给定由N个整数和整数X组成的数组arr [] ,任务是将数组拆分为两个子序列,以使两个数组中总和等于X的对的数量最小。
例子:
Input: arr[] = {1, 2, 3, 4, 5, 6}, X = 7
Output:
The First Array is – 1 2 3
The Second Array is – 4 5 6
Explanation:
The possible 3 pairs in the first array are {(1, 2), (2, 3), (1, 3)}. None of these pairs have sum equal to X (= 7).
The possible 3 pairs in the second array are {(4, 5), (5, 6), (4, 6)}. None of these pairs have sum equal to X (= 7).
Input: arr[] = {3, 3, 3}, X = 6
Output:
The First Array is – 3
The Second Array is – 3 3
方法:请按照以下步骤解决问题:
- 创建两个数组来存储两个拆分的子序列。
- 遍历数组并执行以下操作:
- 如果arr [i] * 2
其插入第一个数组。 - 由于第一个数组当前包含的所有数字均小于X / 2 ,因此,在当前数组中,没有一个对的总和等于X。
- 如果arr [i] * 2> X:将其插入第二个数组。
- 由于第二个数组当前包含所有大于X / 2的数字,因此当前在该数组中没有对的总和等于X。
- 如果arr [i] * 2
分别将元素插入第一个和第二个数组以获取最小对。
- 如果arr [i] * 2
- 最后,在遍历数组后,打印两个数组。
下面是上述方法的实现:
C++
// C++ program for above approach
#include
using namespace std;
// Function to split the array
// into two subsequences
void solve(int arr[], int N, int X)
{
// Stores the two subsequences
vector A, B;
// Flag to set/reset to split
// arrays elements alternately
// into two arrays
int c = 0;
// Traverse the given array
for (int i = 0; i < N; i++) {
// If 2 * arr[i] is less than X
if ((2 * arr[i]) < X) {
// Push element into
// the first array
A.push_back(arr[i]);
}
// If 2 * arr[i] is greater than X
else if ((2 * arr[i]) > X) {
// Push element into
// the second array
B.push_back(arr[i]);
}
// If 2 * arr[i] is equal to X
else {
// Alternatively place the
// elements into the two arrays
if (c % 2 == 0) {
A.push_back(arr[i]);
}
else {
B.push_back(arr[i]);
}
c++;
}
}
// Print both the arrays
cout << "The First Array is - ";
for (int i = 0; i < A.size(); i++) {
cout << A[i] << " ";
}
cout << endl;
cout << "The Second Array is - ";
for (int i = 0; i < B.size(); i++) {
cout << B[i] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 1, 5, 4, 3,
6, 2, 4, 3 };
int X = 7;
// Size of the array
int N = sizeof(arr)
/ sizeof(arr[0]);
// Function Call
solve(arr, N, X);
}
Java
// Java program for the
// above approach
import java.util.*;
class GFG{
// Function to split the array
// into two subsequences
static void solve(int arr[],
int N, int X)
{
// Stores the two subsequences
Vector A =
new Vector(),
B = new Vector();
// Flag to set/reset to split
// arrays elements alternately
// into two arrays
int c = 0;
// Traverse the given array
for (int i = 0; i < N; i++)
{
// If 2 * arr[i] is
// less than X
if ((2 * arr[i]) < X)
{
// Push element into
// the first array
A.add(arr[i]);
}
// If 2 * arr[i] is greater
// than X
else if ((2 * arr[i]) > X)
{
// Push element into
// the second array
B.add(arr[i]);
}
// If 2 * arr[i] is
// equal to X
else
{
// Alternatively place the
// elements into the two arrays
if (c % 2 == 0)
{
A.add(arr[i]);
}
else
{
B.add(arr[i]);
}
c++;
}
}
// Print both the arrays
System.out.print("The First Array is - ");
for (int i = 0; i < A.size(); i++)
{
System.out.print(A.get(i) + " ");
}
System.out.println();
System.out.print("The Second Array is - ");
for (int i = 0; i < B.size(); i++)
{
System.out.print(B.get(i) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {1, 5, 4, 3,
6, 2, 4, 3};
int X = 7;
// Size of the array
int N = arr.length;
// Function Call
solve(arr, N, X);
}
}
// This code is contributed by shikhasingrajput
Python3
# Python3 program for above approach
# Function to split the array
# into two subsequences
def solve(arr, N, X):
# Stores the two subsequences
A = []
B = []
# Flag to set/reset to split
# arrays elements alternately
# into two arrays
c = 0
# Traverse the given array
for i in range(N):
# If 2 * arr[i] is less than X
if ((2 * arr[i]) < X):
# Push element into
# the first array
A.append(arr[i])
# If 2 * arr[i] is greater than X
elif ((2 * arr[i]) > X):
# Push element into
# the second array
B.append(arr[i])
# If 2 * arr[i] is equal to X
else:
# Alternatively place the
# elements into the two arrays
if (c % 2 == 0):
A.append(arr[i])
else:
B.append(arr[i])
c += 1
# Print both the arrays
print("The First Array is - ", end = " ")
for i in range(len(A)):
print(A[i], end = " ")
print()
print("The Second Array is - ", end = " ")
for i in range(len(B)):
print(B[i], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [ 1, 5, 4, 3, 6, 2, 4, 3 ]
X = 7
# Size of the array
N = len(arr)
# Function Call
solve(arr, N, X)
# This code is contributed by mohit kumar 29
C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to split the array
// into two subsequences
static void solve(int []arr,
int N, int X)
{
// Stores the two subsequences
List A =
new List(),
B = new List();
// Flag to set/reset to
// split arrays elements
// alternately into two
// arrays
int c = 0;
// Traverse the given array
for (int i = 0; i < N; i++)
{
// If 2 * arr[i] is
// less than X
if ((2 * arr[i]) < X)
{
// Push element into
// the first array
A.Add(arr[i]);
}
// If 2 * arr[i] is greater
// than X
else if ((2 * arr[i]) > X)
{
// Push element into
// the second array
B.Add(arr[i]);
}
// If 2 * arr[i] is
// equal to X
else
{
// Alternatively place the
// elements into the two arrays
if (c % 2 == 0)
{
A.Add(arr[i]);
}
else
{
B.Add(arr[i]);
}
c++;
}
}
// Print both the arrays
Console.Write("The First Array is - ");
for (int i = 0; i < A.Count; i++)
{
Console.Write(A[i] + " ");
}
Console.WriteLine();
Console.Write("The Second Array is - ");
for (int i = 0; i < B.Count; i++)
{
Console.Write(B[i] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {1, 5, 4, 3,
6, 2, 4, 3};
int X = 7;
// Size of the array
int N = arr.Length;
// Function Call
solve(arr, N, X);
}
}
// This code is contributed by gauravrajput1
输出
The First Array is - 1 3 2 3
The Second Array is - 5 4 6 4
时间复杂度: O(N)
辅助空间: O(N)