要添加到数组 A 中的 N-1 个元素以使数组 B 的元素出现在 A 中的最小可能整数

给定两个长度为 N 的数组A[]和长度为N -1的B[] ，任务是找到最小的正整数X添加到A[]的每个元素中，除了一个元素，使得数组的所有元素B[]存在于数组A[]中。假设找到X总是可能的。

例子：

Input: A[] = {1, 4, 3, 8},  B[] = {15, 8, 11}
Output: 7
Explanation: Adding 7 to every elements of the array A except 3 will give all the elements of array B
 

Input: A[] = {4,  8}, B[] = {10}
Output: 2
Explanation: Adding 2 to 8 gives 10.

方法：这个想法是使用贪婪方法。

• 对数组A[]和B[]进行排序
• 通过观察可以看出， X的值可以是B[0] – A[0]或B[0] – A[1] 。
• 因此，不考虑A[0]或A[1] ，并将X添加到其余元素中。
• 通过遍历数组检查两个值是否有效，如果发现两个X值都有效，则选择最小值并打印。

以下是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the smallest positive integer
// X such that X is added to every element of A
// except one element to give array B
void findVal(int A[], int B[], int N)
{
    // Stores the unique elements of array A
    unordered_set s;
 
    for (int i = 0; i < N; i++) {
        // Insert A[i] into the set s
        s.insert(A[i]);
    }
 
    // Sort array A[]
    sort(A, A + N);
 
    // Sort array B[]
    sort(B, B + N - 1);
 
    // Assume X value as B[0] - A[1]
    int X = B[0] - A[1];
 
    // Check if X value assumed is negative or 0
    if (X <= 0)
 
        // Update the X value to B[0] - A[0]
        X = B[0] - A[0];
 
    else {
 
        for (int i = 0; i < N - 1; i++) {
 
            // If assumed value is wrong
            if (s.count(B[i] - X) == 0) {
 
                // Update X value
                X = B[0] - A[0];
                break;
            }
        }
    }
 
    cout << X << endl;
}
 
// Driver Code
int main()
{
 
    int A[] = { 1, 4, 3, 8 };
    int B[] = { 15, 8, 11 };
    int N = sizeof(A) / sizeof(A[0]);
 
    findVal(A, B, N);
 
    return 0;
}

Java

// Java implementation for the above approach
import java.util.*;
 
public class GFG
{
   
// Function to find the smallest positive integer
// X such that X is added to every element of A
// except one element to give array B
static void findVal(int []A, int []B, int N)
{
    // Stores the unique elements of array A
    HashSet s = new HashSet();
 
    for (int i = 0; i < N; i++) {
        // Insert A[i] into the set s
        s.add(A[i]);
    }
 
    // Sort array A[]
    Arrays.sort(A);
 
    // Sort array B[]
    Arrays.sort(B);
 
    // Assume X value as B[0] - A[1]
    int X = B[0] - A[1];
 
    // Check if X value assumed is negative or 0
    if (X <= 0)
 
        // Update the X value to B[0] - A[0]
        X = B[0] - A[0];
 
    else {
 
        for (int i = 0; i < N - 1; i++) {
 
            // If assumed value is wrong
            if (s.contains(B[i] - X) == false) {
 
                // Update X value
                X = B[0] - A[0];
                break;
            }
        }
    }
     
    System.out.println(X);
}
 
// Driver Code
public static void main(String args[])
{
    int []A = { 1, 4, 3, 8 };
    int []B = { 15, 8, 11 };
    int N = A.length;
     
    findVal(A, B, N);
}
}
 
// This code is contributed by Samim Hossain Mondal

Python3

# Python 3 program for the above approach
 
# Function to find the smallest positive integer
# X such that X is added to every element of A
# except one element to give array B
def findVal(A, B, N):
   
    # Stores the unique elements of array A
    s = set()
 
    for i in range(N):
        # Insert A[i] into the set s
        s.add(A[i])
 
    # Sort array A[]
    A.sort()
 
    # Sort array B[]
    B.sort()
 
    # Assume X value as B[0] - A[1]
    X = B[0] - A[1]
 
    # Check if X value assumed is negative or 0
    if (X <= 0):
 
        # Update the X value to B[0] - A[0]
        X = B[0] - A[0]
 
    else:
 
        for i in range(N - 1):
 
            # If assumed value is wrong
            if (B[i] - X not in s):
 
                # Update X value
                X = B[0] - A[0]
                break
 
    print(X)
 
# Driver Code
if __name__ == '__main__':
    A = [1, 4, 3, 8]
    B = [15, 8, 11]
    N = len(A)
 
    findVal(A, B, N)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# implementation for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the smallest positive integer
// X such that X is added to every element of A
// except one element to give array B
static void findVal(int []A, int []B, int N)
{
    // Stores the unique elements of array A
    HashSet s = new HashSet();
 
    for (int i = 0; i < N; i++) {
        // Insert A[i] into the set s
        s.Add(A[i]);
    }
 
    // Sort array A[]
    Array.Sort(A);
 
    // Sort array B[]
    Array.Sort(B);
 
    // Assume X value as B[0] - A[1]
    int X = B[0] - A[1];
 
    // Check if X value assumed is negative or 0
    if (X <= 0)
 
        // Update the X value to B[0] - A[0]
        X = B[0] - A[0];
 
    else {
 
        for (int i = 0; i < N - 1; i++) {
 
            // If assumed value is wrong
            if (s.Contains(B[i] - X) == false) {
 
                // Update X value
                X = B[0] - A[0];
                break;
            }
        }
    }
     
    Console.Write(X);
}
 
// Driver Code
public static void Main()
{
    int []A = { 1, 4, 3, 8 };
    int []B = { 15, 8, 11 };
    int N = A.Length;
     
    findVal(A, B, N);
}
}
// This code is contributed by Samim Hossain Mondal

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find the smallest positive integer
        // X such that X is added to every element of A
        // except one element to give array B
        function findVal(A, B, N)
        {
         
            // Stores the unique elements of array A
            let s = new Set();
 
            for (let i = 0; i < N; i++)
            {
             
                // Insert A[i] into the set s
                s.add(A[i]);
            }
 
            // Sort array A[]
            A.sort(function (a, b) { return a - b });
 
            // Sort array B[]
            B.sort(function (a, b) { return a - b })
 
            // Assume X value as B[0] - A[1]
            let X = B[0] - A[1];
 
            // Check if X value assumed is negative or 0
            if (X <= 0)
 
                // Update the X value to B[0] - A[0]
                X = B[0] - A[0];
 
            else {
 
                for (let i = 0; i < N - 1; i++) {
 
                    // If assumed value is wrong
                    if (s.has(B[i] - X) == false) {
 
                        // Update X value
                        X = B[0] - A[0];
                        break;
                    }
                }
            }
 
            document.write(X);
        }
 
        // Driver Code
        let A = [1, 4, 3, 8];
        let B = [15, 8, 11];
        let N = A.length;
 
        findVal(A, B, N);
 
     // This code is contributed by Potta Lokesh
 
    

输出

7

时间复杂度： O(N log N)
辅助空间： O(N)