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📜  检查两个排序数组是否可以合并以形成一个排序数组,其中没有来自同一数组的相邻对

📅  最后修改于: 2021-09-04 09:35:20             🧑  作者: Mango

给定两个大小为N 的排序数组A[]B[] ,任务是检查是否可以将两个给定的排序数组合并为一个新的排序数组,使得没有两个连续的元素来自同一个数组。

例子:

处理方法:按照以下步骤解决问题:

下面是上述方法的实现:

C++
// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
bool checkIfPossibleMerge(int A[], int B[], int N)
{
    // Stores index of
    // the array A[]
    int i = 0;
 
    // Stores index of
    // the array  B[]
    int j = 0;
 
    // Check if the previous element are from
    // the array A[] or from the array B[]
    int prev = -1;
 
    // Check if it is possible to merge the two
    // given sorted arrays with given conditions
    int flag = 1;
 
    // Traverse both the arrays
    while (i < N && j < N) {
 
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0) {
 
            // Update prev
            prev = 0;
 
            // Update i
            i++;
        }
 
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1) {
 
            // Update prev
            prev = 1;
 
            // Update j
            j++;
        }
 
        // If A[i] equal to B[j]
        else if (A[i] == B[j]) {
 
            // If previous element
            // are not from B[]
            if (prev != 1) {
 
                // Update prev
                prev = 1;
 
                // Update j
                j++;
            }
 
            // If previous element is
            // not from A[]
            else {
 
                // Update prev
                prev = 0;
 
                // Update i
                i++;
            }
        }
 
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else {
 
            // Update flag
            flag = 0;
            break;
        }
    }
 
    return flag;
}
 
// Driver Code
int main()
{
    int A[3] = { 3, 5, 8 };
    int B[3] = { 2, 4, 6 };
    int N = sizeof(A) / sizeof(A[0]);
 
    if (checkIfPossibleMerge(A, B, N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}


Java
// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
static boolean checkIfPossibleMerge(int[] A, int[] B,
                                    int N)
{
     
    // Stores index of
    // the array A[]
    int i = 0;
 
    // Stores index of
    // the array  B[]
    int j = 0;
 
    // Check if the previous element are from
    // the array A[] or from the array B[]
    int prev = -1;
 
    // Check if it is possible to merge the two
    // given sorted arrays with given conditions
    boolean flag = true;
 
    // Traverse both the arrays
    while (i < N && j < N)
    {
         
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0)
        {
             
            // Update prev
            prev = 0;
 
            // Update i
            i++;
        }
 
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1)
        {
             
            // Update prev
            prev = 1;
 
            // Update j
            j++;
        }
 
        // If A[i] equal to B[j]
        else if (A[i] == B[j])
        {
             
            // If previous element
            // are not from B[]
            if (prev != 1)
            {
                 
                // Update prev
                prev = 1;
 
                // Update j
                j++;
            }
 
            // If previous element is
            // not from A[]
            else
            {
                 
                // Update prev
                prev = 0;
 
                // Update i
                i++;
            }
        }
 
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else
        {
             
            // Update flag
            flag = false;
            break;
        }
    }
    return flag;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] A = { 3, 5, 8 };
    int[] B = { 2, 4, 6 };
    int N = A.length;
 
    if (checkIfPossibleMerge(A, B, N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by akhilsaini


Python3
# Python3 program to implement
# the above approach
 
# Function to check if it is possible
# to merge the two given arrays with
# given conditions
def checkIfPossibleMerge(A, B, N):
     
    # Stores index of
    # the array A[]
    i = 0
 
    # Stores index of
    # the array  B[]
    j = 0
 
    # Check if the previous element
    # are from the array A[] or from
    # the array B[]
    prev = -1
 
    # Check if it is possible to merge
    # the two given sorted arrays with
    # given conditions
    flag = 1
 
    # Traverse both the arrays
    while (i < N and j < N):
 
        # If A[i] is less than B[j] and
        # previous element are not from A[]
        if (A[i] < B[j] and prev != 0):
 
            # Update prev
            prev = 0
 
            # Update i
            i += 1
 
        # If B[j] is less than A[i] and
        # previous element are not from B[]
        elif (B[j] < A[i] and prev != 1):
 
            # Update prev
            prev = 1
 
            # Update j
            j += 1
 
        # If A[i] equal to B[j]
        elif (A[i] == B[j]):
 
            # If previous element
            # are not from B[]
            if (prev != 1):
                 
                # Update prev
                prev = 1
 
                # Update j
                j += 1
 
            # If previous element is
            # not from A[]
            else:
 
                # Update prev
                prev = 0
 
                # Update i
                i += 1
 
        # If it is not possible to merge two
        # sorted arrays with given conditions
        else:
 
            # Update flag
            flag = 0
            break
 
    return flag
 
# Driver Code
if __name__ == '__main__':
 
    A = [ 3, 5, 8 ]
    B = [ 2, 4, 6 ]
    N = len(A)
 
    if (checkIfPossibleMerge(A, B, N)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by akhilsaini


C#
// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
static bool checkIfPossibleMerge(int[] A, int[] B,
                                 int N)
{
     
    // Stores index of
    // the array A[]
    int i = 0;
 
    // Stores index of
    // the array  B[]
    int j = 0;
 
    // Check if the previous element are
    // from the array A[] or from the
    // array B[]
    int prev = -1;
 
    // Check if it is possible to merge
    // the two given sorted arrays with
    // given conditions
    bool flag = true;
 
    // Traverse both the arrays
    while (i < N && j < N)
    {
         
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0)
        {
             
            // Update prev
            prev = 0;
 
            // Update i
            i++;
        }
 
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1)
        {
             
            // Update prev
            prev = 1;
 
            // Update j
            j++;
        }
 
        // If A[i] equal to B[j]
        else if (A[i] == B[j])
        {
             
            // If previous element
            // are not from B[]
            if (prev != 1)
            {
                 
                // Update prev
                prev = 1;
 
                // Update j
                j++;
            }
 
            // If previous element is
            // not from A[]
            else
            {
                 
                // Update prev
                prev = 0;
 
                // Update i
                i++;
            }
        }
 
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else
        {
             
            // Update flag
            flag = false;
            break;
        }
    }
    return flag;
}
 
// Driver Code
public static void Main()
{
    int[] A = { 3, 5, 8 };
    int[] B = { 2, 4, 6 };
    int N = A.Length;
 
    if (checkIfPossibleMerge(A, B, N))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by akhilsaini


Javascript


输出:
Yes

时间复杂度: O(N)
辅助空间: O(1)

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