给定一个整数N ,将前N 个自然数分成两组,使得它们和之间的绝对差最小。任务是打印可以获得的最小绝对差值。
例子:
Input: N = 5
Output: 1
Explanation:
Split the first N (= 5) natural numbers into sets {1, 2, 5} (sum = 8) and {3, 4} (sum = 7).
Therefore, the required output is 1.
Input: N = 6
Output: 1
朴素的方法:这个问题可以使用贪婪技术来解决。请按照以下步骤解决问题:
- 初始化两个变量,比如sumSet1和sumSet2来存储两个集合中元素的总和。
- 从N到1遍历前N 个自然数。对于每个数字,检查当前 set1 中元素的总和是否小于或等于 set2 中元素的总和。如果发现为真,则将当前遍历的数字添加到set1并更新sumSet1 。
- 否则,将当前自然数的值添加到set2并更新sumSet2 。
- 最后,打印abs(sumSet1 – sumSet2)作为所需答案。
下面是上述方法的实现:
C++14
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
int minAbsDiff(int N)
{
// Stores the sum of
// elements of set1
int sumSet1 = 0;
// Stores the sum of
// elements of set2
int sumSet2 = 0;
// Traverse first N
// natural numbers
for (int i = N; i > 0; i--) {
// Check if sum of elements of
// set1 is less than or equal
// to sum of elements of set2
if (sumSet1 <= sumSet2) {
sumSet1 += i;
}
else {
sumSet2 += i;
}
}
return abs(sumSet1 - sumSet2);
}
// Driver Code
int main()
{
int N = 6;
cout << minAbsDiff(N);
}
Java
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
static int minAbsDiff(int N)
{
// Stores the sum of
// elements of set1
int sumSet1 = 0;
// Stores the sum of
// elements of set2
int sumSet2 = 0;
// Traverse first N
// natural numbers
for(int i = N; i > 0; i--)
{
// Check if sum of elements of
// set1 is less than or equal
// to sum of elements of set2
if (sumSet1 <= sumSet2)
{
sumSet1 += i;
}
else
{
sumSet2 += i;
}
}
return Math.abs(sumSet1 - sumSet2);
}
// Driver code
public static void main (String[] args)
{
int N = 6;
System.out.println(minAbsDiff(N));
}
}
// This code is contributed by offbeat
Python3
# Python3 program to implement
# the above approach
# Function to split the first N
# natural numbers into two sets
# having minimum absolute
# difference of their sums
def minAbsDiff(N):
# Stores the sum of
# elements of set1
sumSet1 = 0
# Stores the sum of
# elements of set2
sumSet2 = 0
# Traverse first N
# natural numbers
for i in reversed(range(N + 1)):
# Check if sum of elements of
# set1 is less than or equal
# to sum of elements of set2
if sumSet1 <= sumSet2:
sumSet1 = sumSet1 + i
else:
sumSet2 = sumSet2 + i
return abs(sumSet1 - sumSet2)
# Driver Code
N = 6
print(minAbsDiff(N))
# This code is contributed by sallagondaavinashreddy7
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
static int minAbsDiff(int N)
{
// Stores the sum of
// elements of set1
int sumSet1 = 0;
// Stores the sum of
// elements of set2
int sumSet2 = 0;
// Traverse first N
// natural numbers
for(int i = N; i > 0; i--)
{
// Check if sum of elements of
// set1 is less than or equal
// to sum of elements of set2
if (sumSet1 <= sumSet2)
{
sumSet1 += i;
}
else
{
sumSet2 += i;
}
}
return Math.Abs(sumSet1 - sumSet2);
}
// Driver code
static void Main()
{
int N = 6;
Console.Write(minAbsDiff(N));
}
}
// This code is contributed by divyeshrabadiya07
Javascript
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
int minAbsDiff(int N)
{
if (N % 4 == 0 || N % 4 == 3) {
return 0;
}
return 1;
}
// Driver Code
int main()
{
int N = 6;
cout << minAbsDiff(N);
}
Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
static int minAbsDiff(int N)
{
if (N % 4 == 0 || N % 4 == 3)
{
return 0;
}
return 1;
}
// Driver Code
public static void main (String[] args)
{
int N = 6;
System.out.println(minAbsDiff(N));
}
}
// This code is contributed by sallagondaavinashreddy7
Python3
# Python3 program to implement
# the above approach
# Function to split the first N
# natural numbers into two sets
# having minimum absolute
# difference of their sums
def minAbsDiff(N):
if (N % 4 == 0 or N % 4 == 3):
return 0
return 1
# Driver Code
N = 6
print(minAbsDiff(N))
# This code is contributed by sallagondaavinashreddy7
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
static int minAbsDiff(int N)
{
if (N % 4 == 0 ||
N % 4 == 3)
{
return 0;
}
return 1;
}
// Driver Code
public static void Main(String[] args)
{
int N = 6;
Console.WriteLine(minAbsDiff(N));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出
1
时间复杂度: O(N)
辅助空间: O(1)
高效的方法:为了优化上述方法,该想法基于以下观察:
Splitting any 4 consecutive integers into 2 sets gives the minimum absolute difference of their sum equal to 0.
Mathematical proof:
Considering 4 consecutive integers {a1, a2, a3, a4}
a4 = a3 + 1
a1=a2 – 1
=> a4 + a1 = a3 + 1 + a2 – 1
=> a4 + a1 = a2 + a3
请按照以下步骤解决问题:
- 如果N % 4 == 0或N % 4 == 3 ,则打印0 。
- 否则,打印1 。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
int minAbsDiff(int N)
{
if (N % 4 == 0 || N % 4 == 3) {
return 0;
}
return 1;
}
// Driver Code
int main()
{
int N = 6;
cout << minAbsDiff(N);
}
Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
static int minAbsDiff(int N)
{
if (N % 4 == 0 || N % 4 == 3)
{
return 0;
}
return 1;
}
// Driver Code
public static void main (String[] args)
{
int N = 6;
System.out.println(minAbsDiff(N));
}
}
// This code is contributed by sallagondaavinashreddy7
蟒蛇3
# Python3 program to implement
# the above approach
# Function to split the first N
# natural numbers into two sets
# having minimum absolute
# difference of their sums
def minAbsDiff(N):
if (N % 4 == 0 or N % 4 == 3):
return 0
return 1
# Driver Code
N = 6
print(minAbsDiff(N))
# This code is contributed by sallagondaavinashreddy7
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to split the first N
// natural numbers into two sets
// having minimum absolute
// difference of their sums
static int minAbsDiff(int N)
{
if (N % 4 == 0 ||
N % 4 == 3)
{
return 0;
}
return 1;
}
// Driver Code
public static void Main(String[] args)
{
int N = 6;
Console.WriteLine(minAbsDiff(N));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出
1
时间复杂度: O(1)
辅助空间: O(1)