previous elements of arr1 to the current index of arr2 arr2[i] = arr1[i - 1]; } // Shift the first element of arr2 to arr1 arr1[0] = arr2[0]; // Shift the original last element of arr1 to arr2 arr2[0] = t;}const arr1 = [61, 45, 19, 33, 59, 7, 42, 24, 98, 77];const arr2 = [86, 52, 10, 36, 22, 5, 98, 91, 13, 6];const N = arr1.length;zShift(arr1, arr2, N);// Print the elements of arr1console.log(arr1.join(" "));// Print the elements of arr2console.log(arr2.join(" ")); |
Output
86 52 10 36 22 5 98 91 13 6 77 61 45 19 33 59 7 42 24 98
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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